![Available At: mathcity.org Contact At: qadri86@yahoo.com
18
Then FሬԦ1 + FሬԦ2 + FሬԦ3 = 0 _______________(i)
Taking dot product of eq(i) with FሬԦ2× FሬԦ3, we get
(FሬԦ1 + FሬԦ2 + FሬԦ3).( FሬԦ2× FሬԦ3) = 0
⇒ FሬԦ1.( FሬԦ2× FሬԦ3) + FሬԦ2.( FሬԦ2 × FሬԦ3) + FሬԦ3.( FሬԦ2× FሬԦ3) = 0
⇒ FሬԦ1.( FሬԦ2× FሬԦ3) + 0 + 0 = 0
⇒ [FሬԦ1 FሬԦ2 FሬԦ3] = 0
Which shows that the forces FሬԦ1, FሬԦ2 and FሬԦ3 are
coplanar.
Taking cross product of eq(i) with FሬԦ3, we get
(FሬԦ1 + FሬԦ2 + FሬԦ3) × FሬԦ3 = 0
⇒ FሬԦ1× FሬԦ3 + FሬԦ2× FሬԦ3 + FሬԦ3× FሬԦ3 = 0
⇒ FሬԦ1× FሬԦ3 + FሬԦ2× FሬԦ3 = 0
⇒ FሬԦ1× FሬԦ3 = െ FሬԦ2× FሬԦ3
⇒ หFሬԦ1× FሬԦ3ห = หFሬԦ2× FሬԦ3ห
⇒ F1F3sinβ = F2F3sinα
⇒
F1
sinα
=
F2
sinβ
____________(ii)
Similarly by taking cross product of eq(i) with FሬԦ1, we get
F2
sinβ
=
F3
sinγ
___________(iii)
From eq(ii) and (iii), we get
F1
sinα
=
F2
sinβ
=
F3
sinγ
Which shows that each force has magnitude proportional to the sine of the angle between the
other two.
Question 12
Three forces PሬԦ, QሬሬԦ & RሬሬԦ acting at a point, are in equilibrium and the angle between PሬԦ & QሬሬԦ is
double of the angle between PሬԦ & RሬሬԦ. Prove that R2
= Q(Q – P)
Solution
Let angle between PሬԦ& RሬሬԦ is θ and angle between PሬԦ & QሬሬԦ is 2θ. Since forces are in equilibrium
therefore by Lami’s theorem we have
FሬԦ2
γ
FሬԦ1
α O
β
FሬԦ3](https://image.slidesharecdn.com/ch01compositionofforces-150621075843-lva1-app6892/85/Ch01-composition-of_forces-18-320.jpg)
Ch01 composition of_forces
- 1. Available At: mathcity.org Contact At: qadri86@yahoo.com M th Resultant of Two Forces Acting at a Point Let FሬԦ1 and FሬԦ2 be the two forces acting at a point O. Let α be a angle between FሬԦ1 and FሬԦ2. Let RሬሬԦ be their resultant which makes an angle θ with FሬԦ1. FሬԦ1 FሬԦ2 RሬሬԦ α θ O A CB Then RሬሬԦ ൌ FሬԦ1 FሬԦ2 __________(i) Taking dot product with itself, we get RሬሬԦ . RሬሬԦ ൌ ൫FሬԦ1 FሬԦ2൯. ൫FሬԦ1 FሬԦ2൯ ⇒ R2 ൌ FሬԦ1. FሬԦ1 FሬԦ1. FሬԦ2 FሬԦ2. FሬԦ1 FሬԦ2. FሬԦ2 ൌ F1 2 2FሬԦ1. FሬԦ2 F2 2 ൌ F1 2 2F1F2cosα F2 2 ⇒ R ൌ ටF1 2 F2 2 2F1F2cosα __________(ii) Which gives the magnitude of the resultant. Taking dot product of the eq(i) with FሬԦ1, we get FሬԦ1. RሬሬԦ ൌ FሬԦ1. ൫FሬԦ1 FሬԦ2൯ ⇒ F1R cosθ ൌ F1 2 FሬԦ1. FሬԦ2 ⇒ F1R cosθ ൌ F1 2 F1F2cosα ⇒ R cosθ ൌ F1 F2cosα __________(iii) CHAPTER 1COMPOSITION OF FORCES
- 2. Available At: mathcity.org Contact At: qadri86@yahoo.com 2 Taking cross product of the eq(i) with FሬԦ1, we get หFሬԦ1 ൈ RሬሬԦห ൌ หFሬԦ1 ൈ ൫FሬԦ1 FሬԦ2൯ห ⇒ F1R sinθ ൌ หFሬԦ1 ൈ FሬԦ1 FሬԦ1 ൈ FሬԦ2ห ⇒ F1R sinθ ൌ หFሬԦ1 ൈ FሬԦ2ห FሬԦ1 ൈ FሬԦ1 ൌ 0 ⇒ R sinθ ൌ F2sinθ __________(iv) Dividing eq(iii) by eq(iv), we get Rsinθ Rcosθ = F2sinθ F1 F2cosα ⇒ tanθ = F2sinθ F1 F2cosα ⇒ θ = tanିଵ ൬ F2sinθ F1 F2cosα ൰ ___________(v) Which gives the direction of the resultant. Special Cases Now we discuss some special cases of the above article. Case # 1 From eq(ii) R ൌ ටF1 2 F2 2 2F1F2cosα Which shows that R is maximum when cosα is maximum. But the maximum value of cosα is 1. i.e. cosα = 1 ⇒ α = 00 . Thus Rmax ൌ ටF1 2 F2 2 2F1F2cos00 ൌ ටF1 2 F2 2 2F1F2 ൌ ඥሺF1 F2ሻ2 ൌ F1 F2 Case # 2 From eq(ii) R ൌ ටF1 2 F2 2 2F1F2cosα When shows that R is minimum when cosα is minimum. But the minimum value of cosα is െ1. i.e. cosα = െ1 ⇒ α = π. Thus Rmin ൌ ටF1 2 F2 2 2F1F2cosπ ൌ ටF1 2 F2 2 2F1F2ሺെ1ሻ
- 3. Available At: mathcity.org Contact At: qadri86@yahoo.com 3 ൌ ටF1 2 െ 2F1F2 F2 2 ൌ ඥሺF1 െ F2ሻ2 ൌ F1 െ F2 Case # 3 From eq(ii) R ൌ ටF1 2 F2 2 2F1F2cosα When FሬԦ1 and FሬԦ2 are perpendicular to each other. i.e. α = 90o Then R ൌ ටF1 2 F2 2 2F1F2cos900 ൌ ටF1 2 F2 2 cos900 ൌ 0 From (v) θ = tanିଵ ቆ F2sin900 F1 F2cos900ቇ ⇒ θ = tanିଵ ൬ F2 F1 ൰ sin900 = 1 Question 1 The greatest resultant that two forces can have is of magnitude P and the least is of magnitude Q. Show that when they act at an angle α their resultant is of magnitude: ටP2 cos2 αααα 2 + Q2 sin2 αααα 2 Solution Let FሬԦ1 and FሬԦ2 be two forces and P & Q be magnitude of their greatest and least resultant respectively. Then P ൌ F1 F2 ______________(i) and Q ൌ F1 െ F2 ______________(ii) Adding (i) and (ii),we get 2F1 = P + Q ⇒ F1 ൌ P + Q 2 Subtracting (ii) from (i), we get 2F2 = P െ Q ⇒ F2 ൌ P െ Q 2 Let RሬሬԦ be the resultant of FሬԦ1 and FሬԦ2 when they act an angle α. Then
- 4. Available At: mathcity.org Contact At: qadri86@yahoo.com 4 R ൌ ටF1 2 2F1F2cosα F2 2 ൌ ඨ൬ P + Q 2 ൰ 2 2 ൬ P + Q 2 ൰ ൬ P െ Q 2 ൰ cosα ൬ P െ Q 2 ൰ 2 ൌ ඨ ሺP + Qሻ2 4 2 ቆ P2 െ Q2 4 ቇ cosα ሺP െ Qሻ2 4 ൌ ඨ 1 4 ൣሺP + Qሻ2 ሺP െ Qሻ2 2൫P2 െ Q2 ൯cosα൧ ൌ ඨ 1 4 ൣ2൫P2 Q2 ൯ 2൫P2 െ Q2 ൯cosα൧ ൌ ඨ 1 2 ൣP2 Q2 P2 cosα െ Q2 cosα൧ ൌ ඨ 1 2 ൣP2 ሺ1 cosαሻ + Q2 ሺ1 െ cosαሻ൧ ൌ ඨ 1 2 ቂP2 2cos2 α 2 + Q2 2cos2 α 2 ቃ ൌ ටP2 cos2 α 2 + Q2 cos2 α 2 Which is required. Question 2 The resultant of two forces of magnitude P and Q is of magnitude R. If Q is doubled then R is doubled. If Q is reversed then R is again doubled. Show that P2 : Q2 : R2 = 2 : 3 : 2 Solution Let θ be angle between the forces P & Q. Since R is the magnitude of the resultant of P and Q therefore R ൌ ටP2 2PQcosα Q2 ⇒ R2 ൌ P2 2PQcosα Q2 ____________(i) Since when Q is double then R is double therefore by replacing Q with 2Q and R with 2R eq(i) becomes ሺ2Rሻ2 ൌ P2 2P(2Q)cosα ሺ2Qሻ2
- 5. Available At: mathcity.org Contact At: qadri86@yahoo.com 5 ⇒ 4R2 ൌ P2 4PQcosα 4Q2 ____________(ii) Since when Q is reversed then R is double therefore by replacing Q with െ Q and R with 2R eq(i) becomes ሺ2Rሻ2 ൌ P2 2P2ሺ െ Q)cosα ሺെ Qሻ2 ⇒ 4R2 ൌ P2 െ 2PQcosα Q2 ____________(iii) Multiplying (iii) by 2, we get 8R2 = 2P2 െ 4PQcosα + 2Q2 ____________(iv) Adding (i) and (iii), we get 5R2 ൌ 2P2 2Q2 ⇒ 2P2 2Q2 – 5R2 ൌ 0 _____________(v) Adding (ii) and (iv), we get 12R2 ൌ 3P2 6Q2 ⇒ 4R2 ൌ P2 2Q2 ⇒ P2 2Q2 – 4R2 ൌ 0 __________ ___(vi) Solving (v) and (vi) simultaneously, we get P2 െ8 + 10 = Q2 െ5 + 8 = R2 4 െ 2 ⇒ P2 2 = Q2 3 = R2 2 ⇒ P2 : Q2 : R2 ൌ 2 : 3 : 2 Theorem of resolved parts The algebraic sum of the resolved parts of a system of forces in any direction is equal to the resolved part of the resultant in the same direction. Proof Let RሬሬԦ be the resultant of forces FሬԦ1, FሬԦ2, FሬԦ3, … , FሬԦn and aො be the unit vector in any direction which makes an angle α with RሬሬԦ and α1, α2 , α3 , … , αn with FሬԦ1, FሬԦ2, FሬԦ3, … , FሬԦn respectively. Then RሬሬԦ = FሬԦ1 + FሬԦ2 + FሬԦ3 + … + FሬԦn _________(i) Taking dot product of (i) with aො , we get aො . RሬሬԦ = aො . ൫FሬԦ1 + FሬԦ2 + FሬԦ3 + … + FሬԦn൯ ⇒ aො . RሬሬԦ = aො . FሬԦ1 + aො . FሬԦ2 + aො . FሬԦ3 + … + aො . FሬԦn ⇒ Rcosθ = F1cosθ + F2cosθ + F3cosθ + … + Fncosθ _________(ii)
- 6. Available At: mathcity.org Contact At: qadri86@yahoo.com 6 Similarly by taking cross product of (i) with aො , we get Rsinθ = F1sinθ + F2sinθ + F3sinθ + … + Fnsinθ _________(iii) Eq(ii) and eq(iii) shows that the sum of the resolved parts of a system of forces in any direction is equal to the resolved part of the resultant in the same direction. Question 3 Forces PሬԦ, QሬሬԦ and RሬሬԦ act at a point parallel to the sides of a triangle ABC taken in the same order. Show that the magnitude of Resultant is ටP2 + Q2 + R2 െ 2PQcosC െ 2QRcosA െ 2RPcosB Solution Let the forces PሬԦ, QሬሬԦ and RሬሬԦ act along the sides BC, CA and AB of a triangle ABC taking one way round as shown in the figure. We take BC along x-axis. Let the F be the magnitude of the resultant, then F = ටFx 2 + Fy 2 ______________(i) Now by theorem of resolved parts Fx = Sum of the resolved parts of the forces along x-axis = P cos0 + Qcos(180 – C) + Rcos(180 + B) = P െ QcosC – RcosB Taking square on both sides, we get Fx 2 ൌ ൫P െ QcosC – RcosB൯ 2 ൌ P2 + Q2 cos2 C + R2 cos2 B െ 2PQcosC + 2QRcosBcosC െ 2PRcosB Again by theorem of resolved parts Fy = Sum of the resolved parts of the forces along y-axis = P sin0 + Qsin(180 – C) + Rsin(180 + B) A y RሬሬԦ A QሬሬԦ QሬሬԦ O PሬԦ x B C x-axis RሬሬԦ B PሬԦ C
- 7. Available At: mathcity.org Contact At: qadri86@yahoo.com 7 = QsinC – RsinB Taking square on both sides, we get Fy 2 ൌ ൫QsinC – RsinB൯ 2 ൌ Q2 sin2 C + R2 sin2 B െ 2QRsinBsinC Using values of Rx 2 and Ry 2 in (i), we get F = ඨ P2 + Q2 cos2C + R2 cos2B െ 2PQcosC + 2QRcosBcosC െ 2PRcosB + Q2 sin2 C + R2 sin2B െ 2QRsinBsinC = ඨ P2 + Q2 ൫cos2C + sin2 C൯ + R2 ሺcos2B sin2Bሻ െ 2PQcosC െ 2PRcosB 2QRሺcosBcosC െ sinBsinCሻ = ටP2 + Q2 + R2 െ 2PQcosC െ 2PRcosB 2QRcosሺB Cሻ Since A + B + C = 1800 ⇒ B + C = 1800 – A ⇒ R = ටP2 + Q2 + R2 െ 2PQcosC െ 2PRcosB + 2QRcos൫1800 – A ൯ = ටP2 + Q2 + R2 െ 2PQcosC െ 2PRcosB െ 2QRcosA Question 4 Forces PሬԦ and QሬሬԦ act at a point O and their resultant is RሬሬԦ. If any transversal cuts the lines of action of forces in the points A, B & C respectively. Prove that R OC = P OA + Q OB Solution Let PሬԦ, QሬሬԦ and RሬሬԦ makes angles α, β and γ with x-axis respectively. The transversal LM cuts the lines of action of the forces at points A, B and C respectively as shown in the figure. Since RሬሬԦ is resultant of PሬԦ & QሬሬԦ therefore by theorem of resolved parts Rcosγ = Pcosα + Qcosβ _________(i) From figure cosα = OM OA cosβ = OM OB and cosγ = OM OC y L QሬሬԦ B RሬሬԦ C A PሬԦ α β γ xO M
- 8. Available At: mathcity.org Contact At: qadri86@yahoo.com 8 Using These values in (i), we get R ൬ OM OC ൰ = P ൬ OM OA ൰ + Q ൬ OM OB ൰ ⇒ R OC = P OA + Q OB Which is required. Question 5 If two forces P and Q act at such an angle that their resultant R = P. Show that if P is doubled, the new resultant is at right angle to Q. Solution Let the forces P & Q act at O and makes an angle α with each other. Take Q along x-axis. If R is the magnitude of the resultant then R2 = P2 + 2PQcosα+ Q2 Since R = P therefore P2 = P2 + 2PQcosα+ Q2 ⇒ 2PQcosα + Q2 ൌ 0 ⇒ 2Pcosα + Q ൌ 0 __________(i) By theorem of resolved parts Rx = Sum of the resolved parts of the forces along x-axis = Qcos00 + Pcosα = Q + Pcosα Again by theorem of resolved parts Ry = Sum of the resolved parts of the forces along y-axis = Qsin00 + Psinα = Psinα If P is double. i.e. P = 2P then Rx = Q + 2Pcosα = 0 By (i) And Ry = 2Psinα If the new resultant makes an angle θ with Q then θ = tanି1 ൬ Ry Rx ൰ = tanି1 ൬ 2Psinα 0 ൰ = tanି1ሺ∞ሻ ൌ π 2 Which is required.
- 9. Available At: mathcity.org Contact At: qadri86@yahoo.com 9 Question 6 Forces X, P + X and Q + X act at a point in the directions of sides of a equilateral triangle taken one way round. Show that they are equivalent to the forces P & Q acting at an angle of 1200 . Solution A y Q + X 60 P + X P + X x 60 60 x-axis O X B X C Q + X Let the forces X, X + P and X + Q act along the sides BC, CA and AB of a triangle ABC taking one way round as shown in the figure. We take BC along x-axis. Let the R be the magnitude if the resultant, then R = ටRx 2 + Ry 2 ______________(i) Now by theorem of resolved parts Rx = Sum of the resolved parts of the forces along x-axis = X cos0 + (P + X)cos(1800 – 600 ) + (Q + X)cos(1800 + 600 ) = X – (P + X)cos600 – (Q + X)cos600 = X – 1 2 (P + X) – 1 2 (Q + X) = X – 1 2 P – 1 2 X – 1 2 Q – 1 2 X = – 1 2 ሺP + Qሻ Taking square on both sides, we get Rx 2 ൌ ቆ– 1 2 ሺP + Qሻቇ 2 ൌ 1 4 ൫P2 + Q2 + 2PQ൯ Again by theorem of resolved parts Ry = Sum of the resolved parts of the forces along y-axis = X sin00 + (P + X)sin(1800 – 600 ) + (Q + X)sin(1800 + 600 )
- 10. Available At: mathcity.org Contact At: qadri86@yahoo.com 10 = (P + X)sin600 – (Q + X)sin600 = √3 2 (P + X) – √3 2 (Q + X) = √3 2 P + √3 2 X – √3 2 Q – √3 2 X = √3 2 ሺP െ Qሻ Taking square on both sides, we get Ry 2 ൌ ቌ √3 2 ሺP െ Qሻቍ 2 ൌ 3 4 ൫P2 + Q2 െ 2PQ൯ Using values of Rx 2 and Ry 2 in (i), we get R = ඨ 1 4 ൫P2 + Q2 + 2PQ൯ 3 4 ൫P2 + Q2 െ 2PQ൯ = ඨ P2 + Q2 + 2PQ 3P2 + 3Q2 െ 6PQ 4 = ඨ 4P2 + 4Q2 െ 4PQ 4 = ටP2 + Q2 െ PQ = ඨP2 + Q2 + 2PQ ൬െ 1 2 ൰ = ටP2 + Q2 + 2PQcos1200 This result shows that the given forces are equivalent to the forces P and Q acting an angle of 1200 . Question 7 Forces X, Y, Z, P + X, Q + Y and P + Z act at a point in the directions of sides of a regular hexagon taken one way round. Show that their resultant is equivalent to the force P + Q in the direction of the force Q + Y. Solution P + X E D Q + Y Z F C P + Z Y x – axis A X B
- 11. Available At: mathcity.org Contact At: qadri86@yahoo.com 11 Let the force X, Y, Z, P + X, Q + Y and P + Z act along the sides AB, BC, CD, DE, EF, FA of a regular hexagon taken one way round as shown in figure. Take AB along x-axis. Let the R be the magnitude if the resultant and Rx and Ry be the resolved parts of the resultant, then R = ටRx 2 + Ry 2 ______________(i) Now by theorem of resolved parts Rx = Sum of the resolved parts of the forces along x-axis = X cos00 + Ycos600 + Zcos1200 + (P + X)cos1800 + (Q + Y)cos2400 + (P + Z)cos3000 = X + 1 2 Y െ 1 2 Z െ (P + X) െ 1 2 (Q + Y) + 1 2 (P + Z) = X + 1 2 Y െ 1 2 Z െ P െ X െ 1 2 Q െ 1 2 Y + 1 2 P + 1 2 Z = െ 1 2 ሺP + Qሻ Taking square on both sides, we get Rx 2 ൌ ቆ– 1 2 ሺP + Qሻቇ 2 ൌ 1 4 ሺP + Qሻ2 Again by theorem of resolved parts Ry = Sum of the resolved parts of the forces along y-axis = X sin00 + Ysin600 + Zsin1200 + (P + X)sin1800 + (Q + Y)sin2400 + (P + Z)sin3000 = 0 + √3 2 Y + √3 2 Z െ 0 (P + X) െ √3 2 (Q + Y) െ √3 2 (P + Z) = √3 2 Y + √3 2 Z െ √3 2 Q െ √3 2 Y െ √3 2 P െ √3 2 Z = െ √3 2 ሺP + Qሻ Taking square on both sides, we get Ry 2 ൌ ቌെ √3 2 ሺP + Qሻቍ 2 ൌ 3 4 ሺP + Qሻ2 Using values of Rx 2 and Ry 2 in (i), we get R = ඨ 1 4 ሺP + Qሻ2 3 4 ሺP + Qሻ2 = ඥሺP + Qሻ2
- 12. Available At: mathcity.org Contact At: qadri86@yahoo.com 12 ⇒ R = P + Q If resultant makes angle θ with x-axis then θ = tanି1 ൬ Ry Rx ൰ = tanି1 ൮ െ √3 2 ሺP + Qሻ െ 1 2 ሺP + Qሻ ൲ = tanି1 ቆ െ√3 െ1 ቇ ൌ 2400 Which shows that the resultant is a force P + Q in the direction of Q + Y because Q + Y makes an angle 240 with x-axis. Question 8 Forces P, Q and R act along the sides BC, CA, AB of a triangle ABC. Find the condition that their resultant is parallel to BC and determine its magnitude. Solution Let the forces P, Q and R act along the sides BC, CA and AB of a triangle ABC taking one way round as shown in the figure. We take BC along x-axis. A R c A Q b B a C x-axis B P C Let the F be the magnitude of the resultant, then F = ටFx 2 + Fy 2 ______________(i) Now by theorem of resolved parts Fx = Sum of the resolved parts of the forces along x-axis = Pcos00 + Qcos(180 – C) + Rcos(180 + B) = P െ QcosC െ RcosB And Fy = Sum of the resolved parts of the forces along y-axis = Psin00 + Qsin(180 – C) + Rsin(180 + B) = QsinC – RsinB
- 13. Available At: mathcity.org Contact At: qadri86@yahoo.com 13 If the resultant makes an angle θ with x – axis , then tanθ = Fy Fx Since the resultant is parallel to BC therefore θ must be zero. So tan0 = Fy Fx ⇒ Fy Fx = 0 ⇒ Fy= 0 ⇒ QsinC – RsinB = 0 ⇒ sinC sinB = R Q ______________(ii) Let a, b and c are the lengths of the sides BC, CA and AB respectively. Then by Law of sine a sinA = b sinB = c sinC ⇒ sinC sinB = c b ______________(iii) From (ii) and (iii), we get R Q ൌ c b ⇒ R = Q ቀ c b ቁ ______________(iv) ⇒ Qc = Rb Which is required condition. Using values of Fx and Fy in (i), we get F2 = ሺP െ QcosC െ RcosBሻ2 + ሺQsinC – RsinBሻ2 = P2 + Q2 cos2 C + R2 cos2 B െ 2PQcosC + 2QRcosBcosC െ 2PRcosB + Q2 sin2 C + R2 sin2 B െ 2QRsinBsinC = P2 + Q2 ሺcos2 C sin2 Cሻ + R2 ሺcos2 B sin2 Bሻ െ 2PQcosC െ 2PRcosB + 2QRሺcosBcosC െ sinBsinCሻ = P2 + Q2 + R2 െ 2PQcosC െ 2PRcosB 2QRcosሺB + C) = P2 + Q2 + R2 െ 2PQcosC െ 2PRcosB 2QRcosሺ180 െ A) ∵ A + B + C = 180 = P2 + Q2 + R2 െ 2PQcosC െ 2PRcosB െ 2QRcosA = P2 + Q2 + ቆQ ቀ c b ቁቇ 2 െ 2PQcosC െ 2P ቆQ ቀ c b ቁቇ cosB െ 2Q ቆQ ቀ c b ቁቇ cosA = P2 + Q2 + Q2 ቀ c b ቁ 2 െ 2PQcosC െ 2PQ ቀ c b ቁ cosB െ 2Q2 ቀ c b ቁ cosA
- 14. Available At: mathcity.org Contact At: qadri86@yahoo.com 14 = P2 + Q2 ቆ b2 b2ቇ + Q2 ቆ c2 b2ቇ െ 2PQcosC െ 2PQ ቀ c b ቁ cosB െ 2Q2 ൬ bc bଶ ൰ cosA = P2 + Q2 ቆ b2 b2ቇ + Q2 ቆ c2 b2ቇ െ 2Q2 ൬ bc bଶ ൰ cosA െ 2PQ ൬ b b ൰ cosC െ 2PQ ቀ c b ቁ cosB = P2 + Q2 b2 ൫b2 + c2 െ 2bccosA൯ െ 2P Q b ሺbcosC – c cosBሻ = P2 + Q2 b2 ሺa2ሻ െ 2P Q b ሺaሻ a2 ൌ b2 + c2 െ 2bccosA & a ൌ bcosC – c cosB = ൬P െ Q b a൰ ଶ ⇒ F = ൬P െ Q b a൰ Thus the magnitude of the resultant = ൬P െ Q b a൰ (λλλλ , µµµµ ) Theorem If two concurrent forces are represented by λOAሬሬሬሬሬሬԦ and µOBሬሬሬሬሬሬԦ. Then their resultant is given by (λ +µ)OCሬሬሬሬሬሬԦ where C divides AB such that AC : CB = µ : λ Proof Let R be the resultant of the forces λOA and µOB. Then RሬሬԦ = λOAሬሬሬሬሬሬԦ + µOBሬሬሬሬሬሬԦ ____________(i) Given that AC : CB = µ : λ ⇒ AC CB ൌ µ λ ⇒ λAC = µCB ⇒ λAC െ µCB = 0 ⇒ λAC + µBC = 0 ____________(ii) Again from fig. OAሬሬሬሬሬሬԦ = OCሬሬሬሬሬሬԦ + CAሬሬሬሬሬሬԦ ⇒ λOAሬሬሬሬሬሬԦ = λOCሬሬሬሬሬሬԦ + λCAሬሬሬሬሬሬԦ ____________(iii) From fig. OBሬሬሬሬሬሬԦ = OCሬሬሬሬሬሬԦ + CBሬሬሬሬሬԦ ⇒ µOBሬሬሬሬሬሬԦ = µOCሬሬሬሬሬሬԦ + µCBሬሬሬሬሬԦ ____________(iv) B λ C µ µOBሬሬሬሬሬሬԦ O λOAሬሬሬሬሬሬԦ A
- 15. Available At: mathcity.org Contact At: qadri86@yahoo.com 15 Using eq(iii) and eq(iv) in (i), we get RሬሬԦ = λOCሬሬሬሬሬሬԦ + λCAሬሬሬሬሬሬԦ + µOCሬሬሬሬሬሬԦ + µCBሬሬሬሬሬԦ = (λ + µ)OCሬሬሬሬሬሬԦ + λCAሬሬሬሬሬሬԦ + µCBሬሬሬሬሬԦ = (λ + µ)OCሬሬሬሬሬሬԦ െλCAሬሬሬሬሬሬԦ െ µBCሬሬሬሬሬԦ = (λ + µ)OCሬሬሬሬሬሬԦ – (λCAሬሬሬሬሬሬԦ + µCBሬሬሬሬሬԦ) = (λ + µ)OCሬሬሬሬሬሬԦ – 0 By(ii) Thus RሬሬԦ = (λ + µ)OCሬሬሬሬሬሬԦ Question 9 If forces pABሬሬሬሬሬሬԦ, qCBሬሬሬሬሬԦ, rCDሬሬሬሬሬሬԦ and sADሬሬሬሬሬሬԦ acting along the sides of a plane quadrilateral are in equilibrium. Show that pr = qs Solution Let ABCD be a plane quadrilateral and force pABሬሬሬሬሬሬԦ, qCBሬሬሬሬሬԦ, rCDሬሬሬሬሬሬԦ and sADሬሬሬሬሬሬԦ acting along its sides as shown in figure. By (λ, µ) theorem pABሬሬሬሬሬሬԦ + sADሬሬሬሬሬሬԦ = (p + s) AEሬሬሬሬሬԦ Where E is the point on BD such that BE ED = s p _____________(i) Again by (λ, µ) theorem qCBሬሬሬሬሬԦ + rCDሬሬሬሬሬሬԦ = (q + r) CFሬሬሬሬሬԦ Where F is the point on BD such that BF FD = r q _____________(ii) Since forces are in equilibrium therefore point E & F must coincides. So eq(i) & eq(ii) must equal. Thus BE ED = BF FD ⇒ s p = r q ⇒ pr = sq D rCDሬሬሬሬሬሬԦ C p E sADሬሬሬሬሬሬԦ s q qCBሬሬሬሬሬԦ F r A pABሬሬሬሬሬሬԦ B
- 16. Available At: mathcity.org Contact At: qadri86@yahoo.com 16 Question 10 If forces 2BCሬሬሬሬሬԦ, CAሬሬሬሬሬሬԦ, and BAሬሬሬሬሬሬԦ acting along the sides of triangle ABC. Show that their resultant is 6DEሬሬሬሬሬԦ where D bisect BC and E is a point on CA such that CE = 1 3 CA Solution Let RሬሬԦ be resultant of forces then RሬሬԦ = 2BCሬሬሬሬሬԦ + CAሬሬሬሬሬሬԦ + BAሬሬሬሬሬሬԦ = 2BCሬሬሬሬሬԦ െ ACሬሬሬሬሬሬԦ െ ABሬሬሬሬሬሬԦ = 2BCሬሬሬሬሬԦ െ (ACሬሬሬሬሬሬԦ + ABሬሬሬሬሬሬԦ) _________(i) By (λ, µ) theorem, ACሬሬሬሬሬሬԦ + ABሬሬሬሬሬሬԦ = (1 + 1) ADሬሬሬሬሬሬԦ = 2ADሬሬሬሬሬሬԦ Using this value in (i), we get RሬሬԦ = 2BCሬሬሬሬሬԦ െ 2ADሬሬሬሬሬሬԦ = 2(BCሬሬሬሬሬԦ െ ADሬሬሬሬሬሬԦ) = 2(2DCሬሬሬሬሬሬԦ െ ADሬሬሬሬሬሬԦ) BCሬሬሬሬሬԦ = 2DCሬሬሬሬሬሬԦ = 2(2DCሬሬሬሬሬሬԦ + DAሬሬሬሬሬሬԦ) _________(ii) By (λ, µ) theorem, 2DCሬሬሬሬሬሬԦ + DAሬሬሬሬሬሬԦ = (2 + 1) DEሬሬሬሬሬԦ = 3DEሬሬሬሬሬԦ Using this value in (ii), we get RሬሬԦ = 2ሺ3DEሬሬሬሬሬԦ) = 6DEሬሬሬሬሬԦ A BAሬሬሬሬሬሬԦ 2 CAሬሬሬሬሬሬԦ E 1 B D 2BCሬሬሬሬሬԦ C
- 17. Available At: mathcity.org Contact At: qadri86@yahoo.com 17 Question 11 P is any point in the plane of a triangle ABC and D, E, F are middle points of its sides. Prove that forces APሬሬሬሬሬԦ, BPሬሬሬሬሬԦ, CPሬሬሬሬሬԦ, PDሬሬሬሬሬԦ, PEሬሬሬሬԦ, PFሬሬሬሬԦ are in equilibrium. Solution C F E P A D B Applying (λ, µ) theorem, we get APሬሬሬሬሬԦ + BPሬሬሬሬሬԦ = 2DPሬሬሬሬሬԦ BPሬሬሬሬሬԦ + CPሬሬሬሬሬԦ = 2EPሬሬሬሬሬԦ CPሬሬሬሬሬԦ + APሬሬሬሬሬԦ = 2FPሬሬሬሬԦ Adding above equations, we get APሬሬሬሬሬԦ + BPሬሬሬሬሬԦ + BPሬሬሬሬሬԦ + CPሬሬሬሬሬԦ + CPሬሬሬሬሬԦ + APሬሬሬሬሬԦ = 2DPሬሬሬሬሬԦ + 2EPሬሬሬሬሬԦ + 2FPሬሬሬሬԦ ⇒ 2APሬሬሬሬሬԦ + 2BPሬሬሬሬሬԦ + 2CPሬሬሬሬሬԦ = 2DPሬሬሬሬሬԦ + 2EPሬሬሬሬሬԦ + 2FPሬሬሬሬԦ ⇒ APሬሬሬሬሬԦ + BPሬሬሬሬሬԦ + CPሬሬሬሬሬԦ = DPሬሬሬሬሬԦ + EPሬሬሬሬሬԦ + FPሬሬሬሬԦ ⇒ APሬሬሬሬሬԦ + BPሬሬሬሬሬԦ + CPሬሬሬሬሬԦ െ DPሬሬሬሬሬԦ െ EPሬሬሬሬሬԦ െ FPሬሬሬሬԦ = 0 ⇒ APሬሬሬሬሬԦ + BPሬሬሬሬሬԦ + CPሬሬሬሬሬԦ + PDሬሬሬሬሬԦ + PEሬሬሬሬሬԦ + PFሬሬሬሬԦ = 0 Since the vector sum of all forces is zero therefore the given forces are in equilibrium. Lami’s Theorem If a particle is in equilibrium under the action of three forces then the forces are coplanar and each force has magnitude proportional to the sine of the angle between the other two. Proof Let FሬԦ1, FሬԦ2 and FሬԦ3act at a point O and are in equilibrium.
- 18. Available At: mathcity.org Contact At: qadri86@yahoo.com 18 Then FሬԦ1 + FሬԦ2 + FሬԦ3 = 0 _______________(i) Taking dot product of eq(i) with FሬԦ2× FሬԦ3, we get (FሬԦ1 + FሬԦ2 + FሬԦ3).( FሬԦ2× FሬԦ3) = 0 ⇒ FሬԦ1.( FሬԦ2× FሬԦ3) + FሬԦ2.( FሬԦ2 × FሬԦ3) + FሬԦ3.( FሬԦ2× FሬԦ3) = 0 ⇒ FሬԦ1.( FሬԦ2× FሬԦ3) + 0 + 0 = 0 ⇒ [FሬԦ1 FሬԦ2 FሬԦ3] = 0 Which shows that the forces FሬԦ1, FሬԦ2 and FሬԦ3 are coplanar. Taking cross product of eq(i) with FሬԦ3, we get (FሬԦ1 + FሬԦ2 + FሬԦ3) × FሬԦ3 = 0 ⇒ FሬԦ1× FሬԦ3 + FሬԦ2× FሬԦ3 + FሬԦ3× FሬԦ3 = 0 ⇒ FሬԦ1× FሬԦ3 + FሬԦ2× FሬԦ3 = 0 ⇒ FሬԦ1× FሬԦ3 = െ FሬԦ2× FሬԦ3 ⇒ หFሬԦ1× FሬԦ3ห = หFሬԦ2× FሬԦ3ห ⇒ F1F3sinβ = F2F3sinα ⇒ F1 sinα = F2 sinβ ____________(ii) Similarly by taking cross product of eq(i) with FሬԦ1, we get F2 sinβ = F3 sinγ ___________(iii) From eq(ii) and (iii), we get F1 sinα = F2 sinβ = F3 sinγ Which shows that each force has magnitude proportional to the sine of the angle between the other two. Question 12 Three forces PሬԦ, QሬሬԦ & RሬሬԦ acting at a point, are in equilibrium and the angle between PሬԦ & QሬሬԦ is double of the angle between PሬԦ & RሬሬԦ. Prove that R2 = Q(Q – P) Solution Let angle between PሬԦ& RሬሬԦ is θ and angle between PሬԦ & QሬሬԦ is 2θ. Since forces are in equilibrium therefore by Lami’s theorem we have FሬԦ2 γ FሬԦ1 α O β FሬԦ3
- 19. Available At: mathcity.org Contact At: qadri86@yahoo.com 19 ⇒ Psinθ + Q(3sinθ െ 4sin3 θ) = 0 ⇒ Psinθ + Qsinθ(3 െ 4sin2 θ) = 0 ⇒ P + Q(3 െ 4sin2 θ) = 0 ⇒ P + Q[3 െ 4(1 െ cos2 θ)ሿ = 0 ⇒ P + Q(3 െ 4 + 4cos2 θ) = 0 ⇒ P + Q(4cos2 θ െ 1) = 0 ------- (ii) From (i), we have Q sinθ = R sin2θ ⇒ Qsin2θ = Rsinθ ⇒ Q2sinθcosθ = Rsinθ ⇒ Q2cosθ = R ⇒ cosθ = R 2Q Using value of cosθ in (ii), we get ⇒ P + Q ቈ4 ൬ R 2Q ൰ 2 െ 1 = 0 ⇒ P + Q ቆ R2 Q2 െ 1ቇ = 0 ⇒ P + Q ቆ R2 െ Q2 Q2 ቇ = 0 ⇒ PQ + R2 െ Q2 = 0 ⇒ R2 ൌ Q2 – PQ ⇒ R2 ൌ Q(Q – P) Hence Proved. P sin(2π െ 3θ) = Q sinθ = R sin2θ ⇒ െ P sin3θ = Q sinθ = R sin2θ ____________(i) ⇒ െ P sin3θ = Q sinθ ⇒ െ Psinθ = Qsin3θ y RሬሬԦ θ PሬԦ2π -3θ x 2θ QሬሬԦ
- 20. Available At: mathcity.org Contact At: qadri86@yahoo.com 20 Question 13 Three forces act perpendicularly to the sides of a triangle at their middle points and are proportional to the sides. Prove that they are in equilibrium. Solution Let a, b and c are the lengths of the sides of the triangle ABC. Then given that P a = Q b = R c __________(i) By law of sine a sinA = b sinB = c sinC __________(ii) Comparing (i) and (ii), we get P sinA = Q sinB = R sinC __________(iii) From fig. ∠ QOR = 180 – A, ∠ POQ = 180 – B, ∠ POR = 180 – C ⇒ sin∠QOR = sin(180 – A) = sinA sin∠POQ = sin(180 – B) = sinB sin∠POR = sin(180 – C) = sinC From (iii) P sin∠QOR = Q sin∠POQ = R sin∠POR Thus, by Lami’s theorem forces are in equilibrium. A RሬሬԦ QሬሬԦ A 180-A b c O B a C PሬԦ
- 21. Available At: mathcity.org Contact At: qadri86@yahoo.com 21 Moment or torque of a force The tendency of a force to rotate a body about a point is called moment or torque of that force. Explanation Consider a force FሬԦ acting on a rigid body which tends to rotate the body about O. Take any point A on the line of action of force FሬԦ. Let rԦ be the position vector of A with respect to O. Then moment of force about O is defined as MሬሬሬԦ ൌ rԦ × FሬԦ ⇒ หMሬሬሬԦห ൌ หrԦ × FሬԦห ⇒ M = rFsinθ = F(rsinθ) __________(i) Where θ is the angle between rԦ and FሬԦ. From fig. d = rsinθ Using this in eq(i), we get M = Fd Where d is the perpendicular distance from O to the line of the action of force FሬԦ. Note : 1. Moment will be negative if body rotates in clockwise direction. 2. Moment will be positive if body rotates in anticlockwise direction. Varignon’s Theorem The moment about a point of the resultant of a system of concurrent force is equal to the sum of the moments of these forces about the same point. Proof Let forces FሬԦ1, FሬԦ2, FሬԦ3, … , FሬԦn be concurrent at a point A. Let FሬԦ be their resultant and rԦ be the position vector of A with respect to O. Then FሬԦ = FሬԦ1+ FሬԦ2 + FሬԦ3 + …+ FሬԦn ______ (i) ⇒ rԦ × FሬԦ = rԦ × ൫FሬԦ1+ FሬԦ2 + FሬԦ3 + …+ FሬԦn൯ ⇒ MሬሬሬԦ = rԦ × FሬԦ1+ rԦ × FሬԦ2 + …+ rԦ × FሬԦn ⇒ MሬሬሬԦ = MሬሬሬԦ1+ MሬሬሬԦ2 + MሬሬሬԦ3 + …+ MሬሬሬԦn Where MሬሬሬԦ is the moment of resultant of forces about point O and MሬሬሬԦ1+ MሬሬሬԦ2 + MሬሬሬԦ3 + …+ MሬሬሬԦn is the sum of the moments of forces about the same point. This completes the proof. FሬԦ θ A rԦ θ O d B FሬԦ FሬԦ5 FሬԦ4 FሬԦn FሬԦ3 rԦ A FሬԦ2 O FሬԦ1
- 22. Available At: mathcity.org Contact At: qadri86@yahoo.com 22 Question 14 A system of forces acts on a plate in the from of an equilateral triangle of side 2a. The moments of the forces about the three vertices are G1, G2, and G3 respectively. Find the magnitude of the resultant. Solution C 600 AԢ FሬԦ3 900 FሬԦ2 300 600 A FሬԦ1 B Let FሬԦ1, FሬԦ2 and FሬԦ3 be the forces acting along the sides AB, BC and CA taking one way round. Take AB along x–axis. Let the R be the magnitude of the resultant of the forces, then R = ටRx 2 + Ry 2 ______________(i) Now by theorem of resolved parts Rx = Sum of the resolved parts of the forces along x-axis = F1cos00 + F2cos(180 – 60) + F3cos(180 + 60) = F1 – F2cos60 – F3cos60 = F1 െ F2 ൬ 1 2 ൰ െ F3 ൬ 1 2 ൰ = 2 F1 െ F2 െ F3 2 And Ry = Sum of the resolved parts of the forces along y-axis = F1sin00 + F2sin(180 – 60) + F3sin(180 + 60) = 0 + F2sin60 – F3sin60 = F2 ቆ √3 2 ቇ െ F3 ቆ √3 2 ቇ = √3 2 ሺF2 െ F3ሻ
- 23. Available At: mathcity.org Contact At: qadri86@yahoo.com 23 Using values of Fx and Fy in (i), we get R = ඪ൬ 2 F1 െ F2 െ F3 2 ൰ 2 + ቌ √3 2 ሺF2 െ F3ሻቍ 2 = ඨ 4F1 2 + F2 2 + F3 2 െ 4F1F2+ 2F2F3 െ 4F3F1 4 + 3F2 2 + 3F3 2 െ 6F2F3 4 = ඨ 4F1 2 + 4F2 2 + 4F3 2 െ 4F1F2 െ 4F2F3 െ 4F3F1 4 = ටF1 2 + F2 2 + F3 2 െ F1F2 െ F2F3 െ F3F1 _____________(ii) Take moments of all forces about A. G1 = F2(AAԢ) = F2(ABcos30) = 2aF2 √3 2 = a√3F2 ⇒ F2 = G1 a√3 Similarly by taking moments of all forces about B and C, we get F3 = G2 a√3 and F1 = G3 a√3 Using values of F1, F2 and F3 in (ii), we get R = ඨ G3 2 + G1 2 + G2 2 െ G1G3 െ G1G2 െ G2G3 3a2 = ඨ G1 2 + G2 2 + G3 2 െ G1G2 െ G2G3 െ G3G1 3a2 Which is required.
- 24. Available At: mathcity.org Contact At: qadri86@yahoo.com 24 Couple A pair of forces (FሬԦ, െ FሬԦ) of same magnitude but opposite in direction acting on a rigid body forms a couple. When couple acts on a body it rotates the body. Moment of a couple Let (FሬԦ, െ FሬԦ) be a couple. Let A and B be points on the line of action of FሬԦ and െ FሬԦ respectively. Let rԦ1 and rԦ2 are the position vectors of the points A and B respectively. y B d െFሬԦ rԦ θ FሬԦ rԦ2 A rԦ1 O x z Then the sum of moments of FሬԦ and െFሬԦ about O is GሬሬԦ = rԦ1 × FሬԦ + rԦ2 × (െFሬԦ) = ( rԦ1 െ rԦ2) × FሬԦ From fig. rԦ1 െ rԦ2 ൌ rԦ So GሬሬԦ = rԦ × FሬԦ The vector GሬሬԦ is called the moment of the couple. หGሬሬԦห = ห rԦ × FሬԦห ൌ F rsinθ = Fd ⇒ G = Fd Where d is the perpendicular distance between the line of the action of the forces. Question 15 A couple of moment G acts on a square board ABCD of side a. Replace the couple by the forces acting along AB BD and CA.
- 25. Available At: mathcity.org Contact At: qadri86@yahoo.com 25 Solution Let ABCD be a square board of side a. Let FሬԦ1, FሬԦ2 and FሬԦ3 be the forces acting along AB, BD and CA respectively. Where AB is side and BD and CA are the diagonals of square as shown in fig. D C FሬԦ2 O FሬԦ3 900 450 450 A FሬԦ1 B Take moments of all forces about A. G = F2(AO) = F2(ABsin45) ൌ F2a 1 √2 ⇒ F2 ൌ G√2 a Take moments of all forces about B. G = F3(BO) = F3(ABsin45) ൌ F3a 1 √2 ⇒ F3 ൌ G√2 a Take moment of all forces about D. G = F1(AD) – F3(DO) = F1(a) – F3(ADsin45) ൌ F1a െ F3 a √2 ൌ F1a െ G√2 a a √2 ൌ F1a െ G ⇒ 2G = F1a ⇒ F1 ൌ 2G a
- 26. Available At: mathcity.org Contact At: qadri86@yahoo.com 26 Equivalent couple Any two couples of equal moments lying in the same plane are called equivalent couples. theorem The effect of a couple upon a rigid body is unaltered if it is replaced by any other couple of the same moment lying in the same plane. Proof Let (FሬԦ, െ FሬԦ) be a couple. Let A and B be points on the line of action of FሬԦ and െ FሬԦ respectively. We want to replace the couple (FሬԦ, െ FሬԦ) by any other couple. For this draw two lines AC and BD in the desire direction. Resolve the forces FሬԦ and െFሬԦ at the points A and B respectively into two components. Let QሬሬԦ and SሬԦ are the resolved parts of FሬԦ along CA and BA respectively. Then FሬԦ = QሬሬԦ + SሬԦ __________ (i) Let െQሬሬԦ and െSሬԦ are the resolved parts of െFሬԦ along BD and AB respectively. Since the force SሬԦ and െSሬԦ act along the same lin. Therefore these force balance each other being equal and opposite. Thus we are left with forces QሬሬԦ and െQሬሬԦ acting at A and B along two parallel lines form a couple. So the given couple (FሬԦ, െFሬԦ) has been replaced by the couple (QሬሬԦ, െQሬሬԦ). FሬԦ SሬԦ C QሬሬԦ A B D െQሬሬԦ െSሬԦ െFሬԦ
- 27. Available At: mathcity.org Contact At: qadri86@yahoo.com 27 Now Moment of couple (FሬԦ, െFሬԦ) = BAሬሬሬሬሬሬԦ × FሬԦ = BAሬሬሬሬሬሬԦ × ൫QሬሬԦ + SሬԦ൯ By (i) = BAሬሬሬሬሬሬԦ × QሬሬԦ + BAሬሬሬሬሬሬԦ × SሬԦ = BAሬሬሬሬሬሬԦ × QሬሬԦ + 0 = BAሬሬሬሬሬሬԦ × QሬሬԦ = Moment of couple (QሬሬԦ, െQሬሬԦ) Thus we see that a couple acting on a rigid body can be replaced by another couple of the same moment lying in the same plane. This completes the proof. Compositions of couples Coplanar couples of moments G1, G2, G3, …, Gn are equivalent to a single couple lying in the same plane, whose moment G is given by G = G1 + G2 + G3 + … + Gn Proof Replace couples of moments G1, G2, G3, …, Gn by couples (FሬԦ1, െ FሬԦ1), (FሬԦ2, െ FሬԦ2), (FሬԦ3, െ FሬԦ3), …, (FሬԦn, െ FሬԦn) respectively with common arm d. Then G1 = F1d, G2 = F2d, G3 = F3d, …, Gn = Fnd Now the forces FሬԦ1, FሬԦ2, FሬԦ3, …, FሬԦn act along one straight line and െFሬԦ1, െFሬԦ2, െFሬԦ3, …, െFሬԦn act along a parallel line. So we have a single couple (FሬԦ, െ FሬԦ) with FሬԦ = FሬԦ1 + FሬԦ2 + FሬԦ3 + … + FሬԦn and െFሬԦ = െ(FሬԦ1 + FሬԦ2 + FሬԦ3 + … + FሬԦn) and the moment G of couple is G = Fd = (F1 + F2 + F3 + … + Fn)d = F1d + F2d + F3d + … + Fnd = G1 + G2 + G3 + … + Gn This completes the proof. A force and A couple A force PሬԦ acting on a rigid body can be moved to any point O of the rigid body, provided a couple is added, whose moment is equal to the moment of PሬԦabout O. proof Let the given force PሬԦ act a point A. We want to shift this force PሬԦ to point O of the body. At point O we introduce two equal and opposite force PሬԦ and െPሬԦ. These forces being equal and opposite balance each other. The force PሬԦ act at point A and െPሬԦ act at O form a couple (PሬԦ, െPሬԦ)
- 28. Available At: mathcity.org Contact At: qadri86@yahoo.com 28 and we get a force PሬԦ at point O. Therefore the force PሬԦ acting at the point A is shifted to the point O and a couple has been introduced. PሬԦ A PሬԦ O െPሬԦ Also Moment of couple (PሬԦ, െPሬԦ) = Pd = Moment of the force PሬԦ at A about O This completes the proof. Reduction of a system of coplanar forces to one force and one couple Any system of coplanar forces acting on a rigid body can be reduced to a single force at any arbitrary point in the plane of the forces together with a couple. proof Let FሬԦ1, FሬԦ2, FሬԦ3, … , FሬԦn be a system of forces acting at a points A1, A2, A3, …, An respectively. Let O be a point in the same plane. By shifting the force FሬԦi acting at Ai to point O, we get a force FሬԦi at O together with a couple whose moment GሬሬԦi is equal to the moment of the force FሬԦi about O. Thus by shifting the forces to the point O, we get system of forces FሬԦ1, FሬԦ2, FሬԦ3, …, FሬԦn acting at O together with a system of couples of moments G1, G2, G3, …, Gn. The forces FሬԦ1, FሬԦ2, FሬԦ3, …, FሬԦn acting at O can be replaced by their resultant forces FሬԦ acting at the same point O. Similarly, by the theorem of the composition of couples, all the coplanar couples can be replaced by a single couple of moment G. The force FሬԦ and the couple of the equivalent system are given by FሬԦ = FሬԦi n i=1 and G = Gi n i=1 This completes the proof.
- 29. Available At: mathcity.org Contact At: qadri86@yahoo.com 29 Equation of Line of Action of Resultant y Ai(xi, yi) FሬԦi OԢ x O Let (Xi, Yi) be the component of the force FሬԦi act at the point Ai whose coordinates are (xi, yi) where i = 1, 2, 3, …, n. Let the reduction be made at origin O, we get a single force FሬԦ acting at O together with a couple GሬሬԦ so that FሬԦ = FሬԦi n i=1 = ൫Xiiመ Yijመ൯ n i=1 = Xiiመ n i=1 Yijመ n i=1 = Xi n i=1 iመ Yi n i=1 jመ = Fxiመ + Fyjመ Where Fx and Fy are the component of the resultant FሬԦ. And GሬሬԦ ൌ ൫OAi ሬሬሬሬሬሬሬԦ × FሬԦ i ൯ n i=1 Let the reduction be made at OԢ(x, y). The resultant FሬԦ remains same but moment GሬሬԦ of the couple changes. Let GሬሬԦԢ be the moment of new couple then GሬሬԦԢ = sum of moments about OԢ of FሬԦi ൌ ൫O′Ai ሬሬሬሬሬሬሬሬԦ × FሬԦ i ൯ n i=1
- 30. Available At: mathcity.org Contact At: qadri86@yahoo.com 30 ൌ ቀOAi ሬሬሬሬሬሬԦ – OOԢሬሬሬሬሬሬሬԦቁ × FሬԦi n iൌ1 ൌ OAi ሬሬሬሬሬሬԦ × FሬԦi n iൌ1 െ OOԢሬሬሬሬሬሬሬԦ × FሬԦi n iൌ1 ൌ OAi ሬሬሬሬሬሬԦ × FሬԦi n iൌ1 െ OOԢሬሬሬሬሬሬሬԦ × FሬԦi n iൌ1 ൌ GሬሬԦ െ OOԢሬሬሬሬሬሬሬԦ × FሬԦi n iൌ1 ൌ GሬሬԦ െ ൫xiመ + yjመ൯ × ൫Fxiመ + Fyjመ൯ ൌ GሬሬԦ െ ቮ iመ jመ k x y 0 Fx Fy 0 ቮ ൌ GሬሬԦ െ ൫xFy െ yFx൯k ⇒ GԢ k = Gk െ ൫xFy െ yFx൯k ⇒ GԢ = G െ xFy + yFx If the resultant passes through OԢ then GԢ = 0 ⇒ G െ xFy + yFx = 0 or G – xY + yX = 0 (take Fx = X and Fy = Y) Which is the equation of the line of action of the resultant. Note: A system is in equilibrium if R = G = 0 A system is equivalent to a couple if R = 0 and G ≠ 0 Question 16 A and B are any two points in a lamina on which a system of forces coplanar with it are acting, and when the forces are reduced to a single force at each of these points and a couple, the moments of the couple are Ga and Gb respectively. Prove that when the reduction is made to be a force at the middle of AB and a couple, the moment of the couple is 1 2 ሺGa+ Gbሻ Solution Let the coordinates of A and B are (x1, y1) and (x2, y2). Let C be the midpoint of AB then the coordinates of C are ൬ x1+ x2 2 , y1 + y2 2 ൰ Suppose the given system of forces is reduced to single force acting at O together with a couple G. Let X and Y be the component of the reduced force.
- 31. Available At: mathcity.org Contact At: qadri86@yahoo.com 31 y B C A x O When the same system of forces is reduced to a single force acting at A, the resultant force will remain unchanged, whereas the moment will changed and is given by Ga = G – x1Y + y1X Similarly when the same system of forces is reduced to a single force acting at B, the resultant force will remain unchanged, whereas the moment will changed and is given by Gb = G – x2Y + y2X Suppose that the same system of forces is reduced to a single force acting at C, the resultant force will remain unchanged, whereas the moment will changed and is given by Gc= G െ x1+ x2 2 Y + y1 + y2 2 X= 1 2 ൫2G െ x1Y െ x2Y + y1 X + y2 X൯ = 1 2 ൫G െ x1Y + y1 X + G െ x2Y + y2 X൯ = 1 2 ሺGa + Gbሻ Which is the required. Question 17 Forces P, 2P, 3P, 6P, 5P and 4P act respectively along the sides AB, CB, CD, ED, EF and AF of a regular hexagon of side a, the sense of the forces being indicated by the order of the letters. Prove that the six forces are equivalent to a couple. Solution E 6P D 5P d 3P O F C 4P 2P A P B
- 32. Available At: mathcity.org Contact At: qadri86@yahoo.com 32 Let ABCDEFA be a regular hexagon of side a. Forces P, 2P, 3P, 6P, 5P and 4P act along the sides AB, CB, CD, ED, EF and AF respectively. Take AB along x–axis. Take O is the centre hexagon and d is perpendicular distance of each force from O. Let the R be the magnitude of the resultant of the forces, then R = ටRx 2 + Ry 2 ______________(i) Now by theorem of resolved parts Rx = Sum of the resolved parts of the forces along x-axis = Pcos00 + 2Pcos(1800 + 600 ) + 3Pcos(1800 – 600 ) + 4Pcos(1800 – 600 ) + 5Pcos(1800 + 600 ) + 6Pcos00 = P െ 2Pcos600 െ 3Pcos600 െ 4Pcos600 െ 5Pcos600 + 6P = 7P െ 14Pcos600 = 7P െ 14P ቀ 1 2 ቁ = 7P – 7P = 0 And Ry = Sum of the resolved parts of the forces along y-axis = Psin00 + 2Psin(1800 + 600 ) + 3Psin(1800 – 600 ) + 4Psin(1800 – 600 ) + 5Psin(1800 + 600 ) + 6Psin00 = 0 െ 2Psin600 3Psin600 + 4Psin600 െ 5Psin600 + 0 = 7Psin600 െ 7Psin600 = 0 Using values of Rx and Ry in (i), we get R = 0 Take moment of all forces about point O. G = sum of moments of all forces about O. = Pd – 2Pd + 3Pd – 4Pd + 5Pd – 6Pd = 9Pd – 12Pd = െ3Pd Since R = 0 and G ≠ 0. Therefore the system of the given coplanar forces is equivalent to a couple. Question 18 Forces P1, P2, P3, P4, P5 and P6 act along the sides of a regular hexagon taken in order. Show that they will be in equilibrium if P = 0 and P1 െ P4 = P3 െ P6 = P5 െ P2 Solution
- 33. Available At: mathcity.org Contact At: qadri86@yahoo.com 33 Let ABCDEFA be a regular hexagon. Forces P1, P2, P3, P4, P5 and P6 act along the its sides taken one way round. O is the centre of hexagon and d is perpendicular distance of all forces from O. Take AB along x–axis. E P4 D P5 P3 d F C P6 P2 A P1 B Take moment of all forces about O. G = P1d + P2d + P3d + P4d + P5d + P6d = (P1 + P2 + P3 + P4 + P5 + P6)d = d P Let X and Y be the resolved parts of the resultant of the forces then by the theorem of the resolved parts. X = Sum of the resolved parts of the forces along x-axis = P1cos00 + P2cos60 + P3cos120 + P4cos180 + P5cos240 + P6cos300 = P1 + P2 ൬ 1 2 ൰ െ P3 ൬ 1 2 ൰ െ P4 െ P5 ൬ 1 2 ൰ + P6 ൬ 1 2 ൰ = P1 െ P4 + 1 2 ሺP2 െ P3 െ P5 + P6ሻ And Y = Sum of the resolved parts of the forces along y-axis = P1sin00 + P2sin60 + P3sin120 + P4sin180 + P5sin240 + P6sin300 = 0 + P2 ቆ √3 2 ቇ P3 ቆ √3 2 ቇ െ 0 െ P5 ቆ √3 2 ቇ െ P6 ቆ √3 2 ቇ = √3 2 ሺP2 P3 െ P5 െ P6ሻ System of forces is in equilibrium if and only if X = Y = 0 and G = 0 If G = 0 then d P ൌ 0 ⇒ P ൌ 0 or P1 + P2 + P3 + P4 + P5 + P6 ൌ 0 If Y = 0 then √3 2 ሺP2 P3 െ P5 െ P6ሻ ൌ 0 ⇒ P5 െ P2 ൌ P3 െ P6 ____________(i)
- 34. Available At: mathcity.org Contact At: qadri86@yahoo.com 34 If X = 0 then P1 െ P4 + 1 2 ሺP2 െ P3 െ P5 + P6ሻ ൌ 0 ⇒ P1 െ P4 + 1 2 ൫P2 െ P5 െ ሺP3 െ P6ሻ൯ ൌ 0 ⇒ P1 െ P4 + 1 2 ൫P2 െ P5 െ ሺP5 െ P2ሻ൯ ൌ 0 By (i) ⇒ P1 െ P4 + 1 2 ሺP2 െ P5 െ P5 + P2ሻ ൌ 0 ⇒ P1 െ P4 + 1 2 ሺ2P2 െ 2P5ሻ ൌ 0 ⇒ P1 െ P4 + P2 െ P5 ൌ 0 ⇒ P5 െ P2 ൌ P1 െ P4 ____________(ii) From (i) and (ii), we get P1 െ P4 ൌ P5 െ P2 ൌ P3 െ P6 Hence the system is in equilibrium if P = 0 and P1 െ P4 = P3 െ P6 = P5 െ P2 Question 19 OAB is an equilateral triangle of a side a ; C is the mid-point of OA. Forces 4P, P and P act along the sides OB, BA and AO respectively. If OA and OY (parallel to BC) are taken as x- and y-axis. Prove that the resultant of the forces is 3P and the equation of its line of action is 3y = √3 (3x + a) Solution y B D 4P 900 P 300 600 x O C P A
- 35. Available At: mathcity.org Contact At: qadri86@yahoo.com 35 Let the R be the magnitude of the resultant of the forces, then R = ටRx 2 + Ry 2 ______________(i) Now by theorem of resolved parts Rx = Sum of the resolved parts of the forces along x-axis = Pcos1800 + Pcos300 + 4Pcos60 = െ P + P ൬ 1 2 ൰ + 4P ൬ 1 2 ൰ = െ P + 5P ൬ 1 2 ൰ = 3 2 P And Ry = Sum of the resolved parts of the forces along y-axis = Psin1800 + Psin300 + 4Psin60 = െ P ቆ √3 2 ቇ + 4P ቆ √3 2 ቇ = 3√3 2 P Using values of Fx and Fy in (i), we get R = ඨ൬ 3 2 P൰ 2 + ቆ 3√3 2 Pቇ 2 = ඨ 9 4 P2 + 27 4 P2 = ඨ 36 4 P2 = ඥ9P2 ൌ 3P Let G be the sum of moments of all forces about O. Then G = െP(OD) = െ P(OAcos30) = െ P a√3 2 The equation of line of action of resultant of resultant is G – xRy + yRx = 0 ⇒ െ P a√3 2 െ x ቆ 3√3 2 Pቇ + y ൬ 3 2 P൰ = 0 ⇒ െ a√3 െ 3√3x + 3y = 0
- 36. Available At: mathcity.org Contact At: qadri86@yahoo.com 36 ⇒ 3y = 3√3x a√3 ⇒ 3y = √3ሺ3x a) Which is required. Distance of a point from a line The distance d from the point P(x1, y1) to the line ax + by + c = 0 is given by d = หax1 + by1 + cห ඥa2 + b2 Question 20 Forces of magnitude P, 2P, 3P and 4P act respectively along the sides AB, BC, CD and DA of a square ABCD of side ‘a’ and forces each of magnitude 8√2P act along the diagonals BD and AC. Find the magnitude of the resultant force and the distance of its line of action from A. Solution Y D 3P C 8√2P 4P E 2P 8√2P X A P B Let ABCD be a square of side a. Forces act along its sides according to given conditions taking one way round. Take AB along x-axis and AD along y-axis. Let the R be the magnitude of the resultant of the forces, then R = ටRx 2 + Ry 2 ______________(i) Now by theorem of resolved parts Rx = Sum of the resolved parts of the forces along x-axis = Pcos00 + 2Pcos900 + 3Pcos1800 + 4Pcos2700 + 8√2Pcos45 + 8√2Pcos135 = P + 0 െ 3P + 0 + 8√√√√2P ൬൬൬൬ 1 √√√√2 ൰൰൰൰+ 8√√√√2P ൬൬൬൬െ 1 √√√√2 ൰൰൰൰ = െ 2P + 8P െ 8P ൌ െ 2P
- 37. Available At: mathcity.org Contact At: qadri86@yahoo.com 37 And Ry = Sum of the resolved parts of the forces along y-axis = Psin00 + 2Psin900 + 3Psin1800 + 4Psin2700 + 8√2Pcos45 + 8√2Pcos135 = 0 + 2P + 0 െ 4P + 8√√√√2P ൬൬൬൬ 1 √√√√2 ൰൰൰൰+ 8√√√√2P ൬൬൬൬ 1 √√√√2 ൰൰൰൰ = െ 2P + 8P + 8P ൌ 14P Using values of Rx and Ry in (i), we get R = ඥሺെ2Pሻ2 + ሺ14Pሻ2 = ඥ 200P2 ൌ 10√2P Now G = sum of the moments of the forces about A. = 2P(AB) + 3P(AD) + 8√2P(AE) = 2Pa + 3Pa + 8√2P(ABsin450 ) = 5Pa + 8√√√√2Pa ൬൬൬൬ 1 √√√√2 ൰൰൰൰ ൌ 13Pa The equation of line of action of resultant is G – xRy + yRx = 0 ⇒ 13Pa െ xሺ14Pሻ + yሺെ2Pሻ = 0 ⇒ 13a െ 14x െ 2y = 0 The distance of line of action of resultant from A is: |0 + 0 +13a| ඥሺെ14ሻ2 + ሺെ2ሻ2 = 13a √200 = 13a 10√2 Which is required. Circumcentre of a triangle Circumcentre of the triangle is a point at which right bisector of the triangle meet with one another. Question 21 The three forces P, Q and R act along the sides BC, CA and AB respectively of a triangle ABC. Prove that if P cosA + Q cosB + R cosC = 0 Then the line of the action of the resultant passes through the circumcentre of the triangle.
- 38. Available At: mathcity.org Contact At: qadri86@yahoo.com 38 Solution Let ABC be a triangle and AD, BE and CF are right bisector of the triangle. O be the circumcentre and r be the circumradius of the triangle. Then AO = BO = CO = r A 2A Q F r E 2A OR A r r B D P C Let G be the moment of all forces about O. Then G = P(OD) + Q(OE) + R(OF) ___________(i) From fig. OD = BOcosA = rcosA OE = COcosB = rcosB OF = AOcosC = rcosC Using these values in (i), we get G = P(rcosA) + Q(rcosB) + R(rcosC) = r(PcosA + QcosB + RcosC) The line of the action of the resultant passes through O if G = 0. i.e. r(PcosA + QcosB + RcosC) = 0 ⇒ PcosA + QcosB + RcosC = 0 r ≠ 0 Thus if P cosA + Q cosB + R cosC = 0 Then the line of the action of the resultant passes through O the circumcentre of the triangle. Orthocentre of a Triangle Orthocentre of the triangle is a point at which altitudes (i.e. perpendicular from the vertices to the opposite sides) of the triangle meet with one another.
- 39. Available At: mathcity.org Contact At: qadri86@yahoo.com 39 Question 22 The three forces P, Q and R act along the sides BC, CA and AB respectively of a triangle ABC. Prove that if P secA + Q secB + R secC = 0 Then the line of the action of the resultant passes through the orthocentre of the triangle. Solution A N M 2C O Q R C 90 – C B L P C Let ABC be a triangle and O be the orthocentre. Draw perpendiculars AL on BC, BM on AC and CN on AB. These are also called altitudes. Let a, b and c be the lengths of the sides BC, CA and AB respectively. Let G be the moment of all forces about O. Then G = P(OL) + Q(OM) + R(ON) __________(i) From fig. ∠LBO = 900 – C OL BL ൌ tan(900 െ C) ⇒ OL BL ൌ cotC ⇒ OL ൌ BL cotC __________(ii) In ∆ABL BL AB ൌ cosB ⇒ BL c ൌ cosB ⇒ BL ൌ c cosB Using value of BL in (ii), we get OL ൌ c cosB cosC sinC ⇒ OL ൌ c sinC cosB cosC __________(iii)
- 40. Available At: mathcity.org Contact At: qadri86@yahoo.com 40 By law of sine a sinA ൌ b sinB ൌ c sinC ൌ k __________(iv) From (iii) and (iv), we get OL ൌ kcosBcosC Similarly OM ൌ kcosAcosC ON ൌ kcosAcosB Using values of OL, OM and ON in (i), we get G = P(kcosBcosC) + Q(kcosAcosC) + R(kcosAcosB) = k(PcosBcosC + QcosAcosC) + RcosAcosB) The line of the action of the resultant passes through O if G = 0. i.e. k(PcosBcosC + QcosAcosC + RcosAcosB) = 0 ⇒ PcosBcosC + QcosAcosC + RcosAcosB = 0 Dividing by cosAcosBcosC, we get P cosA + Q cosB + R cosC = 0 ⇒ PsecA + QsecB + RsecC = 0 Thus if P secA + Q secB + R secC = 0 Then the line of the action of the resultant passes through O the orthocentre of the triangle. theorem If three forces are represented in magnitude, direction and position by the sides of a triangle taken in order. They are equivalent to a couple. The magnitude of the moment of the couple is equal to the twice the area of the triangle. Solution CBሬሬሬሬሬԦ A d B C Let ABC be a triangle and let the three forces be completely represented by ABሬሬሬሬሬሬԦ, BCሬሬሬሬሬԦ and CAሬሬሬሬሬሬԦ as shown in figure. Then
- 41. Available At: mathcity.org Contact At: qadri86@yahoo.com 41 BCሬሬሬሬሬԦ + CAሬሬሬሬሬሬԦ + ABሬሬሬሬሬሬԦ = 0 ⇒ CAሬሬሬሬሬሬԦ + ABሬሬሬሬሬሬԦ = െ BCሬሬሬሬሬԦ ⇒ CAሬሬሬሬሬሬԦ + ABሬሬሬሬሬሬԦ = CBሬሬሬሬሬԦ Which shows that the forces CAሬሬሬሬሬሬԦ and ABሬሬሬሬሬሬԦ acting at A are equivalent to a force CBሬሬሬሬሬԦ which acts at A. Thus the given three forces are equivalent to two forces CBሬሬሬሬሬԦ acting at A and BCሬሬሬሬሬԦ along the side BC of the triangle. These two forces form a couple. If d denotes the length of the perpendicular from A to BC. Then Magnitude of moment of the couple = (BC)d = 2 ൬ 1 2 BC.d൰ = 2(Area of the triangle) This completes the proof. Question 23 Forces act along the sides BC, CA and AB of a triangle. Show that they are equivalent to a couple only if the forces are proportional to the sides. Solution Let forces λBCሬሬሬሬሬԦ, (λ + µ)CAሬሬሬሬሬሬԦ and (λ + ν)ABሬሬሬሬሬሬԦ act along the sides of a triangle ABC. Then λBCሬሬሬሬሬԦ + (λ + µ)CAሬሬሬሬሬሬԦ + (λ + ν)ABሬሬሬሬሬሬԦ = λ(BCሬሬሬሬሬԦ + CAሬሬሬሬሬሬԦ ABሬሬሬሬሬሬԦ) + µCAሬሬሬሬሬሬԦ + νABሬሬሬሬሬሬԦ Since the forces BCሬሬሬሬሬԦ , CAሬሬሬሬሬሬԦ and ABሬሬሬሬሬሬԦ are equivalent to a couple whose moment is twice the area of the triangle ABC, we have λBCሬሬሬሬሬԦ + (λ + µ)CAሬሬሬሬሬሬԦ + (λ + ν)ABሬሬሬሬሬሬԦ = a couple + µCAሬሬሬሬሬሬԦ + νABሬሬሬሬሬሬԦ The system is equivalent to a couple only if A (λ + ν)ABሬሬሬሬሬሬԦ (λ + µ)CAሬሬሬሬሬሬԦ B λBCሬሬሬሬሬԦ C
- 42. Available At: mathcity.org Contact At: qadri86@yahoo.com 42 µCAሬሬሬሬሬሬԦ + νABሬሬሬሬሬሬԦ = 0 Which holds only if µ ൌ ν = 0 Thus the forces along the sides of the triangle are λBCሬሬሬሬሬԦ, λCAሬሬሬሬሬሬԦ and λABሬሬሬሬሬሬԦ. Hence forces acting along the sides of a triangle are equivalent to a couple only if they are proportional to the sides of triangle. %%%% End of The Chapter # 1%%%%