Coplanar Non-concurrent Forces
- 1. 1 GOVERNMENT ENGINEERING COLLEGE, PALANPUR Subject :- Mechanics of Solids
- 3. Coplanar Non-concurrent Force System:- This is the force system in which lines of action of individual forces lie in the same plane but act at different points of application. COPLANAR NON - CONCURRENT FORCE SYSTEM F2 F1 F3 F1 F2 F5 F4 F3 3
- 4. Moment Of A Force :- The moment of a force is equal to the product of the force and the perpendicular distance of line of action of force from the point about which the moment is required. M = P x X Here, M = moment P = Force X = Perpendicular distance between the line of action of force and the point about which moment is required. Unit of moment is N.m. or kN,m. Examples of moment are :- i) To tight the nut by spanner ii) To open or close the door Sign for moment:- Generally, clockwise moment is taken as positive and anticlockwise moment is taken as negative.
- 5. Unit of moment is N.m. Or kN. m.
- 6. Couple:- To equal and opposite forces whose lines of action are different form a couple. Resultant or net force of couple is zero. Hence, couple acting on a body do not create any translatory motion of the body. Couple produces only rotational motion of the body. • Types of couple: There are two types of couple. i ) Clockwise couple ii) Anticlockwise couple Clockwise couple:- It rotates the body in clockwise direction. It is taken as positive. Anticlockwise couple:- It rotates the body in anticlockwise direction. It is taken as negative.
- 7. Arm of couple :- The perpendicular distance between the lines of action of two forces forming couple is know as the arm of couple. Moment of couple :- Moment of couple = Forces x arm of couple M = P x a unit of couple is N.M. Or KN.M Example of couple are :- I. Forces applied to the key of a lock, while locking or unlocking it. II. Forces applied on the steering wheel of a car by two hands to steer the car towards left or right. III.To open or close wheel valve of a water supply pipe line.
- 8. CHARACTERISTICS OF A COUPLE :- I. The algebraic sum of the forces, forming the couple is zero. II. The algebric sum of the moment of the forces, forming the couple, about any point is the same and equal to the moment of the couple itself. Moment of couple M = P.a Now, taking moment @ 0, Mo = P ( a + x ) – P.x = P.a + P.x – P.x = P.a III. A couple cannot be balanced by a single force, but can be balanced only by a couple, but of opposite nature. IV. If number of coplanar couples are acting on the body, the they can be reduced to a single couple, whose magnitude will be equal to the algebraic sum of moment of all the couples.
- 9. DIFFRENCE:- MOMENT COUPLE Moment = Force x Perpen- dicular distance M = P.x It is produced by a single Force not passing through C.G. of the body. The force move the body in the direction of force and rotate the body. To balance the force causing moment, equal and opposite forces is required Two equal and opposite forces whose lines of action are different form a couple. It is produced by two equal and opposite parallel, non- collinear forces. Resultant force of couple is zero. Hence, body does not move, but rotate only. Couple can not be balance by a single force. It can be balance by a couple only.
- 10. Example :- Replace the couple and force by a single force – couple applied at a for the lever shown in figure. Also find the distance of a point C from A where only a single force can replace the force-couple System. Solution:- Moment of couple of 250 kN force is = 250 × 120 = 30000N.mm
- 11. Single force to replace the force couple system : Let, 500 N is applied at distance x from A at point C. Moment of force and couple at b from = (250×120) + 500 × (200cos 60̇ ) = 30000 + 50000 = 80000 N.mm …..( i ) Moment of force 500 N at C @ B, = 500× (200 + x) cos60̇ = 250 (200 + x) = 50000 + 250 x …..( ii ) Equating ( i ) and ( ii ) 50000 + 250 x = 80000 250 x = 30000 x = 120 mm.
- 12. VARIGNON’S PRINCIPLE OF MOMENTS It states, ‘If number of coplanar forces are acting simultaneously on a particle, the algebraic sum of the moments of all the forces about any point is equal to the moment of their resultant force about the same point.
- 13. Proof:- Consider two forces P and Q acting at a point a represented in magnitude and direction by AB and AC as shown in figure. let, O be the point, about which the moments are taken. Through O, draw a line CD parallel to the direction of force P, to meet the line of action of Q at C. Now with AB and AC as two adjacent sides, complete parallelogram ABCD. Join the diagonal AD of the parallelogram and OA and OB, diagonal AD represents in magnitude and direction the resultant of two forces P and Q. Now, taking moment of forces P,Q and R about O. Moment of force P about O = 2 × Area of triangle AOB ….. ( i ) Similarly, moment of force Q, about O, = 2 × Area of triangle AOC …..( ii ) O C D R Q A P B
- 14. Moment of resultant R, about O, = 2 × Area of triangle AOD …..( iii ) Now, from the geometry of the figure, Area of ∆ AOD = Area of ∆ AOC + Area of ∆ ACD But area of ∆ACD = Area of ∆ ADB = Area of ∆ AOB Therefore, Area of ∆ AOD = Area of ∆ AOC + Area of ∆ AOB Multiplying both sides by 2, 2 × Area of ∆ AOD = 2 × Area of ∆ AOC + 2 × Area of ∆ AOB Therefore, moment of force P and O = moment of force P and O + moment of force Q and O ( Here, we have considered only two forces. But this principle can be exerted for any number of forces.)
- 15. If the body is acted upon by a number of co-planar non-concurrent forces. It may have one of the following stats: I. The body may move in any one direction. II. The body may rotate about itself without moving. III. The body may move in any one direction, and at the same time it may also rotate about itself. IV. The body may be completely at rest. CONDITION OF EQUILIBRIUM FOR COPLANER NON – CONCURRENT FORCES:-
- 16. Example :- Determine the resultant and locate the same with respect to point ‘A’ of a non-concurrent force system shown in figure. Solution:- ∑H = 0.5 cos60̇ + 1.2 – 2.1 cos45̇ = 0.25 + 1.2 – 1.485 = - 0.035kN (←) ∑V = 1.1 C 0.5 sin60̇ + 2.1 sin45̇ = 1.1 + 0.433 + 1.485 = 3.02kN (↑) R = √(∑H)2 + (∑V)2 = √(-0.035)2+ (3.02)2 = 3.02kN
- 17. tanθ = = = 86.28 ∑V ∑H 3.02 0.035 θ = 89.33̇ ∑H = -Ve ∑V = +Ve Therefore, R will be in 2 quadrant. Location of resultant: Let R is at perpendicular distance x from A. Taking moment @ A. R.x + (1.2×1.5) + 0.5sin60̇ × 2 = 4.0 +2.1cos45̇ × 1.54 3.02 x + 1.8 + 0.866 = 4 + 2.23 3.02 x = 3.564 x = 1.18 m .