Fundamental of Statics (Part 2)
- 1. Department of Mechanical Engineering. Prof. Malay Badodariya +91 9429 158833 Unit no.: 2 Fundamental Of Statics Subject Name: FMD 01ME0305 FundamentalOfMachineDesign (01ME0504) Unit 2: Fundamental of Statics (Part 2)
- 2. Topic will be cover: 1. Equilibrium of rigid bodies: a) Condition of equilibrium of coplanar concurrent forces b) Concept of Free body diagram c) Sine Rule of triangle d) Lami’s Theorem Department of Mechanical Engineering. Prof. Malay Badodariya malay.badodariya@gmail.com Unit no.: 2 Fundamental Of Statics Subject Name: FMD 01ME0305
- 3. a) Condition of equilibrium ofCoplanar Concurrent Forces A system of coplanar concurrent forces is said to be in equilibrium if the following condition are satisfied. 1. ⅀H = 0, i.e. Algebric sum of all the horizontal forces must be zero. 2. ⅀V = 0, i.e. Algebric sum of all the vertical forces must be zero. 3. Since, ⅀H = 0, ⅀V = 0 , “R Must be Zero”
- 4. Topic will be cover: 1. Equilibrium of rigid bodies: a) Condition of equilibrium of coplanar concurrent forces b) Concept of Free body diagram c) Sine Rule of triangle d) Lami’s Theorem Department of Mechanical Engineering. Prof. Malay Badodariya malay.badodariya@marwadie ducation.edu.in Unit no.: 2 Fundamental Of Statics Subject Name: FMD 01ME0305
- 5. b) Concept of Free Body Diagram A scalar is simply a number, a magnitude alone. A force is usually shown as a vector, which includes both magnitude and a direction. “Force (or free-body) diagrams show the relative magnitude and direction of all forces acting upon an object. The object must be isolated and “free” of its surroundings.”
- 6. This is a free-body diagram of the Statue of Liberty. She is represented by a simple box. The forces acting on her are labeled with a magnitude and the arrow shows direction. Notice the surrounding objects are stripped away and the forces acting on the object are shown. ‘W’ here represents the force of the weight of the statue. ‘N’ is the normal force, which represents the force Liberty Island is pushing back up on the statue. The island has a great resistance to compression. The ground is exerting a force upward on the statue. perpendicular, or normal, to the surface.
- 7. Free Body Diagram of the wooden swing (The box represents the wooden swing, W = weight of the swing and the parrot, T represents the ropes that are in tension supporting the weight) Draw a FBD of the wooden swing:
- 8. Free Body Diagram of the Ring at point ‘C’: Free Body Diagram of the ring at point C (T represents the force of the cables that are in tension acting on the ring) Draw a FBD of the ring at point C:
- 9. Free Body Diagram of the traffic light (𝑇𝐶𝐷 represent the force of the cables acting on the light and ‘W’ is the weight acting on the light.) Draw a FBD of theTraffic Light
- 10. Draw a FBD of the pin at point A:
- 11. Topic will be cover: 1. Equilibrium of rigid bodies: a) Condition of equilibrium of coplanar concurrent forces b) Concept of Free body diagram c) Sine Rule of triangle d) Lami’s Theorem Department of Mechanical Engineering. Prof. Malay Badodariya malay.badodariya@marwadie ducation.edu.in Unit no.: 2 Fundamental Of Statics Subject Name: FMD 01ME0305
- 12. c) Sine Rule The Law of Sines (sine rule) is an important rule relating the sides and angles of any triangle (it doesn't have to be right-angled) Consider Δ ABC, Side AB = c Side AC = b Side BC = a If a, b and c are the lengths of the sides opposite the angles A, B and C in a triangle, then: 𝑎 𝑆𝑖𝑛 𝐴 = 𝑏 𝑆𝑖𝑛 𝐵 = 𝑐 𝑆𝑖𝑛 𝐶 We need a right-angled triangle to prove the above equation.
- 13. Draw a perpendicular, CD ⊥ AB.Then CD = h is the height of the triangle. “h” separates the △ ABC in two right- angled triangles, △CDA and △CDB. In the △CDA, Sin A = 𝐶𝐷 𝐶𝐴 = ℎ 𝑏 And in △CDB, Sin B = 𝐶𝐷 𝐶𝐵 = ℎ 𝑎 So fromAbove 2 equation, 𝑆𝑖𝑛 𝐴 𝑆𝑖𝑛 𝐵 = ℎ 𝑏 ℎ 𝑎 = ℎ.𝑎 ℎ.𝑏 = 𝑎 𝑏 𝑎 𝑆𝑖𝑛 𝐴 = 𝑏 𝑆𝑖𝑛 𝐵 Similarly, we can prove, Sin B/ Sin C= b / c and so on for any pair of angles and their opposite sides.
- 14. Topic will be cover: 1. Equilibrium of rigid bodies: a) Condition of equilibrium of coplanar concurrent forces b) Concept of Free body diagram c) Sine Rule of triangle d) Lami’s Theorem Department of Mechanical Engineering. Prof. Malay Badodariya malay.badodariya@marwadie ducation.edu.in Unit no.: 2 Fundamental Of Statics Subject Name: FMD 01ME0305
- 15. d) Lami’s Theorem It states, “If three coplanar forces acting at a point be in equilibrium, then each force is proportional to the sine of the angle between the other two.” Mathematically, 𝑷 𝑺𝒊𝒏 α = 𝑸 𝑺𝒊𝒏 β = 𝑹 𝑺𝒊𝒏 γ where, P, Q, and R are three forces and a, ß, γ are the angles as shown in Fig.
- 16. Proof: Draw the three forces P, Q and R one after the other in direction and magnitude starting from point a. Since the body is in equilibrium (resultant is zero), the last point must coincide with a. Thus, it results in triangle of forces abc as shown in Fig. 2.
- 17. Since ab is parallel to Q, bc is parallel to P, ca is parallel to R. In the triangle of forces abc, ab = Q, bc = P, ac = R
- 18. Applying Sine rule for Δ ABC, 𝑎𝑏 𝑆𝑖𝑛 (180°−β) = 𝑏𝑐 𝑆𝑖𝑛 (180°−α) = 𝑐𝑎 𝑆𝑖𝑛 (180°−γ) 𝑄 𝑆𝑖𝑛 β = 𝑝 𝑆𝑖𝑛 α = 𝑅 𝑆𝑖𝑛 γ
- 19. Example on Lami’sTheorem Example 1: A sphere of weight 100 N is tied to a smooth wall by a string as shown in Fig. 2.33(a). Find the tension T in the string and reaction R of the wall. Solution: Given Data, 15° R T 100 N 15° R T 100 N Free Body Diagram
- 20. Example on Lami’sTheorem Example 1: A sphere of weight 100 N is tied to a smooth wall by a string as shown in Fig. 2.33(a). Find the tension T in the string and reaction R of the wall. Solution: Given Data, 15° R T 100 N Free Body Diagram Applying Lami’sTheorem to the system of force. 𝑻 𝑺𝒊𝒏 𝟗𝟎° = 𝑹 𝑺𝒊𝒏 (𝟏𝟖𝟎° − 𝟏𝟓°) = 𝟏𝟎𝟎 𝑺𝒊𝒏 (𝟗𝟎° + 𝟏𝟓°) 𝑻 𝟏 = 𝟏𝟎𝟎 𝑺𝒊𝒏 𝟗𝟎° 𝑺𝒊𝒏 (𝟗𝟎° + 𝟏𝟓°) T = 103.53 N 𝑹 𝟏 = 𝟏𝟎𝟎 𝑺𝒊𝒏 𝟏𝟔𝟓° 𝑺𝒊𝒏 (𝟗𝟎° + 𝟏𝟓°) R = 26.79 N
- 21. Example on Lami’sTheorem Example 2: An electric light fixture weighting 15 N hangs from a point C, by two strings AC and BC. The string AC is inclined at 60° to the horizontal and BC at 45° to the horizontal as shown in Fig. Solution: Given Data, 𝑻𝒂𝒄 𝑺𝒊𝒏 (𝟏𝟑𝟓°) = 𝑻𝒃𝒄 𝑺𝒊𝒏 (𝟏𝟓𝟎°) = 𝟏𝟓 𝑺𝒊𝒏 (𝟕𝟓°) 𝑻𝒂𝒄 𝟏 = 𝟏𝟓 𝑺𝒊𝒏 (𝟏𝟑𝟓°) 𝑺𝒊𝒏 (𝟕𝟓°) = 𝟏𝟎. 𝟗𝟖 𝑵 𝑻𝒃𝒄 = 𝟏𝟓𝑺𝒊𝒏 𝟏𝟓𝟎° 𝑺𝒊𝒏 (𝟕𝟓°) = 7.76 N
- 22. Example on Lami’sTheorem Example 3: A string ABCD, attached to fixed points A and D has two equal weihts of 1000 N attached to it at B and C. The weights rest with the portions AB and CD inclined at angles as shown in Fig. Find the tensions in the portions AB, BC and CD of the string, if the inclination of the portion BC with the vertical is 120°. Solution: Given Data,
- 23. Example on Lami’sTheorem Example 3: A string ABCD, attached to fixed points A and D has two equal weihts of 1000 N attached to it at B and C. The weights rest with the portions AB and CD inclined at angles as shown in Fig. Find the tensions in the portions AB, BC and CD of the string, if the inclination of the portion BC with the vertical is 120°. Solution: Given Data, 𝑻𝒂𝒃 𝑺𝒊𝒏 (𝟔𝟎°) = 𝑻𝒃𝒄 𝑺𝒊𝒏 (𝟑𝟎°) = 𝟏𝟎𝟎𝟎 𝑺𝒊𝒏 (𝟏𝟓𝟎°) 𝑻𝒂𝒃 𝟏 = 𝟏𝟎𝟎𝟎 𝑺𝒊𝒏 𝟔𝟎° 𝑺𝒊𝒏 (𝟏𝟓𝟎°) Tab = 𝟏𝟕𝟑𝟐. 𝟎𝟓 𝑵 𝑻𝒃𝒄 𝟏 = 𝟏𝟎𝟎𝟎 𝑺𝒊𝒏 (𝟑𝟎°) 𝑺𝒊𝒏 (𝟏𝟓𝟎°) 𝑻𝒃𝒄 = 𝟏𝟎𝟎𝟎 𝑵
- 24. Example on Lami’sTheorem Example 3: A string ABCD, attached to fixed points A and D has two equal weihts of 1000 N attached to it at B and C. The weights rest with the portions AB and CD inclined at angles as shown in Fig. Find the tensions in the portions AB, BC and CD of the string, if the inclination of the portion BC with the vertical is 120°. Solution: Given Data, 𝑻𝒃𝒄 𝑺𝒊𝒏 (𝟏𝟐𝟎°) = 𝑻𝒄𝒅 𝑺𝒊𝒏 (𝟏𝟐𝟎°) = 𝟏𝟎𝟎𝟎 𝑺𝒊𝒏 (𝟏𝟐𝟎°) 𝑻𝒃𝒄 𝟏 = 𝟏𝟎𝟎𝟎 𝑺𝒊𝒏 (𝟏𝟐𝟎°) 𝑺𝒊𝒏 (𝟏𝟐𝟎°) Tbc = 𝟏𝟎𝟎𝟎 𝑵 𝑻𝒄𝒅 𝟏 = 𝟏𝟎𝟎𝟎 𝑺𝒊𝒏 (𝟏𝟐𝟎°) 𝑺𝒊𝒏 (𝟏𝟐𝟎°) Tcd = 𝟏𝟎𝟎𝟎 𝑵
- 25. Example on Lami’sTheorem Example 4: A light string ABCDE whose extremity A is fixed, has weightsW attached to it at B and C. It passes round a small smooth peg at D carrying a weight of 300 N at the free end E as shown in Fig. If in the equilibrium position, BC is horizontal and AB and CD make 150° and 120° with BC, find (i)Tensions in the portion AB, BC and CD of the string and (ii) Magnitudes of W1 andW2. Solution: Given Data,
- 26. Example on Lami’sTheorem Example 4: A light string ABCDE whose extremity A is fixed, has weightsW attached to it at B and C. It passes round a small smooth peg at D carrying a weight of 300 N at the free end E as shown in Fig. If in the equilibrium position, BC is horizontal and AB and CD make 150° and 120° with BC, find (i)Tensions in the portion AB, BC and CD of the string and (ii) Magnitudes of W1 andW2. Solution: Given Data, 𝑻𝒃𝒄 𝑺𝒊𝒏 (𝟏𝟓𝟎°) = 𝑻𝒄𝒅 𝑺𝒊𝒏 (𝟗𝟎°) = 𝑾𝟐 𝑺𝒊𝒏 (𝟏𝟐𝟎°) 𝑻𝒃𝒄 𝑺𝒊𝒏 (𝟏𝟓𝟎°) = 𝟑𝟎𝟎 𝑵 𝑺𝒊𝒏 (𝟗𝟎°) = 𝑾𝟐 𝑺𝒊𝒏 (𝟏𝟐𝟎°) Tbc = 150 N W2 = 259.8 N
- 27. Example on Lami’sTheorem Example 4: A light string ABCDE whose extremity A is fixed, has weightsW attached to it at B and C. It passes round a small smooth peg at D carrying a weight of 300 N at the free end E as shown in Fig. If in the equilibrium position, BC is horizontal and AB and CD make 150° and 120° with BC, find (i)Tensions in the portion AB, BC and CD of the string and (ii) Magnitudes of W1 andW2. Solution: Given Data, 𝑻𝒃𝒄 𝑺𝒊𝒏 (𝟏𝟐𝟎°) = 𝑻𝒃𝒂 𝑺𝒊𝒏 (𝟗𝟎°) = 𝑾𝟏 𝑺𝒊𝒏 (𝟏𝟓𝟎°) 𝟏𝟓𝟎 𝑵 𝑺𝒊𝒏 (𝟏𝟐𝟎°) = 𝑻𝒃𝒂 𝑺𝒊𝒏 (𝟗𝟎°) = 𝑾𝟏 𝑺𝒊𝒏 (𝟏𝟓𝟎°) Tba = 173.2 N W1 = 86.6 N
- 28. Department of Mechanical Engineering. Prof. Malay Badodariya malay.badodariya@marwadie ducation.edu.in