1 5 Static equilibrium of a rigid body

1. 5 Static equilibrium of a rigid body © Manhattan Press (H. K. ) Ltd. 1

1. 5 Static equilibrium of a rigid body (SB p. 64) Static equilibrium if (a) resultant force = 0 (b) algebraic sum of moments of forces about any axis = 0 Sum of all the clockwise moments = Sum of all the anticlockwise moments Principle of Moments © Manhattan Press (H. K. ) Ltd. Go to Example 10 Example 11 2

End © Manhattan Press (H. K. ) Ltd. 3

1. 5 Static equilibrium of a rigid body (SB p. 64) Q: A uniform rod PQ of length 10 m and mass 2 kg is supported horizontally at A, 2 m from one end at B, 4 m from the other end. A mass of 3 kg is placed at P and another mass of 2 kg at Q. What are the normal reactions at the supports A and B? (Assume g = 9. 81 m s− 2. ) Solution © Manhattan Press (H. K. ) Ltd. 4

1. 5 Static equilibrium of a rigid body (SB p. 64) Solution: Suppose R and S are the normal reactions at A and B respectively. Since the rod is in equilibrium, the resultant force in any direction is zero. Therefore, in the direction vertically upwards, R + S – 3 g – 2 g = 0 ( g = acceleration due to gravity) R + S = 7 g Also, algebraic sum of moments about any axis is zero. Return to Text Taking moments about the horizontal axis that passes through A (so that R does not appear in the equation), (3 g × 2) − (2 g × 3) + (S × 4) − (2 g × 8) = 0 4 S = 16 g S = 4 × 9. 81 = 39. 2 N R = 7 g – 39. 2 = 29. 4 N © Manhattan Press (H. K. ) Ltd. 5

1. 5 Static equilibrium of a rigid body (SB p. 65) Q: A ladder of length 10 m and weight 200 N leans against a smooth wall such that it is at an angle of 60° to the horizontal. A boy of weight 500 N stands on the ladder ¼ of the way from its lower end. Calculate the normal reaction at the wall and the magnitude and direction of the resultant force acting on the lower end of the ladder. Solution © Manhattan Press (H. K. ) Ltd. 6

1. 5 Static equilibrium of a rigid body (SB p. 65) Solution: Since the wall is smooth, the reaction R is normal to the wall. The forces acting on the lower end A of the ladder consists of the normal reaction vertically upwards and the friction in the direction AO. Therefore, the resultant force F at A is in the direction as shown. Since the ladder is in equilibrium, the algebraic sum of the moments of forces about any axis is zero. Taking moments about A (so that the force F does not appear in the equation), R (10 sin 60°) − 200 (5 cos 60°) − 500 (2. 5 cos 60°) = 0 10 R sin 60° = 2 250 cos 60° R = 130 N © Manhattan Press (H. K. ) Ltd. 7

1. 5 Static equilibrium of a rigid body (SB p. 65) Solution (cont’d): Also, the algebraic sum of the components of forces along any direction is zero. Therefore, if θ is equal to the angle F makes with the vertical, resolving forces vertically, F cosθ – 500 – 200 = 0 F cosθ = 700. . . . (1) Resolving forces horizontally, F sinθ – R = 0 F sinθ = R = 130. . . . . (2) (1)2 + (2)2 : F 2 cos 2θ + F 2 sin 2θ = 7002 + 1302 F= (2) ÷ (1): tanθ = © Manhattan Press (H. K. ) Ltd. = 712 N θ = 10° 31′ from the vertical Return to Text 8