1© manhattan press (h.k.) ltd. 1.5 static equilibrium of a rigid body

© Manhattan Press (H.K.) Ltd. 1

1.5 Static equilibrium of 1.5 Static equilibrium of a rigid bodya rigid body


Static equilibrium

1.5 Static equilibrium of a rigid body (SB p. 64)

Sum of all the clockwise moments

= Sum of all the anticlockwise moments

Static equilibrium if

(a) resultant force = 0

(b) algebraic sum of moments of forces about any axis = 0

Principle of MomentsGo to

Example 10Example 10

Go to

Example 11Example 11


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Q:Q: A uniform rod PQ of length 10 m and mass 2 kg is supported horizontally at A, 2 m from one end and 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



Solution: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 – 3g – 2g – 2g = 0 ( g = acceleration due to gravity)

R + S = 7g

Also, algebraic sum of moments about any axis is zero.

Taking moments about the horizontal axis that passes through A (so that R does not appear in the equation),

(3g × 2) − (2g × 3) + (S × 4) − (2g × 8) = 0

4S = 16g

S = 4 × 9.81 = 39.2 N

R = 7g – 39.2 = 29.4 N

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Q: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



Solution: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 sin60°) − 200 (5 cos60°) − 500 (2.5 cos60°) = 0

10 R sin60° = 2 250 cos60°

R = 130 N



Solution (cont’d):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 : F2 cos2θ + F2 sin2θ = 7002 + 1302

F = = 712 N

(2) ÷ (1): tanθ = θ = 10° 31′ from the vertical

22 130700

700130 Return to

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1© manhattan press (h.k.) ltd. 1.5 static equilibrium of a rigid body

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