**Problem 7:** Find the magnitude of the forces provided by the supports A and B if shown in balanced condition. Weight of plank is 500 N and it is uniform in shape. Weight of the bloc= 100 N, Weight of the student = 500 N.

**Solution**

**Strategy:** Consider the figure. We have to find the forces due to supports A and B in order to keep the plank in a balanced condition. We apply the conditions of equilibrium, i-e,

**Given data**

**Weights**

Weight of the plank = W = 500 N

Weight of the student = W_{1} = 500 N

Weight of block = W_{2} = 100 N

**Distances**

DC = 2 m

EC = 1.5 m

CG = 1 m

CH = 1.5 m

**Required **

Force by support A = R_{1} =?

Force by support B = R_{2} =?

Now using the first condition of equilibrium

Look at the fig. No force is acting along

x-axis. Therefore,

Similarly

Taking the upward forces as positive and the

downward as negative,

Put the values

R_{1} + R_{2} – 500 – 500 – 100 = 0

R_{1} + R_{2} = 1100

OR R_{1} = 1100 – R_{2 } —–

(A)

Now apply the 2^{nd} condition of

equilibrium

Let the body rotates about point C and the

anti-clockwise torques as positive

Put the values

Put the value from equation (A)

Put this value in equation (A)

R_{1} = 1100 – 250 = 850 N

So the reaction force on the

plank provided by the supports at points A and B are 850 N and 250 N

respectively.

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