Menu Close

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 = W1 = 500 N

Weight of block = W2 = 100 N

Distances

DC = 2 m

EC = 1.5 m

CG = 1 m

CH = 1.5 m

Required

Force by support A = R1 =?

Force by support B = R2 =?

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

R1 + R2 – 500 – 500 – 100 = 0

R1 + R2 = 1100

OR         R1 = 1100 – R   —–    
(A)

Now apply the 2nd 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)

R1 = 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.

3 Comments

  1. Pingback:Numerical on Vectors and Equilibrium, Physics 11 – msa

  2. Pingback:numerical-problem-6-chapter-2-physics-11 – msa

  3. Pingback:numerical-problem-8-chapter-2-physics-11 – msa

Leave a Reply

Your email address will not be published. Required fields are marked *