**Question 10:** A uniform plank AB of length 4 m and weight 500 N is suspended by a vertical rope at each end. A girl of weight 300 N is standing at a distance of 1.2 m from the end A, calculate the tension in the rope supporting end B. Would you expect the tension in the rope at A to be larger or smaller than the rope at B. State a reason for your answer.

**Solution**

Given data

**Distances **

Length of the plank = AB = 4 m

Midpoint C = 2m from A and B

Distance of the girl from A = 1.2 m

**Forces**

Weight of the girl = W_{1} = 300 N

Weight of the plank = W = 500 N

Required

T_{1} and T_{2}

Now there is no force acting in the horizontal direction, therefore, applying first law of equilibrium (only for vertical directions)

Taking upward forces as positive and downward as negative,

Now apply second law of equilibrium by assuming point A to be the point of rotation and anti-clockwise torques as positive and clockwise as negative.

This gives the tension in the rope B.

Putting this value in (A)

This is tension in rope at A.

Comparing (i) and (ii) we see the tension in the rope at A is greater than the tension in the rope at B. This represents an interesting situation. When the girl stands on the plank at the side closer to the side A, it shifts the center of gravity of the whole system a little towards side A. So the distance between the center of gravity and side A decreases, thereby, decreasing the moment arm. Now for the system to be in equilibrium, the force must increase to cancel exactly the torque produced by the other rope on side B. So this is the reason for the increased tension in the rope at side A.

Pingback:Numerical on Vectors and Equilibrium, Physics 11

Pingback:numerical-problem-9-chapter-2-physics-11 – msa

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