Question

A 10 N board of uniform density is 5 meters long. It is supported on the left by a string bearing a 3 N upward force. In order to prevent the string from breaking, a person must place an upward force of 7 N at a position along the bottom surface of the board. At what distance from its left edge would they need to place this force in order for the board to be in static equilibrium?
a. 3/7 m
b. 5/2 m
c. 25/7 m
d. 30/7 m
e. 5 m

Answer 1

Answer:

C.$\frac{25}{7}$m

Explanation:

We are given that

Weight of board=w=10 N

Length of board=L=5 m

Tension in the string=T=3 N

Applied upward force=F=7 N

We have to find the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

Let r be the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

The board is uniform therefore, the center of board is the mid- point of board.

Therefore, the lever arm of weight=$r_1=\frac{L}{2}=\frac{5}{2}m$

Now, the torque exerted by the weight of the board

$\tau_1=Force\times perpendicular\;distance=10\times \frac{5}{2}=25 N$

The torque exerted  by applied force=$\tau_2=7\times r=7r$

In static equilibrium

The sum of rotational forces=0

$\tau_1+\tau_2=0$

The two rotational force act in opposite direction therefore,

$\tau_2=-7r$

Substitute the values

$25-7r=0$

$7r=25$

$r=\frac{25}{7}$m

Hence, option C is true.