Question

Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

Answer 1

Answer:

0.9 meters

Solution:

In these type of questions there are two equations:

First one is Force Balance and second one is Torque Balance.

Force Balance

First equation is generally used to know the magnitude of the force applied.

Since they have to lift wooden board in order to carry it, therefore they have to apply a counter force to overcome its weight, which is 160 N.

So the net force applied by the two men in upward direction is 160 N.

The force applied by person 1 (as in the figure attached) is 60 N and by person 2 is F N. So,

60 + F = 160

∴ F = 100 N

Torque Balance

(Torque is nothing but a twisting force that tends to cause rotation.

To calculate torque multiply the force applied with the perpendicular distance between axis of rotation and line of application of force)

The second equation is generally used to know the position of the force applied.

Since they are carrying wooden board in horizontal position, therefore there is no net torque on the wooden board.

To calculate torque first of all choose the axis of rotation. Here, there is no rotation so you can choose any axis of rotation which can help you to calculate the distance of application of force. Here the axis of rotation is chosen as the axis perpendicular to the wooden board passing through the center of it.

Since torque of two person are opposing each other as one is in clockwise direction and other one is in anticlockwise direction, therefore, if they are equal then the net torque will become zero. So,

60 $\times$ 1.5 = F $\times$ x

60 $\times$ 1.5 = 100 $\times$ x

90 = 100 $\times$ x

x = $\frac{90}{100}$

x = 0.9 meters

( NOTE: If you apply force in the same half side then the torque won't oppose each other rather it will support each other)