Question

two people are carrying a uniform wooden board, oriented horizontally, that is 3.00 m long and weighs 160 n. if one person applies a 60-n upward force at one end, at what position along the board is the other person supporting it?

Answer 1

To maintain static equilibrium, the other person must support the 3.00 m long and 160 N board 2.4 m from the left end, countering the 60 N upward force applied at one end.

Step-by-step explanation:

To calculate the position where the other person is supporting the board, we'll use the principles of static equilibrium. The sum of the forces and the sum of the moments (torques) about any point must be zero.

Let's take moments about the left end where the 60 N force is applied. The board is 3.00 m long and weights 160 N, which means the weight acts at the center of the board (1.5 m from each end). The person applying the 60 N force does not create a moment since it's at the point we are taking moments about.

Setting up the equation, we have:

1. 60 N x 0 m (distance from the pivot point) + F x d = 160 N x 1.5 m, where F is the force the other person applies (which must be 100 N since the total weight is 160 N and one person applies 60 N), and d is the distance from the left end.
2. Solving for d, we get d = (160 N x 1.5 m) / 100 N = 2.4 m from the left end where the other person is supporting the board.

The person must support the board 2.4 m from the left end to maintain equilibrium.