Physics Levers Question

Question

A uniform board is carried by Man A and Man B. Man A holds up the left end of the board. Where should Man B hold the board so that he supports twice as much weight as Man A when the board is horizontal?

Answer 1

I'll show how to do this in 2 different ways:

First, let's figure out how much weight each of them will be carrying. we'll call the weight of the board W.

In order for the total weight of the board to be lifted, the weight the two men carry needs to add up to W.

W_a + W_b = W

And W_b needs to be two times W_a.

W_b = 2*W_a

We have two equations and two unknowns. Solving for Wa, we get 3*W_a = W, or

W_a = 1/3 * W

so W_b = 2/3 * W

Next we need to solve for the location. If we create a pivot point where man A is holding the board, we can look at the torques being applied by the weight of the board and the force of man B holding it.

Let's say that where man B is holding the board is 0, and the other end of the board is L. The wight of the board, W, is located at L/2. So, the torque being applied by the weight of the board is W*L/2.

So to cancel out our torques, we can write the equation as:

W*L/2 = 2/3*W * x*L

Where X is the how far from the pivot man B should be holding the board.

The W's and L's cancel out, leaving us with:

1/2 = 2/3 * X

X = 3/4

So, person B should hold the board 3/4 of the way down the board from person A.

THE EASY WAY

Because person B is carrying twice as much weight, they need to be half as far from the center point to cancel out the torques.

If person A is half the board length from the center (meaning they're at the end of the board), person B needs to be 1/4 of the way from the center on the other side.