What are the forces that are holding up the plank?

Question

Two painters are standing on a plank while painting a building. The plank is uniform, weighs 72.0 lb, and is 16.0 ft long. The plank is being held up by two forces that are 1.25 ft in from each end. The first painter weighs 155 lb, and he is standing 1.00 ft from the left end. The second painter weighs 198 lb, and he is standing 5.75 ft from the right end.

Answer 1

So this is what's called a statics problem, meaning that forces are being applied, but nothing is moving. That tells us that all the forces cancel, so there is no linear acceleration, and all the torques cancel so that there is no rotational acceleration.

I will call the forces in the problem the weight of the beam W, the weight of first painter w₁, second painter w₂, and the two supporting forces F₁ and F₂. The weights try to push the beam down while the supporting forces hold the beam up. There's no acceleration, so these must balance: F₁+F₂=W+w₁+w₂.

Since we know the weights, we now know the sum of the two supporting forces. To solve further, we need another equation, which we get from considering the torques. A torque trying to rotate a rigid object at a perpendicular angle like this will have the value τ=rF, where r is the distance to the pivot point and F is the magnitude of that force.

If I consider the center of the beam as my pivot point, then I can start calculating my torques. I will call the painters' distances from my pivot point x₁ and x₂. For the distance to the supporting forces, I will call that d for both as they are both d=6.75ft from the center.

Finally, we'll call a torque positive if it wants to rotate the beam counter-clockwise and negative for clockwise.

Again, because nothing is rotating, all these torques must balance, i.e. they combine to make zero net torque. That gives us, from left to right, x₁w₁ - d F₁ - x₂w₂ +d F₂ = 0.

Note that the weight of the beam doesn't appear in this equation because it is applied at the center of the beam, which is my origin so r=0 for it.

Now, we have a system of two equations to solve for our two variables.

F₁+F₂ - W - w₁ - w₂ = 0

and

- d F₁ + d F₂ + x₁w₁ - x₂w₂ = 0 , where I've done some light rearranging.

You learn several methods to solve these kinds of equations in high school algebra. I'll show just one way here. To solve this, I'll multiply my first equation by d so that adding the equations together cancels out F₁.

d F₁+d F₂ - dW - dw₁ - dw₂ = 0

+

- d F₁ + d F₂ + x₁w₁ - x₂w₂ = 0

Gives us

2dF₂ - dW +(x₁-d)w₁-(x₂+d)w₂ = 0

Solving, we get

F₂ = [dW -(x₁-d)w₁+(x₂+d)w₂]/2d.

Plugging in the numbers given in the problem, with d=6.75ft, W=72lbs, w1=155lbs, w₂=198lbs, x₁=7ft (they're 1 foot from the edge, which is 7ft from my origin at the middle), x₂=2.25ft, I get F₂=165.13lbs. Plugging that back into our first equation lets us calculate that F₁=259.87lbs. To three significant figures that's F₁=260.lbs and F₂=165 lbs.