This question already has an answer here:

I came to this problem and solved it but I don't fully understand how this works.

The problem is:

A 200lb man climbs to the top of a 20ft ladder that is leaning against a smooth wall at an angle of 60 degrees with the horizontal. The ladder weights 100lb and its center of gravity is 6ft from the bottom end. What must be the coefficient of static friction at the ground, if the ladder is not to slip?

Using trig we find the wall is 17.3ft and the floor under the ladder is 10ft. I'll use the bottom point where the ladder meets the floor as the axis for the ladder.

Finding the ladder's torque would just be the sum of the two torques: (200lb)(10ft) + (100lb)(6ft)

And it makes sense that the walls torque must be equal to the ladder since the wall is pushing back on the ladder with the same force and therefore:

(200lb)(10ft) + (100lb)(6ft) = (Wlb)(17.3ft)

Solving: W = 133lb Where W is the force of the wall on the ladder.

I follow up to this point but after this is where I get a bit confused.

The book then claims H = W = 133lb, where H is the frictional force opposing the ladder at the floor.

If two equal forces oppose each other then there will be no change in velocity. So if the ladder is pushing on the wall with the same force that the wall pushes on the ladder, why would friction need to oppose the bottom with that same force? Wouldn't the push of the ladder on the wall and the push of the wall on the ladder cancel each other out? As in, why would H = W?

I get that torque must be symmetric about the axis if in equilibrium, but the axis is at the bottom of the ladder, where the friction is acting. What is pushing the ladder away from the wall? I know it's going to be the wall, but why is that not canceled by the push of the ladder?

I may have a bad or incomplete understanding for torque so any help would be greatly appreciated! Thanks!


marked as duplicate by sammy gerbil, John Rennie newtonian-mechanics Jan 3 at 11:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It would be nice if you added a picture of the problem. Use Mathjax for formulae too. It's about writing them in between of dollar symbols, and laTex commands inside. See here: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – FGSUZ Jan 1 at 20:55

I don't think your main problem, at least with this question, is with torque. I think you're confused about Newton's third law.

Look at your statement: Wouldn't the push of the ladder on the wall and the push of the wall on the ladder cancel each other out?

They would cancel out, but not in the way you're thinking. You answered your own question in the line right before. Look at what you said: ...So if the ladder is pushing on the wall with the same force that the wall pushes on the ladder...

The ladder is pushing ON THE WALL, and the wall is pushing ON THE LADDER. They aren't forces on the same object; the force ON THE LADDER by the wall has to be balanced by another force ON THE LADDER, in the same way the force ON THE WALL by the ladder is balanced out by a force from the floor ON THE WALL in the opposite direction (as for why there's not a net torque on the wall, that's a story for another time...)


As in, why would H = W?

You need to look at a free body diagram of the ladder. See figure below. The free body diagram shows all the external forces acting on the ladder. As you can see, there are only two horizontal forces acting on the ladder, the reaction of the wall, $W$ and the friction force on the floor, $H$. In order for the ladder to be in equilibrium, the sum of the horizontal forces must be zero. Therefore, $H= W$.

Hope this helps. enter image description here


Not the answer you're looking for? Browse other questions tagged or ask your own question.