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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!