Recent content by kffd

= 0.5 mg cotθ I think that's all. Thanks a lot, ehild and haruspex; I've been out of school sick recently, going through this problem with me really did help me catch up. Once I'm a bit more learned when it comes to physics, I'll do my diligence to help out some other people on here, too. :)

Well, net torque must be 0 if it's in equilibrium. So, mg 2.5cosθ - 5sinθ Ff = 0 I used a minus just to indicate that one direction is negative. (I arbitrarily picked clockwise.) Going off of that: mg 2.5cosθ = 5sinθ Ff Ff = (mg 2.5cosθ)/(5sinθ) Ff = FNwall, so FNwall = (mg 2.5cosθ)/(5sinθ)...

Ok, good news is I knew that about torque, at least. That's what I didn't understand, how a "torque" could work if it were simply a line. (We have only discussed it as clockwise/counterclockwise.) |\-LOA-> |..\...| |...\...| |...\.| |...θ\ I think that is the line of action, and the vertical...

Another go for FNwall: 5sin θ x Ff Here's a diagram to help explain what I'm not getting: ||\ ||..\ ||...\--> ||...\ v|...θ\ Vertical arrow is what I now think torque for FNwall should be, 5sin θ x Ff. Horizontal is torque for g. Is that correct? If so, is one positive and the other negative...

Ok, I'll pick the bottom of the ladder, where it meets the ground. For force of g: T = mg x 2.5cos θ For force of FNwall: T = ? x 5cos θ For the other two forces, T = 0 I'm not sure where that brings me, though.

Homework Statement There is a 5m ladder leaned against a wall, making an angle of θ with the ground. The ground is rough (getting at friction), and the wall is frictionless. Here's an ASCII attempt: |\ |..\ |...\5 |...\ |...θ\ We want to make a force diagram for the ladder, then find the...