M2 Help

Hey guys.

I'm having real trouble understanding a few questions on this paper (should be attached to this thread).

4ii) I'm fairly certain the reactant force at P is perpendicular to AB. I also know there must be a reactant force at A going vertically up. But now.. What I would do is see that the weight of the rod must be = to the frictional force at P?

So 30sin(x) = uR = 0.6R

Now I would go to resolve using moments, but I don't know the length of AP, so I guess I'd have to resolve horizontally instead, which is the same as above?

If someone could shine some light on this one, I'd highly appreciate it.

I'm also stuck on the one after, but little steps at a time haha!

The weight of the rod would equal the reaction force at A + the vertical component of the perpendicular reaction force at the block + the vertical component of the frictional force at the block.



Don't understand how you got this.



If you just resolve horizontally for the rod, that's all you need.

Isn't 30sin(x) the horizontal component of of the weight? And I have no idea how to find the horizontal component of the Friction .

Edit: Not horizontal component, but parallel to the frictional force?

I see way you're saying now. You've resolved parallel to the bar.

So yes, 30sin(x) is the component of the weight along the bar, and muR is the frictional force at the point of contact with the block, BUT there is also a component of the reaction at A that is parallel to the bar, and you don't know that the reaction at A is.

Resolving horizontally is the way to go.

The frictional force at the block is at what angle to the horizontal? Draw a detailed diagram is necessary and mark on all the angles and right-angles.

Would the frictional force be Fcos(x) to the horizontal?