UKC

I can't get my forces to balance in a trivial moments problem.

There is uniform rod of weight W, leaning at an angle a from the vertical. It is fixed at the top and at the bottom.

Resolving W perpendicularly (p) and parallel (||) to the rod,

Wp = Wsin(a), W|| = Wcos(a)

Take moments about the bottom, then I get a force at the top acting perpendicularly to the rod of W/2 . sin(a).

Similary taking moments about the top, I get a force acting perpendicularly at the bottom, of W/2 . sin(a).

The || components at top and bottom at W/2 . cos(a).

Resolving vertically at the top & bottom, I get
Ftopv = W/2 . sin(a). sin(a) + W/2 . cos(a) .cos (a) = W/2
And the same for the bottom, which both add up neatly to W, as expected.

When I look at the horizontal components, I don't get a neat answer.

What should I get?

Please.

In reply to lowersharpnose:

All you can say about the components parallel to the rod at the top and they total Wcos(a) (The component of the rod's weight parallel to the rod). Their combined horizontal component is then Wcos(a)sin(a), which balances the combined horizontal components of your perpendicular components at each end.

In reply to Robert Durran:

I get those terms, but the signs are the same, so I can't get them to cancel out as I think they should.

n reply to lowersharpnose:


I get:
FtopH=W/2.Sin(a).Cos(a)-W/2.Cos(a).Sin(a)=0
(and the same at the other end)

In reply to Jack B:

Surely, the horizontal force at the top is non-zero and of opposite sign to that at the bottom?

In reply to lowersharpnose:

That would imply the rod was under net tension or compression... I see no reason for that to be the case.

In reply to Jack B:

I think I was getting confused with 'ladder leaning against rough wall' type problems. Thanks.