# Simple Moment Question

1. Mar 20, 2006

### John O' Meara

A uniform rod AB, of mass 5lb, is hinged at upper end A and hangs down freely. What horizontal force applied at B will be required in order to deflect the rod through 30 degrees from the vertical. Thanks.
My attempt: taking moments about A, Fcos30xAB = wxAC = wcos30xAB/2 therefore F = w/2. F= applied horizontal force, w = weight.A 2nd question what does "a gauge of 56.5 ins" mean in rhe context of a railway line?

2. Mar 20, 2006

### Staff: Mentor

Careful. Note that the weight acts vertically, so its moment should have a sin30, not a cos30.

The inner distance between the rails.

Last edited: Mar 20, 2006
3. Mar 20, 2006

### civil_dude

I would take a second look and the Fcos30xAB. Shouldn't it be sin to get the correct length of the normal distance?

4. Mar 20, 2006

### Staff: Mentor

Actually, that part is correct, since the force is applied horizontally. But the moment due to the weight should be Wsin30xAB/2, not Wcos30xAB/2. My bad, for not reading carefully. I'll revise my comments in the original post.

Last edited: Mar 20, 2006