# Solid Mechanics, calculating reaction forces and M(X) for a cantilever thing

1. Apr 27, 2010

### PenTrik

1. The problem statement, all variables and given/known data

2. Relevant equations
M = F * D

3. The attempt at a solution

For a, I need to find the reaction force on the right side. So what I first did was make the sum of the moment on the left equals zero.
$$0 = M_0 + r_y * L.$$
$$r_y = \frac{-M_0}{L}$$
I think this is right.

When you take net sum of the moments, then there should be no moment imparted by the wall correct? I'm not sure.

For my M(X) term, assuming that I have my reaction forces on the right correct, it should be
$$M_0 - \frac{M_0}{L} * (L-X)$$

As for the boundary conditions, I'm not really sure what they should be. I'm pretty sure they're supposed to mean v(0) = v(L) = 0

I'm not sure if this is correct at all.

2. Apr 28, 2010

### pongo38

The support on the left is unusual, if not impractical, but I interpret it as being a normal encastre support resting rx and moment, but with no vertical restraint. Therefore ry must be zero. Your ry=M0/L is correct for a left hand support resisting ry but not rx. Are you sure the diagram is correct?
If the left support is capable of resisting moment then your assumption that it is zero is not correct. If the left support resists only rx, then the whole thing is a mechanism, with rotation about the right support.