# Find the bending moment of a pole attached to a moving block

I'm having trouble with the following problem.

What I've done so far:

x-y is the usual coordinate system.

$a=\frac{F}{m}=\frac{800}{60}$ and the y component of this is $a_y=a\sin{60^\circ}$.

To figure out the force acting on the rod and contributing to the momentum, I take the sum of all forces in the y direction. Note that from now on, I am thinking about the rod as being separated from the rest of the body. The weld is represented by forces ($O_y, O_x$), and the rod is moving upwards with the acceleration calculated above.

$\Sigma F_y= O_y-20g = 20a_y\implies O_y=20a_y+20g$

Now, the momentum equation about the center of mass of the rod should be

$\Sigma M_G=0.7O_y+\bar{I}\omega_G=\bar{I}\alpha=0$

Because the rod is not rotating. Is this equation right? (Why not?)

If it is right, it implies that $\bar{I}\omega_G=-0.7O_y$. Where do I go from here? The things I've tried from here on don't yield the right answer, and I do not have much confidence in those strategies.

The correct answer, if anyone cares, is 196 Nm.

-
Always start from a Free Body Diagram. In fact, break the weld and make 2 FBDs. – ja72 Apr 10 '13 at 20:11
Does gravity act also? – ja72 Apr 10 '13 at 20:18
Yes, gravity acts as well. Your answer looks very good, I will try it out later. – user714 Apr 11 '13 at 4:27

The weld acts on the rod with force components $B_x$ and $B_y$ as well as a moment $M$. The equations of motion for the rod are

$$B_x = m_{rod} \left( - \cos\theta \, \ddot{q} \right)$$ $$B_y = m_{rod} \left( \sin\theta \, \ddot{q} + g \right)$$ $$M - \frac{L}{2} B_y = I_{rod} \dot{\omega}_{rod} = 0$$

where $q$ is the distance along the guide of travel and the last equation is the sum of moments about the center of gravity of the rod. Gravity is included above as $g$.

The equations of motion for the block are

$$-F\,\cos\theta + N \sin\theta - B_x = m_{block} \left(- \cos\theta\, \ddot{q} \right)$$ $$F\,\sin\theta + N \cos\theta - B_y = m_{block} \left(\cos\theta\, \ddot{q} + g\right)$$

where $N$ is the contact normal force.

Combined the above is

$$m_{rod} \left( \ddot{q} + g\sin\theta \right) - F = -m_{block} \left( \ddot{q} + g\sin\theta \right)$$ $$N-m_{rod} g \cos\theta = m_{block} g \cos\theta$$

with solution

$$N = \left(m_{rod}+m_{block}\right)\, g \cos\theta$$ $$\ddot{q} = \frac{F}{m_{rod}+m_{block}} - g \sin \theta$$

Now going back to moment, $M=\frac{L}{2} B_y$ which you can solve now.

-