# Moment of weight of leaning beam

Having trouble with a FBD moment. The problem is:

How big can the force couple C (looks like a G in the pic) be in order for the disk to not spin? The disc has mass m and the beam also has mass m. All other variables are given.

Let point A = bottom the disc and point B = where the beam touches the ground. Also let $$\ell = R/\tan(\alpha)$$ = distance from point A to point B.

In the solution sheet, the professor stated that the counter-clockwise moment about point A due to the beam is $$\frac{\ell}{2} m g$$, implying that the center of mass of the beam is located halfway between A and B.

How can this be? If the beam was parallel to the ground, surely that would have been the case. But now it's tilted $$\alpha$$ degrees - should that not shift the center of gravity to the right?

For reference, this is his equation of moment equilibrium about A:

$$M_a = N_b \ell - m g \frac{\ell}{2} + C = 0.$$

• Is the beam rigidly attached to the disk, or just pinned? Commented Jan 3, 2014 at 19:15
• It is pinned... Commented Jan 3, 2014 at 19:46
• Surely rotating the bar (clockwise) would shift the COG to the left.
– DWin
Commented Jan 3, 2014 at 22:17
• That's reasonable to think IMO, but the solution sheet does not agree. Commented Jan 4, 2014 at 8:29

So you have two rigid bodies in contact to the ground. Let us call $N_a$ and $N_b$ the contact force (normal to ground) and $F_a$, $F_b$ the frictional force at A and B (arbitrarily chosen to act in a positive x direction). We can call the pin forces $P_x$ and $P_y$ acting on the bar (an reacting on the disk).

The sum of moments about the disk center is

$$r F_a + C = I \ddot{\theta}$$

$$\frac{\ell}{2} \cos\theta (N_b-P_y) - \frac{\ell}{2} \sin\theta (N_b+P_x) = 0$$

couple with the motion in the x and y axes of the disk

$$m \ddot{x} = m R \ddot{\theta} = F_a -P_x \\ m \ddot{y} = N_a -P_y - m g = 0$$

and the bar

$$m \ddot{x} = F_b + P_x \\ 0 = N_b+P_y - m g$$

For the condition of motionless with $\ddot{x}=\ddot{\theta}=0$ the above is 6 equations for 6 unknowns ($P_x, P_y, N_a, N_b, F_a, F_b$) which is solved by elimination.

Next find which is the smallest $C$ value that makes $|F_a| = \mu_s N_a$ and $|F_b| = \mu_s N_b$

If you get the correct result you will have $P_x<0$ and $F_a<0$ and $C \propto \mu_s R m g$