Principle of impulse and momentum

[Queren Suriano]

Homework Statement

In the gear arrangement shown, gears A and C are attached to rod ABC,
which is free to rotate about B, while the inner gear B is fixed. Knowing
that the system is at rest, determine the magnitude of the couple M which
must be applied to rod ABC, if 2.5 s later the angular velocity of the rod is
to be 240 rpm clockwise. Gears A and C weigh 2.5 lb each and may be
considered as disks of radius 2 in.; rod ABC weighs 4 lb

Homework Equations

Syst. Momenta 1 + Syst. Ext. Imp. 1→2 = Syst. Momenta

The Attempt at a Solution

: I draw external drives like this.[/B]

However in Figure 2 corresponds to the solution of the problem, they draw only two forces: P and Q , but I don't really understand how they simplify all other forces in only one force Q (weight and reactions on the pin to get to that)

 

[nilesh_pat]

Any help?A:The rod ABC will be acted on by the couple $\mathbf{M}$ and also by the forces of gravity acting on the two gears $\mathbf{F_1}$ and $\mathbf{F_2}$. The rod ABC will also be supported by two reactions at B, one pointing left $\mathbf{P}$, and one pointing right $\mathbf{Q}$. These two reactions balance each other out, so that $\mathbf{P}=-\mathbf{Q}$. That is, they form a couple $P=Q$ that is equal and opposite to the applied couple $\mathbf{M}$. Therefore the net external force acting on the rod ABC is simply $\mathbf{F_1}+\mathbf{F_2}-\mathbf{Q}$.The net external torque is then $\mathbf{M}+(\mathbf{F_1}+\mathbf{F_2}-\mathbf{Q})\times r$, where $r$ is the length of the rod. You can now use the equation of motion to solve for the required magnitude of $\mathbf{M}$.