Question

The 100[N] pole below is non uniform. It is leaning on a pipe on the right and the floor on the left. The center of mass of that pole is at point C.
lul --1.5m- 2.5m 42°

1. Draw a rigid body diagram for the pole
2. Fill the table below
3. What is the Normal (upwards) force by the floor on the pole?

Answer 1

This is the FBD of the pole:

F_N B 1.5 mg sin(42) F_N 2.5 mg cos(42) 42 mg A F_friction

The red lines indicate the components of the weight split into parallel and perpendicular to the length of the rod.

Note that eventhough friction is not explicitly stated in the problem, the only way for the pole to be in this state is if there were a friction force between the pole and the ground. Otherwise, it will slip right off.

Also note that the pipe will offer a normal reaction force that will be at 90 degrees to the rod. (the pipe is cicular and the rod is resting on it. THis means that the rod is a tangent to a circle. The reactive force from the pipe is along the line connecting the center of the pipe to the point of contact. Now we know that the tangent is always perpendicular to the radius)

Resolving the normal reaction force into the components, we have

$F_{NA}+F_{NB}\cos\theta -mg=0\implies F_{NA}+F_{NB}\cos\theta =mg$

$F_{fric}=\frac{mg}{\cos\theta}\sin\theta\implies F_{fric}=mg\tan\theta$

Now, the torques taken about the point where the rod makes contact with the floor is:

(taking counter clockwise to be positive)

FNB(4) – mg cos 0(2.5) = 0 3 FNB 5mg cos 0 8

The above equations are necessary if the rod has to stay in place.

Thus, for the table given in the question:

the vertical forces are $F_{NA},F_{NB}\cos\theta,-mg$

the horizontal forces are: $-F_N\sin\theta,F_{fric}$

To find the normal force from the floor, from the equations

$F_{NA}+F_{NB}\cos\theta =mg\implies F_{NA}=mg-F_{NB}\cos\theta$

$F_{NB}=\frac{5mg\cos\theta}{8}$

We have:

$F_{NA}=mg\left(1-\frac{5}{8}\cos^2\theta \right )$