Pole Propped in Room

Tags:
1. Feb 17, 2015

minimario

1. The problem statement, all variables and given/known data
A uniform pole is propped between the floor and the ceiling of a room. The height of the room is 7.80 ft, and the coefficient of static friction between the pole and the ceiling is 0.576. The coefficient of static friction between the pole and the floor is greater than that. What is the length of the longest pole that can be propped between the floor and the ceiling?

2. Relevant equations
Torque = F * d

3. The attempt at a solution
We consider the forces:

We first have the x-force balance: $0.576 n_1 = \mu_2 n_2$

And the y-force balance: $mg = n_2-n_1$

And the torque balance: $\frac{L}{2} mg \cos \theta + L n_1 \cos \theta = L 0.576 n_1 \sin \theta$. We can cancel the L and multiply by 2: $mg \cos \theta + 2 n_1 \cos \theta = 1.152 n_1 \sin \theta$. ($\theta$ is the bottom angle the pole makes with the ground)

But from the torque equation, we only have $mg = n_2-n_1$ so how do we make the $n_2$s cancel?

2. Feb 17, 2015

DEvens

What you calculated is the condition required such that the pole does not fly upward or sink through the floor. Good.

Now calculate the condition required such that the upper end does not slide across the ceiling.

If the rod was perfectly vertical there would be zero horizontal force, and so it will not slide. The question is claiming that at some point the rod's weight will cause it to slide. You have set things up in terms of the angle theta. How does theta relate to the length of the rod?

3. Feb 17, 2015

minimario

If $L$ is the length of the rod, then $L \sin \theta = 7.08$. How do you set up the condition required such that the upper end does not slide across the ceiling?

4. Feb 22, 2015

minimario

Anyone can answer the above question?

5. Feb 23, 2015

haruspex

I would assume all available poles are of the same mass per unit length.