# (Mechanics) A Plank in a Circular Trough

1. May 25, 2006

### Saketh

I'm trying to solve a problem from Richard P. Feynman's Tips on Physics that I was assigned for homework. It's problem **I-5 in the book, if that helps at all.

Here is the problem:
A plank of weight W and length $$R\sqrt{3}$$ lies in a smooth circular trough of radius R. At one end of the trough is a weight W/2. Calculate the angle that the plank makes with the horizontal. ​

I tried to use equilibrium tactics, creating two normal forces and one gravity force at the plank-weight system's center of mass. I then used net force equations and net torque equations to solve for the angle. But this failed. Since it is a circle, the normal forces are not vertical or horizontal.

How am I supposed to do this?

P.S. Is it possible to use Latex to format stuff on these forums?

Last edited: May 25, 2006
2. May 25, 2006

### Staff: Mentor

Using torque and force equations should work just fine. While the normal forces are neither vertical nor horizontal, they do have a fixed relationship to each other. (Hint: What is the angle between normal force and plank?)

3. May 25, 2006

### Saketh

I figured that both angles between each normal force and the plane are equal, but that it wouldn't help me. Is there any method to solve this problem besides torques and forces? I seem to be missing something in my work.

Here's my premise:
$$\Sigma \tau=F_{N_{2}}sin(\alpha)-(3W/2)sin(\alpha)$$
$$\Sigma F_{y} = F_{N_{1}}sin(\alpha-\theta)+F_{N_{2}}sin(\alpha-\theta)-3W/2$$
$$\Sigma F_{x} = F_{N_{1}}cos(\alpha-\theta)-F_{N_{2}}sin(\alpha-\theta)$$

where $$\theta$$ is the angle that we are looking for, and $$\alpha$$ is the difference between the plank-normal angle and $$\theta$$.

This should work, but I fear the geometry of it.

Last edited: May 25, 2006
4. May 25, 2006

### Staff: Mentor

Sure it will help you. Find that angle and use it in your equations--it will simplify things greatly.
That's the most straightforward method that I can see.

You're closer than you think. In addition to the above tip, all you need are these two equations:
(1) Sum of torques about the center of mass = 0.
(2) Sum of horizontal forces = 0

That's all you need. (The geometry seems more difficult than it is because you are trying to solve the problem in general; instead, make use of the specific plank length that you are given.)

I see forces and angles, but no lengths.

5. May 25, 2006

### Saketh

The lack of lengths in torques was a mistake in typing. It should be $$\Sigma \tau=F_{N_{2}}sin(\alpha)(R\sqrt{3})-(3W/2)sin(\alpha)(5R\sqrt{3}/6)$$

6. May 25, 2006

### Staff: Mentor

Where is the center of mass of the system?

7. May 25, 2006

### Saketh

Of the rod, I thought that the center of mass is at $$5R\sqrt{3}/6$$. But now I am thinking otherwise.

Here is my work for it:
$$x_{com}=\frac{(W)(R\sqrt{3}/2)+(W/2)(R\sqrt{3})}{W + W/2}$$
$$x_{com}=\frac{R\sqrt{3}}{3}$$

This makes no sense to me! How can the center of mass be $$\frac{R\sqrt{3}}{3}$$ from the left end of the rod, assuming the weight is on the right end?

8. May 25, 2006

### Staff: Mentor

Good.

Good.

You're right, it doesn't make sense. (You made an error.)

9. May 25, 2006

### Saketh

$$x_{com}=\frac{2R\sqrt{3}}{3}$$
A stupid math error. Thanks for pointing it out, or I may have spent hours poring over the physics.

Previous, I had a trigonometric equation equal to $$\frac{5}{6}$$, and I wished that it instead equaled $$\frac{2}{3}$$, because that would yield the correct answer of 30 degrees. Now that I have pinpointed the error, I get the trigonometric equation equal to $$\frac{2}{3}$$, and I can rest easy.

Thanks for the assistance!