# Rotational Statics problem

1. Jun 9, 2006

### µ³

"One end of a uniform meter stick is placed against a vertical wall View Figure . The other end is held by a lightweight cord that makes an angle $\theta$ with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.400 ."

I can't for the life of me get the equation for this.
Part A:
What is the maximum value the angle $\theta$ can have if the stick is to remain in equilibrium?
So far I've tried using the last end as a pivot point.
so that would give me
Balancing torque:
$$\mu T cos(\theta) = mg(1-x)$$
and balancing total forces
$$\mu T cos(\theta) + T sin(\theta) = mg$$
combining the two
produces
$$T sin(\theta) = mgx$$
dividing the third by the first gives me
$$tan(\theta) = \mu \frac{x}{1-x}$$
and there is no way I can get a "max theta". Naturally I can't get Part B and C but I figured if you guys can help me with A it will be easy to get the rest.

Last edited: Jun 9, 2006
2. Jun 9, 2006

### Hootenanny

Staff Emeritus
What is the mass of the meter rule? Is it light?

3. Jun 9, 2006

### µ³

negligible I presume.

4. Jun 9, 2006

### Gokul43201

Staff Emeritus
Can you draw the forces in that figure?
This is the end of the ruler not touching the wall, right?

I see a mistake here. What torque does the string exert about the point of attachment of the string? And are there no other forces on the ruler that exert a torque about this point?

Again, never do a force problem without a free body diagram.

Last edited: Jun 9, 2006
5. Jun 9, 2006

### µ³

Yeah that's the end of the ruler not touching the wall. I thought about the string itself exhibiting torque but since I chose it as a pivot point it wouldn't make much sense if it did. Can you suggest an alternate pivot point? I also considered using the end touching the wall as pivot point and that gave me the equation
$$T sin(\theta) = mgx$$
which was already derived from the first two. I also considered using the ruler's center of mass a pivot point and that didn't give me a sensical answer either.

Pretty sure that's right.

Last edited: Jun 9, 2006
6. Jun 9, 2006

### Hootenanny

Staff Emeritus
Perhaps taking moments about the centre of the rule since we don't know its mass? I haven't worked through this problem so don't take my word for it.

7. Jun 9, 2006

### µ³

$$1/2 \mu T cos(\theta) + mg(x-1/2) = 1/2 T sin(\theta)$$
$$\mu T cos(\theta) + T sin(\theta) = mg$$

so...
$$\mu T cos(\theta) + 2mgx -mg= T sin(\theta)$$
$$\mu T cos(\theta) + 2mgx -\mu T cos(\theta) - Tsin(\theta)= T sin(\theta)$$
$$mgx = T sin(\theta)$$
$$\mu T cos(\theta) + T sin(\theta) = mg$$
$$\mu T cos(\theta) + mgx = mg$$
$$\mu T cos(\theta) = mg(1-x)$$