# Torque Problem Involving a Block Hanging from an Unanchored, Friction-assisted post

1. Nov 10, 2008

### DD31

1. The problem statement, all variables and given/known data

Alright, so here's the problem I've got. I'll write it out first verbatim, then say what I've tried and failed with.

One end of a uniform meter stick is placed against a vertical wall (Fig. 11.40). The other end is held by a lightweight cord that makes an angle with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0.60.

(a) What is the maximum value the angle can have if the stick is to remain in equilibrium?

(b) Let the angle be 15°. A block of the same weight as the meter stick is suspended from the stick, as shown, at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium?

(c) When = 15°, how large must the coefficient of static friction be so that the block can be attached 20 cm from the left end of the stick without causing it to slip?

3. The attempt at a solution (relevant equations included within)

OK, so I'm still stuck on a, and I'll be fine with an answer for just that one, cuz I think I can figure out the others if I can get that one...

Alright, so I know it's an equilibrium problem. We've got variables Mm (which I'm using for the meter stick), Mb (the box), θ, T (the tension in the wire), and x (the distance from the wall to the box). It's a meter stick, so r=1.

So I figure the total force diagram first:
Pulling down, we have Mbg + Mθmg
Pulling up: Tsinθ + Friction from wall
This friction is Fnμ, and Fn comes from the tension's horizontal component, so the total pulling up is:
Tsinθ + Tcosθμ

So, the force setup:
Mbg + Mmg = Tsinθ + Tcosθμ

Now, since it's in equilibrium, torque is also gonna be equal, so I made up that equation:
Pulling clockwise (down, basically): Mbgx + Mm(1/2) <--r = 1/2 because the center of mass should be halfway down the meterstick.

Pulling counterclock: Tsinθ (r=1)

So:

I then tried to put those two together using Tsin(theta):

Shifting everything over to one side, I get
Mbg (1-x) + Mmg/2 = Tcos(θ)μ

And that's where I'm at. They give μ, but Mb, Mm, θ, x, and T are all variables here, and I don't know how to get rid of them. I think that, for a), I could possibly get rid of x by extending Mb all the way to the end of the meterstick or putting it up against the wall (whichever would give the largest angle), but that still leaves Mb, Mm, T, and θ. I'm solving for θ, but 3 variables is still too many.

I feel like there's gotta be some way I can relate Mb, Mm, and T together to get rid of them, but any time I try to fix it, that μ from the wall keeps messing things up.

Anybody have any ideas?

Thanks.

2. Nov 10, 2008

### LowlyPion

Re: Torque Problem Involving a Block Hanging from an Unanchored, Friction-assisted po

Welcome to PF.

For a) you have 2 equations to satisfy don't you? Sum of the vertical forces is 0 and sum of the moments are 0.

For vertical forces the vertical component of tension T*Sinθ + Ffriction = Weight of ruler? And since it doesn't rotate about the center of mass they are equal and hence each are equal to half the weight?

T*Sinθ = W/2

Now the normal force is T*Cosθ so Ffriction = u*T*Cosθ = W/2

So ... we have T*Sinθ = W/2 = u*T*Cosθ

Simplifying then Sinθ/Cosθ = u = Tanθ

3. Nov 10, 2008

### DD31

Re: Torque Problem Involving a Block Hanging from an Unanchored, Friction-assisted po

Thanks for the welcome. And your help was awesome. I got the right answer, so let me just reiterate what you said to make sure I get what's going on 100%.
So, for finding the maximum angle, we can effectively disregard the weight of the hanging block, since this maximum angle would only occur when the block's weight was very, very small.
Since the weight of the ruler is in the middle of the ruler, and the ruler isn't rotating(a point I totally overlooked, but makes total sense since it's in equilibrium), both the friction on one end and the tension on the other end are equal to half the weight.
This also lets us set T*cos$$\theta$$*u equal to T*sin$$\theta$$. Then, things just cancel out, and we get that tan$$\theta$$ = u.
Fantastic. I got the answer, I get what happened...thanks so much.

Additionally, while I was messing around for the last little while, I managed to figure out the answers for b and c...so now I've got the whole thing done. I may come back here and post the solutions to them a little later, for the benefit of anybody else with the same question.

Once again, thanks a ton.

4. Nov 10, 2008

### LowlyPion

Re: Torque Problem Involving a Block Hanging from an Unanchored, Friction-assisted po

No. We can effectively disregard the hanging weight because it isn't introduced into the problem until part b).

5. Nov 10, 2008

### DD31

Re: Torque Problem Involving a Block Hanging from an Unanchored, Friction-assisted po

ohhhhhhhhhhh...That makes sense. Thanks.