Simple Mechanics Problem: Finding Minimum Mass on an Inclined Plane

[steelphantom]

I had this problem on a test I took last week, which our professor recently returned to us. Everyone has been getting the answer 3.33, but I'm not sure if it's correct. I got it wrong on the test. Here is the problem:

Given:
M1: 5kg
Theta: 60*
Coefficient of static friction: 0.4

http://files.nicktiberi.com/files/inclinedplane.gif

Excuse the crude diagram, but it was either that or my sloppy handwriting from my test. Anyway, without further ado, the objective of the problem is: Find the minimum mass M2 such that the blocks do not move.

Here's my force analysis.

M2 vertical forces: T - M2g = 0
M1 horizontal forces: M1g[sin(theta)] - T - fs = 0
M2 vertical forces: N - M1g[cos(theta)] = 0

fs = mu static * N
fs = (.4)(24.5) = 9.8

Then, setting the two tensions equal to each other, I get:
M2g = M1g[sin(theta)] - fs

M2 = (M1g[sin(theta)] - fs) / g

After subbing all the numbers in, I get M2 = 3.33; according to one kid, he put 3.33 on his test but got it marked wrong. I really don't know any other way of doing this problem. Any help would be great.

 

[finchie_88]

The error is that it says the least mass required. That tells you that the frictional force is acting down the slope, not up it. you should get something like...
$$m_1sin\theta + m_1acos\theta = m_2$$, then solve that for $$m_2$$.

I think that I'm right.

P.S. a means the static co-efficient of friction between the block and the surface.

 

[steelphantom]

Why would the friction act down the slope? Wouldn't it still be acting up the slope to keep block M1 from falling down?

 

[topsquark]

Why would the friction act down the slope? Wouldn't it still be acting up the slope to keep block M1 from falling down?

I think we can all agree that with the way the question is posed that the static friction force is at its maximum value, yes? Now, if m2 is at its minimal value to keep the system in equilibrium we can infer that there is also a maximum value of m2 that would keep the system in equilibrium. In this case fs disappears.

So let's see what happens in either case of the direction of fs if we add just a touch more mass to m2.

If fs points down the slope, then adding mass to m2 will tend to cause m1 to accelerate up the slope. So we would need to increase the size of fs to compensate. The problem is that we are already at max fs.

If fs points up the slope, then adding mass to m2 will again tend to cause m1 to accelerate up the slope. But this time fs decreases because it is only the size it needs to be to stop the "intended" motion. So fs is reduced in size which is the result we want.

So I agree with steelphantom. fs should point up the slope in this case. (I also agree m2 = 3.33 kg.)

-Dan

 

[steelphantom]

So I agree with steelphantom. fs should point up the slope in this case. (I also agree m2 = 3.33 kg.)

-Dan

Thanks, Dan. I'm glad to see that some others are coming up with the same answer. I was working on this problem with a few classmates today and we all ended up with 3.33 kg. This HAS to be the answer.

 

[topsquark]

I think we can all agree that with the way the question is posed that the static friction force is at its maximum value, yes? Now, if m2 is at its minimal value to keep the system in equilibrium we can infer that there is also a maximum value of m2 that would keep the system in equilibrium. In this case fs disappears.

So let's see what happens in either case of the direction of fs if we add just a touch more mass to m2.

If fs points down the slope, then adding mass to m2 will tend to cause m1 to accelerate up the slope. So we would need to increase the size of fs to compensate. The problem is that we are already at max fs.

If fs points up the slope, then adding mass to m2 will again tend to cause m1 to accelerate up the slope. But this time fs decreases because it is only the size it needs to be to stop the "intended" motion. So fs is reduced in size which is the result we want.

So I agree with steelphantom. fs should point up the slope in this case. (I also agree m2 = 3.33 kg.)

-Dan

For the record, a slight correction to my post...I had mentioned:

Now, if m2 is at its minimal value to keep the system in equilibrium we can infer that there is also a maximum value of m2 that would keep the system in equilibrium. In this case fs disappears.

This statement is not correct. The maximum possible m2 will be for fs,max acting down the slope. For this we get a value of m2=5.33 kg, obviously larger than the 3.33 kg using fs up the slope.

-Dan