# Simple little motion problem(at least for you guys)

1. Oct 14, 2009

### Melchoire

The mass m3 is tied to m2 with an idealized string and m2 rests on m1. There is a rough
interface between m2 and m1 with a coefficient of friction μ. The mass m1 rests on a frictionless
surface. The pulley is “ideal,” that is, massless and frictionless. All three masses move together
as if m1 and m2 were solidly attached.
(a) Draw free body diagrams for the three masses showing all forces and directions of acceleration.
Find expressions for the following in terms of g, m1, m2 and m3:
(b) the acceleration of each mass.
(c) the tensions T2 and T3.
(d) the smallest value of the coefficient of friction μ to permit the above motion
http://img407.imageshack.us/img407/9299/m1m2m3.jpg [Broken]

I drew the free body diagrams that wasn't really complicated. But I'm stuck on all the rest. For 'b' I wrote out newtons second law for each mass where 'a' was equal in all but that didn't get me anywhere. When I think about it I know that the larger the masses of m1 and m2 the slower the acceleration of system. But I don't know how to get that in my equations.

Any help would be appreciated. =]

Last edited by a moderator: May 4, 2017
2. Oct 14, 2009

### rl.bhat

Write down Newtons second law for m1, m2 and m3.
Let us see whether you done it properly.
Then we will provide further hint.

3. Oct 15, 2009

### Melchoire

m3:
Fnet = T3-m3*g = m3*a
m2:
Fnet = Fn + Fg + Ff + T2= Ff = us*m2*g + T2 = m2*a
m1:
Fnet = Fn + Fg + T2 = T2 = m1*a

the magnitude of T2 and T3 are equal.

Do I have those right? Did I calculate the static friction force properly?

4. Oct 15, 2009

### rl.bhat

m3:
If you take -g then acceleration must be -a.
m2.
frictional force acts in the opposite direction to T
So m2a = ....?
m1
On m1 only reaction to the frictional force act in the forward direction.
So m1a = ...?

5. Oct 15, 2009

### Melchoire

You're kinda confusing me a little, you probably didn't mean to but 'a' can't be equal to 'g'...or is that not what you meant?

But from the rest I got this:
m2*a=T2+Ff

m1*a = T2

Am I right?

6. Oct 15, 2009

### rl.bhat

What I meant is that T - m3g = - m3a
Second one
T -μm2g = m2a.

7. Oct 15, 2009

### Melchoire

Ok so my final f=ma equations are:

-m3*a = T - m3*g
m2*a = T - us*m2*g
m1*a = T? (am I right for this one?

But from the first one I know that T = -m3*a+m3*g = m3(g-a). So then I can just use an equivalency to solve 'a' in terms of m1,m2,m3 and g.

8. Oct 15, 2009

### rl.bhat

Last one is wrong.
m1a = μs*m2*g

9. Oct 15, 2009

### Melchoire

Why is that? I thought that since m1 and m2 are moving together then the force acting on m1 should be the tension force.

10. Oct 15, 2009

### rl.bhat

In this problem a is same for all the masses. T is not connected to m1.
Force on m1 is reaction of the frictional force acting on m2.

11. Oct 15, 2009

### Melchoire

Alright that makes sense. I hope I can remember this stuff for the mid term on friday. Anyways I'll try the other questions on my own and if I need help I'll come back.

12. Oct 15, 2009

### rl.bhat

OK. All the best for mid term examination.

13. Oct 15, 2009

### Melchoire

I'm still stuck on the first one after using equivelancy I got these equations:
a = us*m2*g/m1
a = T-us*m2*g/m2
a = m3*g-T/m3

But the question is asking to express a in terms off m1,m2, m3 and g. the closest thing I have is the first equation, but it's missing m3.

14. Oct 15, 2009

### rl.bhat

First of find the tension T. Then find a.

15. Oct 15, 2009

### Melchoire

T = m3*g right?

16. Oct 15, 2009

### rl.bhat

No.
Equate eq. 2 and 3 from your post# 13 and solve for T.

17. Oct 15, 2009

### Melchoire

Ok I get
T = m2*m3*g(1+us)/(m3+m2)

Which equation do I plug it into?

18. Oct 15, 2009

### rl.bhat

In either eq. 2 or 3

19. Oct 15, 2009

### Melchoire

K got it. How would I approach D?

20. Oct 15, 2009

### rl.bhat

m1 and m2 will start moving when μm2g = (m1 + m2)a
Find μ.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook