# Finding acceleration of a mass being pulled by another mass over two pulleys

1. Nov 25, 2012

### cener

1. The problem statement, all variables and given/known data
A mass m1 is pulled along a frictionless horizontal surface by a rope of tension T1, which passes over a pulley of mass mP1 to a hanging pulley of mass mP2, after which it is fixed to the ceiling, as shown in Fig. 1. Labeling the
tension between pulleys 1 and 2 T2, and that between pulley 2 and the ceiling as T3, find the acceleration of the mass
m1. Note that the acceleration of pulley 2 is half the acceleration of m1
The acceleration of m1 is:
(a) 0
(b) $\frac{m(P2)g}{2m1+mP1+(3/4)mP2}$
(c) $\frac{m(P2)g}{2m1+mP2}$
(d) g
(e) $\frac{m(P2)g}{2m1+mP1+mP2}$

2. Relevant equations
I have no clue

3. The attempt at a solution
when I did it, I got a=$\frac{m2g}{2m1+(m2)/2}$
But, this is not right. Does it depend on if pulley1 has a mass?
http://www.freeimagehosting.net/wph3q
(if for some reason the diagram doesnt show)
http://www.freeimagehosting.net/wph3q

2. Nov 26, 2012

### grzz

Yes! You noticed the trouble in your solution.

The masses of the two pulleys MUST be taken into account. The reason for this is that the pulleys are assumed to be rotating. That was the reason that the tensions in all parts of the string were named differently.The string is not just slipping over smooth motionless pulleys.

Hence one must use
F = ma ..........................................................for the linear motion and
Torque = moment of inertia x angular acceleration..for the angular motion.

3. Nov 26, 2012

### cener

So how do I start?

What I have:
τmp1=Iω
τmp2=Iω
Fm1=ma
-where all of the accelerations are the same
Fmp2=ma
-where the acceleration is half of the other acceleration

Is this correct? If so, how do I put them together?

4. Nov 26, 2012

### haruspex

Try to stick to the given variable names where appropriate: m1, T1, T2, T3, mp1, mp2. Of course you'll also need new variables for acceleration: maybe a for the mass m1 (and therefore a/2 for the vertical accn of pulley 2), and α for the angular accn (ω is usually for angular velocity).
I find it surprising that the radii of the pulleys don't come into it. I would let those be r1 and r2 and see what happens.
With all that in place, list the forces acting on each component and write out corresponding acceleration equations.

5. Nov 27, 2012

### grzz

The moment of inertia for the pulley may be assumed to be that for a uniform disc
i.e. I = $\frac{M_{p}R^{2}}{2}$.

The angular acceleration $\alpha$ is best put in terms of the linear acceleration a
using a = R$\alpha$

Then when the equation for the torque τ is used it will be noticed that the radius R of the pulley will be eliminated from the equation.