# Find the mass of m1 on an inclined pulley system

1. Feb 21, 2013

### IcyDuck

1. The problem statement, all variables and given/known data
Two masses $m_1$ and $m_2$ are connected by a massless cord and a pulley, on a rough ramp, tilted at angle ∅ The pulley is massless and frictionless. The coefficient of kinetic friction between the ramp and m1 is $μ_k.$ Derive an expression for $m_1,$ given that $m_2$ accelerates downward with acceleration of magnitude $a.$

Diagram:

2. Relevant equations
$F=ma$
$g=9.8 m/s^2$

3. The attempt at a solution
I drew my free body diagrams for each mass.
For $m_1,$ the axes are aligned with the ramp, not the ground. $m_1g$ is broken up into its $x$ and $y$ components: $m_1gsin∅$ and $m_1gcos∅,$ respectively. Tension $T$ is larger than frictional force $F_f$ since the object is accelerating up the ramp despite the frictional force in the opposite direction.

For $m_2,$ the axes are aligned with the ground with the up direction being the positive $y$ direction. The weight $mg$ is larger than the tension $T$ due to its downward acceleration.

I think that the tension will be the same everywhere in the rope, so $T_{m_1}=T_{m_2}.$

With regard to $m_1:$
$T=m_2a$
$F_{net}=T-F_f,$
or
$F_{net}=m_2a-μ_km_1gcos∅.$

With regard to $m_2:$
$F_{net}=m_2g-T,$
or
$F_{net}=m_2g-m_2a.$

That's about as far as I got. It's not much of a start. I don't know what steps I need to take towards isolating $m_1.$

Last edited: Feb 21, 2013
2. Feb 21, 2013

### ap123

Where did you get T = m2a from?
For the net force on m1, you forgot the component of gravity acting down the ramp.

3. Feb 21, 2013

### IcyDuck

I derived tension $T$ by applying $F=ma$ and getting $T=ma$ since $T$ is a force. I saw that since the mass of $m_2$ is given and its downward acceleration $a$ is given and on the same axis as $T$ (the $y$ axis), I could plug those into the force equation and get the tension.

And yeah, I can see what you're saying about the gravity. Since $m_1$'s motion is completely horizontal, it should actually be $F_{net_{m_1}}=m_2a−μ_km_1gsin∅.$ Not sure how I managed to mix up sin and cos, sorry.

4. Feb 21, 2013

### ap123

Have another look at your free-body diagrams.
The one for m2 should have the tension pointing up and the weight down.
The sum of both of these forces will be m2a.
So, T = m2a is not correct.

The motion of m1 may be horizontal with respect to the ramp, but it's not horizontal with respect to the ground, so there is a component of the weight acting to pull the mass back down the incline.

5. Feb 21, 2013

### IcyDuck

Ok, so $T+m_2g=m_2a,$
or
$T=m_2a-m_2g.$

So, correct me if I'm misunderstanding what you're saying, does that mean that $mg_x$ should actually be subtracted, making $F_{net_{m_1}}=T-F_f-mg_x$?

6. Feb 21, 2013

### ap123

Remember that T and the weight are in opposite directions, so
m2g - T = m2a
( my y-axis points down here )

Your expression for Fnet for m1 is correct.
When you equate this with m1a, be careful with the signs.