# An inclined plane, a pulley, and three masses

1. Dec 10, 2012

### smashd

1. The problem statement, all variables and given/known data
A system comprising blocks, a light frictionless pulley, a frictionless incline, and connecting ropes is shown. The 9 kg block accelerates downward when the system is released from rest. The acceleration of the system is closest to:

A.) 1.9 $m/s^2$
B.) 2.1 $m/s^2$
C.) 1.7 $m/s^2$
D.) 1.5 $m/s^2$
E.) 2.3 $m/s^2$

2. Relevant equations

$F = ma$

3. The attempt at a solution
1. First

$m_{1} = 6 kg$
$m_{2} = 4 kg$
$m_{3} = 9 kg$

$\theta = 30°$

$a = a_{x} = a_{y}$

2. Then, the sum of forces on the three masses

$\sum F_{x1} = T_{2}-m_{1}g\sin\theta = m_{1}a$
$\sum F_{y1} = 0$

$\sum F_{x2} = T_{1}-T_{2}-m_{2}g\sin\theta = m_{2}a$
$\sum F_{y2} = 0$

$\sum F_{x3} = 0$
$\sum F_{y3} = T_{1}-m_{3}g = m_{3}a$

3. Combine $F_{x1}$, $F_{x2}$, & $F_{y3}$ and isolate $a$...

$a = \frac{2T_{1} - g (m_{1}\sin\theta + m_{2}\sin\theta + m_{3})}{(m_{1} + m_{2} + m_{3})}$

4. Solve for $T_{1}$

$\sum F_{y3} = T_{1}-m_{3}g = m_{3}a$
$\sum F_{y3} = T_{1}-m_{3}g = 0$
$T_{1} = m_{3}g = (9 kg)(9.81 m/s^2) = 88.29 N$

5. Plug $T_{1}$ into $a$ and solve

$a = 2.1 m/s^2$

Is this the correct solution and answer? Did I solve correctly for T_1? I don't think it's right because the tension should not equal weight of m_3 because technically the block IS accelerating downward at this instant, isn't it?

Last edited: Dec 11, 2012
2. Dec 11, 2012

### WillemBouwer

smashd, ja if the tension in the rope was the same as the weight no acceleration will take place: Very important note: Define your axis for the entire system, I see you take the x axis for your first 2 masses as parallel with the surface so the y axis should be perpendicular to this surface for the entire system you cannot change the axis for the 3rd mass... After defining your axis, draw the 3 FBD's for the masses, you are on more or less the right track, lets see if we can get to solution here... Do the FBD first...

3. Dec 11, 2012

### smashd

Thanks for the input, WillemBouwer.

So $\sum F_{3y}$ should be:

$\sum F_{3y} = m_{3}g - T_{1} = m_{3}a$

Then $a$ would become after combining the forces on the system:

$\frac{m_{3}g - m_{2}g\sin\theta - m_{1}g\sin\theta}{m_{1} + m_{2} + m_{3}}$

Which is still 2.1 m/s^2, but this is the proper solution to the problem?

4. Dec 11, 2012

### WillemBouwer

yes, that looks better, ja as you take the acceleration as positive downward the weight force should be positive aswell... and it cant be 0 as you stated...