An inclined plane, a pulley, and three masses

[smashd]

Homework Statement

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 $m/s^2$

Homework Equations

$F = ma$

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
   
   $a = 2.1 m/s^2$
   Or, answer B.

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?

 

[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, let's see if we can get to solution here... Do the FBD first...

 

[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?

 

[WillemBouwer]

yes, that looks better, ja as you take the acceleration as positive downward the weight force should be positive aswell... and it can't be 0 as you stated...

 

[Nickie]

Your solution and answer are correct. The tension in the rope attached to m3 is equal to its weight because the block is accelerating downward, but the tension in the other rope is greater than the weight of m3, causing the block to accelerate downward. In this case, the tension in the rope attached to m3 is acting as a counterbalance to the weight of m3, allowing it to accelerate downward. So, your solution is correct.