Cable car and counterweight forces problem. (Tension)

[Sentience]

Homework Statement

The 2060 kg cable car shown in the figure descends a 200-m-high hill. In addition to its brakes, the cable car controls its speed by pulling an 1890 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

The hill that has the 2060 kg cable car descending is 30 degrees. The tip of the triangle is 200 m high, and then the hill that has the ascending 1890 kg counterweight is angled at 20 degrees.

Question 1 : How much braking force does the cable car need to descend at constant speed?

Question 2: One day the brakes fail just as the cable car leaves the top on its downward journey. What is the runaway car's speed at the bottom of the hill?

I'm not sure if this image link will work : http://session.masteringphysics.com/problemAsset/1073561/3/knight_Figure_08_39.jpg

Homework Equations

F = ma

The Attempt at a Solution

Ok, there are 3 forces acting on the cable car. Sin30*mg being the force of gravity, with direction down the incline, the force of the breaks pointed up the incline, and the Tension on the cable from the counterweight on the other side of the hill.

There are only 2 forces acting on the counterweight. They are Sin20*mg and the force of tension on the cable. Because the mass of the cable car is greater, it's pulling the counterweight over and down the hill.

First I found (I think) the force of the breaks.

Total forces on cable car = F_t+F_b-sin30*m*g

Total forces on counterweight = F_t-sin20*m*g

To find the the force on the breaks I set equation 2 as: Tension = sin20*m*g and I get approx. 6335 N. Then I inserted it into the equation for the cable car and set acceleration = 0 since the problem mentioned constant speed.

I then subtract that from the force of gravity on the cable car (sin30*m*g = 10094 N) and get Force of breaks = 3759 N

When I submitted that answer I was correct.

However, I'm completely lost on how to find the acceleration so I can then find the velocity using a kinematic equation. Please help.

 

[Yuqing]

Part 2 is pretty analogous to part 1. You did the right thing in setting both accelerations to 0. But now in part two, that acceleration is not 0 but it is still the same for both cars (since they are connected by rope). That should allow you to connect the two equations.

 

[Sentience]

a is longer zero, but F_b is. So is this the right equation?

F_t-sin30*m1*g +T - sin20*m*g = m1a + m2a

 

[Yuqing]

Some of the signs don't look right. The tension should cancel.

 

[Sentience]

I was inconsistent with my signs on my axis for each mass, I found the right answer. Thanks for the second set of eyes!

It should have been -T + sin20*m*g on that second equation.

 

[Sentience]

For future reference, if you have multiple forces in different directions acting on a mass do you set your equations as Larger force minus smaller force?

That way you always end up with a positive net force.