(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The 2000 kg cable car shown in Figure P8.39 descends a 200 m high hill. In addition to its brakes, the cable car controls its speed by pulling an 1500 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

(Image attached)

(a) How much braking force does the cable car need to descend at constant speed?

________ N

(b) 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?

________ m/s

2. Relevant equations

Kinematics equations

2nd Law statements

Fk= (Mu)(N)

3. The attempt at a solution

A) Well, since it says a constant speed, acceleration equals 0. So I made a 0 in my 2nd law statement:

(2000)(0)= T- (2000)(9.8)sin30

T=9800

This was not the answer. Where did I go wrong?

B) I began by drawing force diagrams and writing out the 2nd law statements. I then combined the two equations (for the counterweight and the car) by solving one of them for T. I ended with this equation:

Aa= Ba + Bgsin@ - Agsin@

Then I realized that the counterweight basically fails right? But then this problem wouldn't make much sense. So I continued on as if just the car's brakes fail. Solving for a. My a turned out to be really high (23.65), but, I continued on my way. I drew a kinematics table, trying to find the final velocity and used the 30 degrees and the height to find the distance traveled, which turned out to be 400. So, I used the equation Vf^2= V0^2 + 2ax:

Vf^2= 0^2 + 2(-23.65)(-400)

Vf= 97.3

This was wrong as well. Anyone know what I'm doing wrong?

Thanks in advance!

~Phoenix

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# Homework Help: Cable Car and Counterweigh- A Problem of Tension, Friction, and Kinematics

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