Cable Car and Counterweigh- A Problem of Tension, Friction, and Kinematics

[Phoenixtears]

Homework Statement

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

Homework Equations

Kinematics equations
2nd Law statements
Fk= (Mu)(N)

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

 

[Doc Al]

Post the diagram.

 

[Phoenixtears]

Yikes, looks like that attachment didn't work. I'm posting it now. Thanks!

 

[Doc Al]

Homework Statement

The coefficient of kinetic friction between the m = 2.6 kg block in Figure P8.35 and the table is 0.28. What is the acceleration of the 2.6 kg block? (Image attached)

This doesn't match the diagram.

Also, it's a little hard to decipher equations like

Aa= Ba + Bgsin@ - Agsin@

..since it's not immediately obvious what A & B are. (Forces? masses?)

 

[Phoenixtears]

This doesn't match the diagram.

Oh, great- I think my computer's blowing up. Sorry about that. Figures..

..what A & B are. (Forces? masses?)

A= mass of car
B= mass of counterweight

The force (being tension) was substituted out.

 

[Doc Al]

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

It looks like you are analyzing the forces on the cable car. But you left out the braking force, which is what you need to solve for.

You also need a similar equation for the counterweight. Solve them together to get the braking force.

 

[Phoenixtears]

It looks like you are analyzing the forces on the cable car. But you left out the braking force, which is what you need to solve for.

You also need a similar equation for the counterweight. Solve them together to get the braking force.

I see what you're saying. For some reason I considered the tension force to be enough (from the 3rd law). So now I have the tension force of the counterweight... could I not use this to solve for the total force, which would be the answer (because, as you said, I need to analyze the braking force, not the cable car force. Which, of course, I would do incorrectly )

Thanks, I'll try that right now!