# Cable Car - Tension and Force

## Homework Statement

The 2020 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 1850 kg counterweight up the other side of the hill. The rolling friction of both the cable car and the counterweight are negligible.

http://session.masteringphysics.com/problemAsset/1000050974/6/knight_Figure_08_39.jpg

If the cable car is descending at a constant speed, then what is the tension in the cable?

F=ma
newton's 2nd law

## The Attempt at a Solution

Since is moving at a constant speed acceleration is zero.

I found the tension equal to the y component of the weight of the cable car (i have the y component being parallel to the ramp).

2020(9.80)sin30 = 9898N

I thought this would be the tension but it's not so i considered the tension in the counterweight as well.

1850(9.8)sin(20) = 6200 N

Would i maybe subtract these to get the tension in the cable?

Thank you

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## Answers and Replies

Hello mybroshi,
for the cable car there are three relevant forces parallel to the ramp the third force being the force due to the brakes.For the counterweight there are two relevant forces mgsin20 down the slope and tension up the slope.

Thanks for explaining that but i cant seem to figure out how that helps me find the tension in the cable :(

Find the tension by considering the forces on the counterweight.On the counterweight there are two forces only and if the pulley is smooth the tension will be the same all along the cable.

Thank you i got it.

Now if i was finding that 3rd force (the braking force) for the cable car would I....

T + B_f + x_c = T + x_w

T=tension

B_f = braking force

x_c = the x force on the cable car ----> 2020(9.8)sin30 = 9898 N

x_w = the x force on the counterweight -----> same as the tension 6200 N

so 6200 + B_f + 9898 = 6200 + 6200

B_f = -3698

Does that look correct?

Thank you

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I had trouble following your symbols but yes it looks good.I got the same answer..3698N up the slope.

Yeah that is kind of confusing haha sorry.

Im having trouble with this next part.

One day the brakes fail (causing the braking force to immediately go to zero) just as the cable car leaves the top on its downward journey. What is the resulting tension in the cable?

It says as a hint: you will use 2 equations with 2 unknowns.

So now wouldnt each side just have 2 forces.

the force parallel to the ramp for the cable car and the tension

and then the force parallel to the ramp for the counter weight and the tension?

Thats right.Continue with your method by writing Newton's second law for the car and for the counterweight.You now have two equations with two unknowns...tension and acceleration.After that its maths.

Just to make sure would my 2 equations be:

T + Fc = m(a)

T + Fw = m(a)

T = tension

Fc = force parallel to the ramp for the cable car

Fw = force parallel to the ramp for the counterweight

Make sure you get the directions right.For the car mgsin30-T =ma,for the counterweight the tension is bigger.I have to go now.Good luck.

How can the tension be bigger for the counterweight if it only asks for

What is the resulting tension in the cable?

shouldn't the tensions be the same?

I also need help with the last part of this question if anyone can direct me in the right direction.

what is the runaway car's speed at the bottom of the hill?

thank you

How can the tension be bigger for the counterweight if it only asks for

What is the resulting tension in the cable?

shouldn't the tensions be the same?

Sorry mybroshi my reply was ambiguous.I meant that for the counterweight the tension was bigger than the component of the weight.The tension is the same all along the cable.
For the last part of ther question use an equation of motion.

so do these 2 equations look right?

mgsin30 - T =ma and i will use m as the mass of the cable car.

T - mgsin20 = ma and i will use m as the mass of the counterweight?

so i solved for a from the first equation...

a = (9898 - T)/2020

then i plugged that into the 2nd equation for a and solved for T and got

T = 7968 N

Does that look correct?

and i do not believe we have learned motion yet in my physics class. is there any other way to solve this last part?

thank you

ok so that was correct.

now can anyone help me find the what is the runaway car's speed at the bottom of the hill?

any help would be greatly appreciated.

thank you

ok so that was correct.

now can anyone help me find the what is the runaway car's speed at the bottom of the hill?

any help would be greatly appreciated.

thank you

Yes it was correct.If you're not familiar yet with equations of motion I suppose you could use PE lost=KE gained.

we havent learned about energy yet.

i dont know why i said we havent learned about motion. i mean we have but we have just covered it briefly.

are there some equations you could point me to that would maybe help me solve that last part of the question.

It is due today!!!! :(

1.Look again at your post 14.Plug the tension in your equation to find the acceleration(a)
2.Use trig to find the distance the car travels down the slope(s)
3.With your question the initial velocity of the car(u)at the top of the slope is zero.
4.Look up equations of motion.There is one that links final velocity(v) with u,a and s.

I am getting 27.64 m/s for my final velocity.... this is what i did

a = 0.955 m/s^2

the distance traveled is 400m : i got this from using the law of sines on the right triangle to solve for the hypotenuse x (the part that the cable car is going down on)

200/sin30 = x/sin90

x=400

then i used the equation vf^2 = vi^2 + 2ad

vf^2 = 0 + 2(0.955)(400)

vf = 27.64

does this look correct?

thanks for all of the help

yay that was right!

thank you Dadface for all of the help!