Pulley with Multiple Masses

[MDwannabe7]

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Blocks A, B, and C are connected to each other on an inclined plane (theta = 30 degrees) by a massless pulley. Blocks A and B are connected with one rope on the inclined plane, while Block C (which is connected to block B by a second rope) hangs freely on the other side of the pulley. Block A has a mass of 15 kg, Block B has a mass of 10 kg and Block C has a mass of 30 kg. The coefficient of kinetic friction is 0.1. What is the tension on the rope between Blocks A and B?

I can get within 15 N of the correct answer, and am wondering if that is just an issue of rounding or if I'm still doing something wrong. I've attached an image of the problem - please let me know what you get.

 

[MDwannabe7]

Suppose that the incline on the plane is now 90 degrees. What is the magnitude and direction of acceleration of Block C?

If I use the same equations that I figured out for the first problem and plug in 90 instead of 30 for theta, I get the correct answer. However, if I treat this as a separate problem with new diagrams and equations, I get a completely different answer and can't figure out what I am doing wrong. When the angle is 90, doesn't that basically turn this problem into a pulley system where two blocks are basically sitting on a table instead of an inclined plane?

Therefore, Block A would have Tension from the rope pulling it in one direction, while friction is pushing in the opposite direction. Block B is basically the same, except that it has two different tensions, one from each side, in addition to the friction force. But Block C only has it's weight counteracting the Tension that is pulling it up. The weights of Blocks A and B only affect the force of friction because they are equal to the normal force - correct? So is it possible that the book answer is wrong - again?

help!

I just realized that if I eliminate the coefficient of kinetic friction and switch the signs in my new equations, I get the right answer. But if the blocks are on a frictionless plane, how does their weight act in the x direction?

 

[lixxx669]

My method is to find out the acceleration of the whole system (direction along the rope) and then analyze block A. what I came up with is the T on block A is around 120N.
Are you sure the answer is 15N? because only the weight of those blocks are already 150N, 100N, 300N respectively.

Also there is one tricky point: the whole system is accelerating, which means T(rope ab on a) doesn't not equal to T(rope ab on b)

 

[MDwannabe7]

My method is to find out the acceleration of the whole system (direction along the rope) and then analyze block A. what I came up with is the T on block A is around 120N.
Are you sure the answer is 15N? because only the weight of those blocks are already 150N, 100N, 300N respectively.

Also there is one tricky point: the whole system is accelerating, which means T(rope ab on a) doesn't not equal to T(rope ab on b)

No - the answer is not 15N, I was saying that my answer and the correct answer were only 15N apart. Your answer is only 7N from the correct answer. That was my method too - so not sure why you ended up closer than me, but I'll check again.

Can you please explain what you mean by the tension on the same rope not being equal for both blocks A and B. How can one rope not have the same tension on it?

 

[lixxx669]

I used calculator and get a finally answer of 129.8N

sorry about the "T(rope ab on a) doesn't not equal to T(rope ab on b)" thing, they are actually equal, because we count the rope is massless...

 

[MDwannabe7]

I used calculator and get a finally answer of 129.8N

sorry about the "T(rope ab on a) doesn't not equal to T(rope ab on b)" thing, they are actually equal, because we count the rope is massless...

Did you have g = 9.8 or g=10 and what did you get for the total acceleration? The answer is 127.2 N.

 

[lixxx669]

oh, I used g=10
if I use g=9.8, then, it would be 129.8/10*9.8=127.2
yup~ that's the answer~

 

[CrimsonMirage]

Hi,
Wait...what exact equations did you use to get to that answer?

 

[johnson]

Hi,
Wait...what exact equations did you use to get to that answer?

Hi.

How is one to go about solving this problem.

So far I have:

Forces on mass A:
T1 right
75N left (mgsin30)
12.75N left (.1mgcos30)


Forces on mass B:

8.5N left (.1mgcos30)
50N left (mgsin30)
T1 left
T2 right



Forces on mass C:

T2 up
300N down

 

[Kaustikos]

oh, I used g=10
if I use g=9.8, then, it would be 129.8/10*9.8=127.2
yup~ that's the answer~

I'm not pointing fingers, but you definitely will be using 10 m/s2 for g. Just in case anyone is wondering and doesn't want to ask.

 

[johnson]

Please help.

 

[johnson]

bump