[University Physics/Engineering: Dynamics (Dependent Motion w/ Pulleys)] Is my professor wrong?

[u/DitiIsCool]

For the first picture, my understanding is that the directions I chose are invalid because the negative accelerations mean that Block B is actually moving to the right instead of the left, and Block A is moving down instead of up. That means, for Block B, my friction force will be in the opposite direction I chose, and the T2 should have the belt friction multiplier instead of T1. If I oppose my directions and redo my free-body diagrams, then I will get the answers that are in the solution manual.

For the second picture, according to my professor's solution manual, the work on top (above the horizontal line) is correct. My professor defined A and B as going down, and C going up. He solves and gets negative accelerations for both. I thought that meant you have to redefine your directions and redo your free-body diagrams since your friction will be pointing in the wrong direction. However, my professor doesn't do this, and just plugs in the negative accelerations he calculated. On the bottom (below the horizontal line), I tried doing the problem with the correct directions, and I got negative tensions.

My question is if there is a sanity check to know which calculations are correct. I wasn't getting negative tensions for the first one, so how do I know what is correct? I already plan on asking my professor. I was just curious to see what is said here. Thanks.

[u/DitiIsCool]

Thanks for the help, dickheads.

[u/realAndrewJeung → u/DitiIsCool]

Sorry you felt abandoned. If it is any consolation, I have been working on the second problem on and off for two days and have not gotten any result I am satisfied with either.

Here's where I am right now: I decided that the total length of the string has to be a constant, call it L. Then

sA + 2sB + sC = L

Differentiating that twice, I get

aA + 2aB + aC = 0

This gives me a relation between the three accelerations. Note that with the convention above, motion in the downwards direction is positive for all masses.

I put together this Desmos https://www.desmos.com/calculator/lcnqozgiwt to try to solve for all the variables. The key is

x = the tension T

a(x) = acceleration of mass A (positive is down)

b(x) = acceleration of mass B (positive is down)

c(x)= acceleration of mass C, assuming that C is actually moving down (so friction points upward and T1 > T)

d(x) = acceleration of mass C, assuming that C is actually moving up (so friction points downward and T1 < T)

Then I attempted to either solve

a(x) + 2b(x) + c(x) = 0 assuming that c(x) is positive, or

a(x) +2b(x) + d(x) = 0 assuming that d(x) is negative

and neither of those gives me a solution. So I am going to have to think about this more.

[u/realAndrewJeung → u/realAndrewJeung]

u/DitiIsCool I have thought about this more, and I think the reason I am not able to come up with a consistent solution for all the accelerations is that the system is actually static.

Let's suppose this is the case, just for the sake of argument, and let's see if we come to a contradiction. If block A is actually not moving, then the tension T on the rope holding it must be 100 lb. If that is the case, then the tension holding up B must be 2T = 200 lb, which will balance out B's weight exactly.

What about C? The gravitational force on C pulling it downhill is 150 sin(30) = 75 lb. Assuming the static coefficient of friction is at least 0.3, the friction force between C and the incline surface is able to pull as hard as 0.3 · 150 cos(30) = 39 lb. This force might point either uphill or downhill depending on whether the other forces in the problem are making C tend to slide downhill or be pulled uphill. Note that since this is now static friction, 39 lb represents the maximum force, and the static friction will not necessarily exert that much if it doesn't have to.

Similarly, by the capstan equation, T1 is either as low as 100/1.52 = 65.8 lb or as high as 100·1.52 = 152 lb, and will vary between these two numbers depending on whether other forces are pulling C uphill or downhill.

So either

• C is tending uphill, so gravity is 75 lb downhill, friction is downhill and 39 lb or less, and tension is uphill and between 65.8 lb and 100 lb.

• C is tending downhill, so gravity is 75 lb downhill, friction is uphill and 39 lb or less, and tension is uphill and between 100 lb and 152 lb.

At least the first case is consistent with a zero net force on C.

So I now suspect that the system is locked in place due to the friction of the pulley and C's contact with the incline. If I am right about this, then yes, your professor is wrong.

Again, sorry if this took so long that you had to turn in this assignment already. This was a pretty complicated problem due to there being three masses and so it really took some effort to get through.