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Homework Help: Static/Kinetic Friction Question. Is my reasoning correct?

  1. Apr 28, 2013 #1
    I'm reviewing my physics and it's been over a year since I took a course in it. My textbook does have the solutions, but I have a tendency to do it my way. My solution is quite different, but the answer matches up, yet would my answer be correct?

    1. The problem statement, all variables and given/known data
    Two blocks are in static equilibrium. (I'll describe the picture.) One block (A) is 15 kg on the table, and it has a string going through a pulley which is connected to a block (B) that is dangling and is of unknown weight.

    a) Determine the maximum mass of block B. Coefficient of static friction mu_s = 0.20

    b) If an extra 5 kg are added to B, find the acceleration of A and the tension T in the rope. mu_k = 0.14.

    2. Relevant equations

    F = muF_N
    F = ma

    3. The attempt at a solution

    a) Determine the maximum mass of block B. Coefficient of static friction mu_s = 0.20

    The way I see it is that gravity is going to pull block B which will exert tension on the pulley to pull block A to the right. But friction is working against it in the negative direction.

    Now, if we consider that the force block B is exerting on block A should sum to zero, we have:

    F(due to gravity pulling on B) + F(friction working against the force of B) = 0
    mg + (mu)F_N = 0
    10m - 0.20(15*10) = 0
    10m = 0.20*150
    10m = 30
    m = 3

    The book says my answer is correct, but I want to make sure the way I understand it is correct.

    b) If an extra 5 kg are added to B, find the acceleration of A and the tension T in the rope. mu_k = 0.14.

    Nearly similar reasoning as above in that block B will exert some rightwards force on A while friction is working against it (but now we're moving), I'll consider just the movement of A because the acceleration of A and B should be the same magnitude by different direction.

    BUT it would come in useful here to use F = ma to solve for a. Considering the force in terms of one dimension only (horizontal) reduces our problem greatly to few variables which will allow us to simply solve for a.

    F(due to gravity pulling on B) + F(fraction working against the force of B) = ma
    10(3 + 5) - 0.14(15)(10) = (15 + 8)a
    10(8) - 21 = 23a
    59 = 23a
    a = 2.56

    Which matches up with the answer key.

    I can find the tension on my own, I just want to know if I'm understanding the underlying process correctly.

    Thank you.
  2. jcsd
  3. Apr 28, 2013 #2


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    Overall, your reasoning looks good, but there is one problem. Let's look a little more closely at each part.

    In the first part, you're right that block A will feel a force to the right from the tension in the rope, and a force to the left from the static friction from the ground. You set the tension equal to weight of block B, but realize that that is only the case because block B is not accelerating in the up or down direction. This tells us that the tension force on B is up and the weight force is down, and since they have to sum to 0 for B to not accelerate, they must be equal.

    In the second part, I'm surprised that your textbook matches your answer. I calculated it out myself (assuming that the 3 kg that you got in the first part was correct) and got an answer of 2.51 m/s2. Here's where your reasoning was off: remember what I said earlier, the tension applies a force on block A, and the tension will not always equal the weight of block B. In the second case, block B is going to accelerate, so they can't be equal. Instead, let's set up some equations to describe the forces on this system. I'll help you out with one of the blocks: looking at block B, we know from Newton's Second Law (assuming down is positive), mB*g - T = mB*a. If you write out a similar equation for block A, you should have a system of two equations and two unknowns, a and T. Does that make sense?
  4. Apr 28, 2013 #3
    Thank you! I see what you're saying. The textbook did exactly that. This is what it says:

    ƩFy = mbg - T = mbab

    ƩFx = T - μkFN = T - μkFN = mAaA

    Since the blocks are connected by the same string, the magnitude of the acceleration for both of them is the same..."

    Then it goes on, but the math I did for part b exactly matches up with it (the method of constructing my equations to get there was different, but the math itself matched up).

    Sorry it might sound confusing, I can't post the picture of it up on here (don't want to violate the rules).


    I think you got 2.51 because you used g = 9.8. Sorry for the confusion, that is also what the book got. My book used 9.8 (so it doesn't exactly match up per se, but the only essential difference is that I use 10 as my constant for gravity. The book gives 2.5 but using a calculator for more decimal places gives 2.51. Other than my using 10 instead of 9.8, it exactly matches up).

    I used 10 for my gravity because the test I'm writing won't allow calculators.
    Last edited: Apr 28, 2013
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