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Coefficient of kinetic friction between table and floor

  1. Oct 1, 2009 #1
    1. The problem statement, all variables and given/known data

    When the three blocks in Fig. 6-33 are released from rest, they accelerate with a magnitude of 0.800 m/s^2. Block 1 has mass M, block 2 has 2M, and block 3 has 2M. What is the coefficient of kinetic friction between block 2 and the table?


    2. Relevant equations

    F_f = mu_k*F_N

    3. The attempt at a solution

    So far I have:

    The x-component of the force on the second block is equal to the sum of the y-component of 1, the y-component of 3, and the friction between the table and 2 :

    F_2,x = 2*M(0.8) = F_1,y + F_3,y + F_f

    For friction, we have:

    F_f = mu_k*F_N = mu_k*2*M*g

    Since the normal force is just gravity in this case. Substituting, we have

    F_2,x = 2*M*g - (M*g + mu_k*2*M*g) = (1 - 2*mu_k)*M*g

    Since the acceleration is 0.8 then we solve for mu_k

    F_2,x = (0.8)*(2.0)*M = 1.6*M = (1 - 2*mu_k)*M*g

    1.6 = (9.8)*(1 - 2*mu_k)

    (-1/2)*((1.6/9.8) - 1) = mu_k

    (-1/2)*(-8.2/9.8) = 0.418

    But this answer isn't right. I have a hunch I messed up the beginning by assuming that

    F_2,x = 2*M(0.8) = F_1,y + F_3,y + F_f

    But I'm not sure exactly.
  2. jcsd
  3. Oct 1, 2009 #2


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    You need to draw three separate free body diagrams for each block. An important item that you forgot to include is the tension in the ropes. Note that the tension in the right rope is not the same as the tension in the left rope. In fact the difference in the tensions minus the force of friction is the net force on the middle block. Sorry, you have to start all over again.
  4. Oct 1, 2009 #3
    Thank you for responding. I had a feeling that equation was wrong. I started to ask myself the question, what are the tensions in the ropes, but I realized that each time I attempted to solve it, I would require the tension in the other rope. Is that correct? Aren't they dependent on one another?

    I should probably draw up another diagram, but I was just doing another, much simpler problem, and got it wrong because I don't fully understand the way these two forces interact. Here's the problem I got wrong:

    (a) An 11.0 kg salami is supported by a cord that runs to a spring scale, which is supported by a cord hung from the ceiling (Fig. 5-35a). What is the reading on the scale, which is marked in weight units? (b) In Fig. 5-35b the salami is supported by a cord that runs around a pulley and to a scale. The opposite end of the scale is attached by a cord to a wall. What is the reading on the scale? (c) In Fig. 5-35c the wall has been replaced with a second 11.0 kg salami, and the assembly is stationary. What is the reading on the scale?


    Now the first two are fairly obvious, which gives 108 N for both the freely hanging body and the body attached to a cord, by a pulley. However, for the third question, I assumed that the two forces would combine to produce a total of 216 N on the scale. However, the book says it's also 108 N. This is kind of lost on me because there should be a force in both directions, away from the scale, causing it to read double. I've looked through the book and I don't quite see an example that explains this phenomenon.
  5. Oct 1, 2009 #4


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    Think about the freely hanging body. There are two forces on the scale in that scenario: the gravity of the hanging body pulling down, and the force of the rope above the scale pulling up. The net force has to be zero, or else the scale would accelerate. There same thing happens in (c): there's the force of one salami pulling at the scale, and a corresponding force from the other salami that balances it out to produce a net force of 0 on the scale.
  6. Oct 1, 2009 #5


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    They are, but that's not a problem. In the equations y=x^2+4 and y=x+4, x and y depend on each other, but the system of equations can be solved to yield unique answers for both x and y. Similarly, the tensions depend on each other, but writing out Newton's second law for each body will give you a system of equations that you can solve to get unique answers for each tension.
  7. Oct 1, 2009 #6
    Okay, I'm sorry to say it, but that makes me even more confused :( So you're saying that since F_net = 0, then the scale should read only one and not both bodies? ... um.. ? Still lost on that..

    What would happen if we put two scales there, one facing each direction.. Would each read 108 or would each read 54?
  8. Oct 1, 2009 #7
    Ok, I actually meant to say they were dependent in a non-solvable way, but if they are solvable, then I'll run with that. Thanks.
  9. Oct 1, 2009 #8


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    Let's get really specific: suppose the scale is one of those involving a spring. If there's only one force on the spring, it would accelerate and won't stretch a millimeter. Every time a scale is used, there has to be another force holding on to the spring's other end to stretch the spring and get a reading.
  10. Oct 1, 2009 #9
    It seems to me this is just something I'll have to get used to, even though I don't complete understand it. Thanks for the help.
  11. Oct 2, 2009 #10
  12. Oct 2, 2009 #11


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    Look at it this way. Suppose you are looking at one of the salamis wondering when it will be time to eat them and I sneak unseen over to the other side of the table, hold the string with one hand and cut it with the other then tie it to the wall so that you get picture (b). I have stolen your salami but as far as you are concerned, nothing has changed. It doesn't matter whether the other side of the string is tied to the wall, is held by a person or another salami. As long as the salami does not accelerate, the net force on it is zero which means that the tension is equal to the weight.
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