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Homework Help: Friction Ramp Problem (I'm challenging a score received on a test)

  1. Oct 26, 2009 #1
    Hello! First time, long time!

    This is kind of a long post. I did what I could to keep it clear. TYIA!

    I was marked totally wrong on a test question and I think I may have been correct. I'm trying to get my ducks in a row before I ask the professor to review this with/for me. Everyone in the class seems to have gotten different answers so I don't have anything to compare it to.

    I believe that in this system, there is no acceleration because the friction is too great. This calculator I found online agrees with my answers:

    Any thoughts or advice would be helpful.

    Please, help me stick it to the man! Or, keep me from making a jerk out of myself.

    1. The problem statement, all variables and given/known data
    The hanging 300g mass is connected to a 500g mass on a 35 degree downwards incline by an ideal string/pulley arrangement. Calculate the acceleration of the masses and the tension in the string when the system is released. The friction coefficient between the 500g mass and the ramp is Mk=.150.

    Here is a picture of the situation:

    2. Relevant equations
    N=mg(cos@) <----- @=theta
    Force parallel to ramp = mg(sin@)

    3. The attempt at a solution
    For the 500g block:
    W = (.5)(9.8) = 4.9N
    N = (.5)(9.8)(cos35) = 4.01N
    Force parallel to ramp = (.5)(9.8)(sin35) = 2.81N
    Friction on ramp = (4.01)(.150) = .601N

    For the 300g (hanging) block:
    W = (.3)(9.8) = 2.94N

    OK, if we were totally frictionless we would get: Fnet = 2.94 - 2.81 = .13N towards the hanging 300g block. This means:
    .13N = (.8kg)a
    a = .1625 m/s^2

    But, we have friction and friction works against any motion. I believe that in this system, the friction is too great to overcome with these masses. We are in the zone where the blocks will not move.

    Since the system wants to move towards the 300g mass, friction opposes it:
    Fnet = 2.94N – 2.81N - .601N = -.471N

    Therefore, it will not accelerate towards the 300g hanging mass because the friction is too great.

    AND, I just don’t get to add the friction as a force going down the hill and say:
    Fnet = 2.81N + .601N – 2.94N = .471N

    This due to the fact that the friction would then oppose the downhill motion and bring me back to the first situation.

    Therefore, we have too much friction in this system.

    Am I correct in my logic and reasoning?

    Since there is no movement, the Tensions are equal to the weights of the masses.
    T1 = 2.81N
    T2 = 2.94N
    Last edited: Oct 26, 2009
  2. jcsd
  3. Oct 26, 2009 #2


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    Homework Helper

    Mass 1

    down the plane

    [tex]m_1a = m_1 gsin\theta - \mu mgcos\theta -T[/tex]

    hanging mass

    [tex]m_2 a = T-m_2 g[/tex]

    Solve. I don't think there would be too much friction. But I don't have a calculator at hand to check it.
  4. Oct 26, 2009 #3
    Hello! Thanks for responding.

    I get that. But, I don't think this is a problem where you can just plug the numbers into the formula. I think it's one of those special cases where you have intermediate values that prevent acceleration.

    With no friction:
    a = [(m2)g-(m1)(g)(sin@)] / [m1+m2]

    with numbers
    a = [(.3)(9.8) - (.5)(9.8)(sin35)] / [.5 +.3]
    a = .1618

    When we add friction:
    a = [(m2)g-(m1)(g)(sin@) - (Mu)(m1)(g)(cos@)] / [m1+m2]

    With numbers
    a = [(.3)(9.8) - (.5)(9.8)(sin35) - (.15)(.5)(9.8)(cos 35)] / [.5 +.3]
    a = -.591

    The magnitude of the acceleration increases when you add friction. Plus the acceleration changes directions. I don't see how that is possible. That's why I think this is in the intermediate range where the weights don't overcome friction.

    Do you see the point I'm trying to make? Is this correct?

    Or, am I just crazy?
  5. Oct 27, 2009 #4


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    Homework Helper
    Gold Member

    No, you are quite sane. I tended to agree with rockfreak until I cranked out the numbers. In problems such as these, depending on the values, the mass on the ramp could move up the plane, down the plane, or stand still. You have to work it out, as you did. In this case, there is just enough static friction force available to keep the system still (in equilibrium), where the static friction force on the mass on the ramp is less than (mu_s)N. Your tension calculation is wrong, however; the tension on both sides of the pulley must be the same, whether the masses are moving or still. Don't forget that there is still some static friction acting on the mass on the ramp.
  6. Oct 28, 2009 #5
    Thanks for your help!

    I talked to the professor. She said that the only time there is zero movement is when the frictional force is equal to (not greater than) the forces creating movement. If there is any difference, there is movement acceleration.

    So, no extra points for me.

    Honestly, I don't buy this explanation because it entirely negates the idea that friction could hold something in place. I was under the impression that friction always against any motion. Not just against the direction of the initial tug.

    For example, by her logic, if I have my initial situation but with Mk=.95 (say it's covered in velcro), the block would fly down the the ramp at 32.23 m/s^2 (I did the math). Obviously, this does not happen.

    Hmmm... maybe I'll try her one more time with this example.
  7. Oct 28, 2009 #6
    You're entirely correct. There's no need to solve rock.freak667's equations if the friction force is too great for movement, nor is it valid because the maximum of the friction force is [itex] -\mu m g cos \theta [/itex], the force can be smaller.
  8. Oct 28, 2009 #7

    My professor is essentially saying that the force of friction between the block and ramp is exceeding the other force on the block. Therefore, the force of friction is causing the block to move. This is impossible; the frictional force cannot do this.

    At least I feel vindicated even if my grade didn't change.

    I think I'll let this rest now... that is unless at the end of the semester, I'm 1% from the next letter grade.

    Thanks for your input.
  9. Mar 30, 2010 #8
    lol i feel for ya shankman...

    you got a pretty stupid prof lol

    sound like my high school physics teacher....
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