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Homework Help: Friction and tension problem

  1. Oct 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the the acceleration, a[/i, and the Tension, T, in the system shown below if (a) m1 = 5 kg and (b) m1 = 2 kg. assume the coefficient of kinetic fricton on the incline plane is µk = 0.1.

    Other relevant information:
    Mass 1 is a 5 or 2 kg block on a 30 degree incline plane (based so that Fy is Fsin(30) ) connected to a massless rope that holds Mass 2, a 5 kg block, over a pulley.

    2. Relevant equations
    Ff = (µk)(Fn)
    w = mg (acceleration due to gravity is assumed to be 10 m/s^2 for easy calculation according to the professor)
    Fnet = ma

    3. The attempt at a solution

    I solved for the weight of mass 1 as being:

    m1 = 5 kg: W = 50 N * sin(30) = 25 N
    m2 = 2 kg: W = 20 N * sin(30) = 10 N

    And found force of friction:
    Ff = 0.1 * 25 N = 2.5 N

    The weight of block m2:
    w = 10 m/s^2 * 5 kg = 50 N

    I tried creating a FBD for block m1, where the following forces were applied:
    W = 25 N
    Fn = 25 N
    Ff = 2.5 N
    F = Wm2 (Wm2 = 50 N)

    And a FBD for block m2, where:
    W = 50 N

    Except, I'm confused where to go from here now. I know I have to calculate tension still. I'm thinking that the total force being applied on m1 according to Newton's 2nd Law is:
    Fnet = ma - Ff = 50 N - 2.5 N = 47.5 N

    And acceleration is: a = 47.5 N / 5 kg = 9.5 m/s^2

    2nd attempt:
    I know Fnet = ma - T. and Fnet is not 0 N, it's 50 N. Since the rope is massless the tension is 2T = 100 N.

    I therefore arrive at the following answers of:
    a = 9.5 m/s^2
    T = 100 N

    Does this look right?
    Last edited: Oct 10, 2008
  2. jcsd
  3. Oct 10, 2008 #2


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    Welcome to PF, Sagekilla.

    Do you mean the block is 30 degrees from the vertical direction?

    I see some issues here, mainly with terminology.

    How about calling these mA and mB instead, since you say earlier that Mass 2 is fixed at 5 kg. It would make sense then to let m2 be the mass of Mass 2.

    Also, the weight is simply m times g. It looks like you're calculating the weight force's y-component (because of the trig terms). Is this supposed to be the normal force--I see you use these numbers for calculating friction force next?

    True if 25 N is the normal force, which is true if 30 degrees is with respect to vertical.

    Okay, so now m2 is for Mass 2?


    W = m1 g = ____?
    Fn -- looks okay.
    Ff -- looks okay.
    F (rope tension?) is not necessarily the weight of Mass 2, since there is an acceleration.

    The rope tension should appear here too.

    Not quite.
    Fnet = ma, period

    I'm gonna stop here for now, and give you a chance to work through these comments before going further.

  4. Oct 10, 2008 #3
    Thanks for your help :)

    Here's the exact diagram if it'll help you any

    http://img100.imageshack.us/img100/3443/diagramzv7.png [Broken]

    I'll read over your comments and rework the problem once more and reply once I've finished.

    I did the problem once more, this time only trying to solve for (a) to make things less confusing:

    I found the Force in the x direction for mass 1:
    F = 5 kg * 10 m/s^2 * cos(30) = 43.3 N

    I solved for weight of mass m1:
    w1 = 5 kg * 10 m/s^2 * sin(30) = 25 N

    The normal force is the same magnitude as the weight for mass m1:
    Fn = w1 = 25 N

    And solved for the Frictional force, Ff, with coefficient of friction 0.1:
    Ff = 0.1 * 25 N = 2.5 N

    I also solved for weight of mass m2:
    w2 = 5 kg * 10 m/s^2 = 50 N

    The FBD for m1 was this:
    w = -25 N
    Fn = 25 N
    Ff = -2.5 N
    F = -43.3 N
    T = ?

    The FBD for m2 looks like this:
    w = -50 N
    T = ?
    Last edited by a moderator: May 3, 2017
  5. Oct 11, 2008 #4


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    Thanks, yes, that is helpful.

    Not sure what you mean by "the Force in the x direction". There are several forces on m1 that act in the x-direction. Do you mean the sum of all the forces along x? Or something else?

    Also, can you confirm that the x direction is the same as the vector a that is drawn in the figure, i.e. along the sloped surface?

    The weight of anything is it's mass, multiplied by g=10m/s2. So the weight of mass m1 is m1*g. No trig is involved with finding the weight.

    Uh, the normal force is different than the weight because of the slope. You will need trig for that; it equals the weight only when the angle is zero degrees.
    Last edited by a moderator: May 3, 2017
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