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Homework Help: Basic Friction Problem

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    A block of mass m1 = 21.5 kg is at rest on a plane inclined at θ = 30.0° above the horizontal. The block is connected via a rope and massless pulley system to another block of mass m2 = 17.7 kg, as shown in the figure. The coefficients of static and kinetic friction between block 1 and the inclined plane are μs = 0.109 and μk = 0.086, respectively. If the blocks are released from rest, what is the displacement of block 2 in the vertical direction after 1.47 s? Use positive numbers for the upward direction and negative numbers for the downward direction.


    2. Relevant equations

    D = v_i*t + (1/2)a*t^2
    F = ma

    3. The attempt at a solution

    For mass 2, Positive is going upwards. For mass 1, Positive is going down the ramp.
    (sum) 1. F_m1_x = [(m_1)g*sin(30)] - [(m_1)g*cos(30)] - T = (m_1)a
    (sum) 2. F_m2_y = T - (m_2*g) = (m_2)a --> T = (m_2*a) + (m_2*g)

    Plugging equation 2 into 1 for T, we get:

    [(m_1)*g*sin(30)] - [(m_1)g*cos(30)] - [(m_2*a) + (m_2*g)] = m_1*a --->
    [(m_1)*g*sin(30)] - [(m_1)g*cos(30)] - (m_2*g) = m_1*a - [(m_2*a) --->
    g(m_1*sin(30) - m_1*cos(30) - m_2) = a(m_1 - m_2)

    a = [g(m_1*sin(30) - m_1*cos(30) - m_2)] / (m_1 - m_2)

    plugging in the values, we get a = -66.01 m/s^2

    cool, now we can find how far it travels in 1.47 s.

    D = v_i*t + (1/2)a*t^2
    d = 0 + (1/2)(-66.01 m/s^2)*(2.1609)
    d = -71.32m

    That's not right. Unsure what I'm doing wrong :/
    I also tried the block going up the ramp, but that didn't work either.

    This is my first post here, any help is appreciated, thanks!
    Last edited: Sep 17, 2010
  2. jcsd
  3. Sep 17, 2010 #2
    I'm getting a lot of views but no replies. Does that mean you guys agree with my solution? Or are you just not sure?
  4. Sep 17, 2010 #3


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    You did not attach any figure.

  5. Sep 17, 2010 #4
    Sorry, figure now attached
  6. Sep 17, 2010 #5


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    You forgot the coefficient of friction. And be careful, first check if the static friction is not too big to block motion. Decide if block 1 moves downhill or uphill. The force of friction acts opposite to the velocity.

  7. Oct 8, 2010 #6
    You're exactly right. I did forget the coefficient of friction. so plugging that in we get:

    (sum) 1. F_m1_x = [(m_1)g*sin(30)] - [uk(m_1)g*cos(30)] - T = (m_1)a
    (sum) 2. F_m2_y = T - (m_2*g) = (m_2)a --> T = (m_2*a) + (m_2*g)

    Plugging equation 2 into 1 for T, we get:

    [(m_1)*g*sin(30)] - [uk(m_1)g*cos(30)] - [(m_2*a) + (m_2*g)] = m_1*a --->
    [(m_1)*g*sin(30)] - [uk((m_1)g*cos(30)] - (m_2*g) = m_1*a - [(m_2*a) --->
    g(m_1*sin(30) - uk(m_1*cos(30) - m_2) = a(m_1 - m_2)

    a = [g(m_1*sin(30) - ukm_1*cos(30) - m_2)] / (m_1 - m_2)

    I am not getting a to be 22.67 m/s2
    Because the acceleration is positive, this indeed means the mass 1 is sliding down the ramp, while mass 2 is being pulled up?

    D = v_i*t + (1/2)a*t^2
    d = 0 + (1/2)(22.67 m/s^2)*(2.1609)
    d = 24.49 m.

    This is still not right. (for reference, the correct answer is (-1.45m)
  8. Oct 8, 2010 #7


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    You used the wrong sign when putting (m_2*a) to the right hand side of the equation. So the acceleration would be
    a = [g(m_1*sin(30) - ukm_1*cos(30) - m_2)] / (m_1 + m_2)

    As this is negative, your initial assumption about the motion of the system was wrong (you assumed that m1 moved down the slope, in the same direction the component of gravity pointed at.) So m1 will move up on the slope and m2 will move down. Rewrite your equations.

  9. Oct 8, 2010 #8
    I'll rewrite the equations, but do I have to? As long as I know it moves in the negative direction (up the slope), won't that still give me the right answer?

    New equations:
    Fm1x = T- [µk(m1)g*cos(30)] - [(m1)g*sin(30)] = (m1)a
    Fm2y = m2g - T = m2a --> T = (m2g) - (m2a)

    (m2g) - (m2a)- [µk(m1)g*cos(30)] - [(m1)g*sin(30)] = (m1)a
    (m2g) - [µk(m1)g*cos(30)] - [(m1)g*sin(30)] = (m1)a + (m2a)
    g(m2-[µk(m1) cos(30)] - [(m1)sin(30)] = a(m1+ m2)
    a= g(m2-[µk(m1) cos(30)] - [(m1)sin(30)]/ (m1+ m2)

    a = 1.3386 m/s^2

    d = vi*t + 1/2at2
    d = 0 + 1/2(1.3386)(2.1609)
    d = 1.446m (mass 1 in the negative direction)

    ahhh, ehild, you are a lifesaver.
    I did this problem, only because my teacher disagreed with this solution. She tried to explain to me that even though the two masses are connected by a pulley, the two masses can't have their own axis. In this example, I have two seperate axis. For mass 1, it's up the ramp in the x direction. For mass 2, it's downward, in the y direction. She explained that you can't do that, and threw lots of extra sins and cosines to the block on the ramp trying to put it in the direction of the upward-downward axis. Regardless of if it works or not, this is much, much easier. I'm glad to see the pulley can, in fact, change the axis.
  10. Oct 8, 2010 #9


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    This is a one-dimensional problem basically. The motions happen along the rope, determined by the constrains (the surface of the slope and the constant length of the rope). The problem could be solved by splitting the motion of m1 into horizontal and vertical components, but the resultant should be along the slope and you would get the same equation as you used. It is not correct to
    assign x to the forces acting on the slope, but it did not matter really.
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