1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravity Pulling Blocks

  1. Feb 6, 2008 #1
    1. The problem statement, all variables and given/known data
    M1 and M2 are two masses connected as shown. The pulley is light (massless) and frictionless. Find the mass M1, given that M2 (3.5 kg) accelerates downwards at 3.35 m/s^2, θ is 35o, and μk is 0.25.

    [​IMG]


    2. Relevant equations



    3. The attempt at a solution
    The total force (which is the total mass times the acceleration) is equal to the sum of the forces.

    Forces:
    Gravity = M1g
    Normal = M1g(sinθ)
    Applied = M1a
    Kinetic Friction = μkM1a

    g=9.81m/s^2

    a(M1 + M2) = -M1g +M1g(sinθ) +M1a -μkM1a
    M1a +M2a = M1g((sinθ)-1) +M1a -μkM1a
    M2a = M1(g((sinθ)-1) -μka)
    M1 = M2a/(g((sinθ)-1) -μka)
    M1 = (3.5*3.35)/(9.81((sin35)-1) -0.25*3.35)
    M1 = 11.725/(9.81(-.426)-0.8375)
    M1 = -2.34 kg

    A negative answer isn't even possible, but I tried putting in "2.34 kg" is an answer and that didn't work. I don't know what I'm doing wrong.
     
  2. jcsd
  3. Feb 6, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    I get M1=2.058kg.
    If that is correct then I got it by simply using the tension as a force in the system, then found the two resultant forces acting. Then solved.
     
  4. Feb 6, 2008 #3
    Yeah it's 2.056 kg. Can you show me the math for that?
     
  5. Feb 6, 2008 #4

    rock.freak667

    User Avatar
    Homework Helper

    The tension(T) in the strings are also forces in the system. (T is up the plane in the string at m1 and vertically opposite to the weight of m2)

    The components of the weight of m1 is m1gsin35(down the plane) and m1gcos35(perpendicular to the plane).

    The motion is downward so that the resultant force up the plane is

    [itex]m_1a=T-m_1gsin35-\mu m_1gcos35...(*)[/itex]

    the resultant motion of m2 is

    [itex]m_2a=m_2g-T...(**)[/itex]

    Now you have the value of a(the downward acceleration),g,m2.
     
  6. Feb 6, 2008 #5
    Thanks a lot! This is a big help. I understand it now.
     
  7. Feb 13, 2008 #6
    this is a silly question but how do you factor out the m1 so that you solve for m1. my problem looks like this

    M1 and M2 are two masses connected as shown. The pulley is light and frictionless. Find the mass M1, given that M2 (7.00 kg) accelerates downwards at 2.19 m/s2, that theta is 20.0°, and that muk is 0.500.

    i have m1*2.19 = 53.27N-m1*3.35m/s/s-m1*4.60m/s/s, but i do not know how to get it to a m1= ____ kg.
     
    Last edited: Feb 13, 2008
  8. Feb 13, 2008 #7

    rock.freak667

    User Avatar
    Homework Helper

    Put everything with m1 on one side,then factor it out.
     
  9. Feb 13, 2008 #8
    i will try that thanks
     
  10. Feb 14, 2008 #9
    thank you soo much it clicked with me you are my favorite people in the world right now!
     
  11. Feb 2, 2010 #10
    I was just wondering if you could help me with the same problem. My prof is useless. He's completely conceptual.
    M1=?
    M2=7.0 kg
    /mu_k=.38
    A=3.05 m/s^2
    Theta=25

    If possible can you show the problem as you worked it? I'm completely lost and need to know how to do this for a test at the end of the week.
     
    Last edited: Feb 2, 2010
  12. Feb 2, 2010 #11

    rock.freak667

    User Avatar
    Homework Helper

    Just follow the way I showed in post #4.
     
  13. Feb 2, 2010 #12
    I have tried, but I'm unsure as to what I am getting from the equations. I don't know what I am supposed to do with the values acquired...
     
  14. Feb 2, 2010 #13

    rock.freak667

    User Avatar
    Homework Helper

    Show the equations you got and we will see from there.
     
  15. Feb 2, 2010 #14
    That is just it, I have no clue where to start.
     
  16. Feb 2, 2010 #15

    rock.freak667

    User Avatar
    Homework Helper

    Start with the free body diagram shown in post #1 and then write the components of the weight of the mass M1. (Were you also given the angle the plane is at?)
     
  17. Feb 2, 2010 #16
    FBD for m1 would consist of the contact force pushing up (parallel to the ramp), the tension force pulling m1 up the ramp, and gravity pulling down. But how can I solve for any of that if I don't have the mass?
     
  18. Feb 2, 2010 #17

    rock.freak667

    User Avatar
    Homework Helper

    I think you mean pulling down. Assuming you have the angle θ, and assuming M2 moves downwards. Consider the two systems separately. Knowing that the sum of the forces in any direction is equal to ma,

    For T and M2, can you write an expression for the resultant force? Do the same for T and M1

    (If you still aren't too sure how to do it, look back at the equations in post #4)
     
  19. Feb 2, 2010 #18
    Post #4 confuses me. My prof is absolutely worthless, and I am unsure of what each equation you've stated, is solving for.
     
  20. Feb 2, 2010 #19

    rock.freak667

    User Avatar
    Homework Helper

    Ok let's just consider everything to the left of the pulley, parallel to the plane. What are the forces acting? (state whether up or down the plane as well)
     
  21. Feb 2, 2010 #20
    The horizontal component of gravity pulling on the block(so down the plane), tension pulling the block up the plane. Then of course the contact force of the plane pushing up on m1.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Gravity Pulling Blocks
  1. Pulling a block (Replies: 4)

  2. Pulling a block (Replies: 6)

  3. Block pulled upward (Replies: 9)

Loading...