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Multi-Mass Force/Acceleration problem

  1. Oct 10, 2013 #1
    1. The problem statement, all variables and given/known data
    tfNDWB3.jpg
    Box 1 of mass m1 is attached by a thin rope to box 2 of mass m2. Box 1 is pulled by force F. What is the acceleration of box 1 given:
    m1= 3kg
    m2= 2kg
    V(initial)= 4m/s
    F= 50N
    Friction is present between boxes and floor
    μ = 0.2
    θ = 20 degrees
    ∅ = 35 degrees
    gravity(g) = 9.8

    And what value would F need to be so the velocity is constant at 4.0 m/s
    3. The attempt at a solution
    I know that Fnet is going to be the x component of the force F minus the friction between box 1 and the ground and the x component of the tension on box 1 by box 2
    Fnet = Fx-(Fμ+Ftx)
    By plugging into equations I found
    Fx = cos∅*F = 40.9576
    Fy = sin∅*F = 28.6788
    Fg = m*g = 29.4
    N(normal) = Fg+ Fy+Fty=50.0788+Fty
    Fμ = N*μ=μ(Fg+Fy+Fty)=0.2(50.0788+Fty)
    Fnet =Fx-(Fμ+Ftx)=40.9576-(0.2(50.0788+Fty)+Ftx)
    I just can't find Ft
    I initially tried that Fty=m2gsinθ and Ftx=m2gcosθ bur this was marked wrong
     
  2. jcsd
  3. Oct 10, 2013 #2
    You know that the velocity is constant, what does that tell you???........it tells that net force is 0.....If you correctly balance the forces then you should be able to get the answer, using the above fact. Balance force on both the object (m1 and m2), making a free body diagram should simplify the things.
     
    Last edited: Oct 10, 2013
  4. Oct 10, 2013 #3
    Yes, I know that much. But that's only for the second half of the problem. I still need the force block 2 pulls on block 1 (Ft, Ftx, Fty) to solve it, and to solve the original problem which is for the acceleration of box 1 given the force of 50 Newtons
     
  5. Oct 10, 2013 #4
    You need to write Newton's second law for each block.
    And then solve the coupled equations. You have two unknowns: acceleration and tension in the rope. You can eliminate the tension between the equations and solve for a.
     
  6. Oct 10, 2013 #5
    just like nasu said, and I previously mentioned.

    You will have four equations and four unknowns:
    Part A - 2 normal forces, tension and of course the required acceleration
    Part B - 2 normal forces, tension and of course the required force F

    It should be fairly easy to solve

    -----------------------------

    don't you think there are four unknowns?
     
  7. Oct 10, 2013 #6
    Yes, you are right if you consider the components as separate equations. Which you should. Sorry for the vague formulation.
     
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