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Find the Force to move 2 objects, and force between the 2

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data
    You are moving furniture in your room as shown in the figure below. What force, F, is needed to make both pieces of furniture (M1=20kg and M2=15kg) move? Assume the coefficient of kinetic friction is 0.40 and the coefficient of static friction is 0.60 between the floor and furniture but there is no friction on the vertical sides where the furniture pieces touch each other. Once the furniture starts moving, if you continue applying the same force, F, what will be the force between the two pieces of furniture? The angle, θ, is 25 degrees.

    Part 1: Find the amount of force to move the furniture at f_staticmax

    Part 2: What will be the force of the two furniture if we continue applying the same force in part 1?

    2. Relevant equations
    ƩFxnet=m*ax
    ƩFynet=m*ay=0 (ay=0)
    f_static=μ_static*N


    3. The attempt at a solution
    FBD: View attachment 58524

    Part 1: Solve for Fx that moves the furniture

    M1+M2:
    ƩFx=m_(1+2)*ax
    m_(1+2)*ax=Fcos(25)-fs_(g on 1+2)
    fs_(g on 1+2)=Fcos(25)-m_(1+2)*ax

    ƩFy=m_(1+2)*ay=0 (ay=0)
    0=N-m(1+2)*g-Fsin(25)
    N=m_(1+2)*g+Fsin(25)

    M1:
    ƩFx=m1*ax
    m1*ax=Fcos(25)-fs(g on 1)-F(2 on 1)
    ax=[Fcos(25)-fs(g on 1)-F(2 on 1)]/m1

    ƩFy=m1*ay=0 (ay=0)
    0=N-m1*g-Fsin(25)
    N=m1*g+Fsin(25)

    M2:
    ƩFx=m2*ax
    m2*ax=Fcos(25)-fs_(g on 2)+F(1 on 2)
    ax=[Fcos(25)-fs(g on 2)+F(1 on 2)]/m2

    ƩFy=m2*ay=0 (ay=0)
    0=N-m2*g-Fsin(25)
    N=m2*g+Fsin(25)

    Solve for ƩFx of m1+m2, using the acceleration of one of the furniture. The ax of m1 is equal to the ax of m2.

    ƩFx=(m1+m2)*ax
    I use ax of m1 and m2 respectively.

    m1:
    ƩFx=(m1+m2)*[Fcos(25)-fs(g on 1)-F(2 on 1)]/m1
    =(m1+m2)/m1 * [Fcos(25)-fs(g on 1)-F(2 on 1)]

    m2:
    ƩFx=(m1+m2)*[Fcos(25)-fs(g on 2)+F(1 on 2)]/m2
    =(m1+m2)/m2 * [Fcos(25)-fs(g on 2) + F(1 on 2)]

    m1 should equal m2, but I don't know how to solve it since I don't have F.
    (m1+m2)/m1 * [Fcos(25)-fs(g on 1)-F(2 on 1)] = (m1+m2)/m2 * [Fcos(25)-fs(g on 2) + F(1 on 2)]

    This is my first time using physics forums, and I'm not sure if I can ask this, but I am lost on how to solve for this problem, and if I am solving for the correct Force. Not sure if I'm allowed to ask these questions though. :(

    All help is much appreciated
     
  2. jcsd
  3. May 6, 2013 #2
    The link to your attachment do not work?
     
  4. May 6, 2013 #3
  5. May 7, 2013 #4
    Static friction

    The static friction force grows to maximum if F is increased steadily.
    Just before the object starts moving it is given by

    f = μS N

    This means that the acceleration will be zero just before the furniture starts to move.
    Only at this point can you evaluate the static friction.
     

    Attached Files:

    Last edited: May 7, 2013
  6. May 8, 2013 #5
    Just before the furniture will start to move we can then say that

    F = f

    or

    F - f = m[itex]\:[/itex]a
    [itex]\ [/itex]= 0
     
  7. May 8, 2013 #6
    This means that in you formula

    M1:
    ƩFx=m1*ax
    m1*ax=Fcos(25)-fs(g on 1)-F(2 on 1)

    you can set ax to zero. Also F(2 on 1) will be zero since the friction balances the horizontal component of the pushing force out. This means that the 1st piece of furniture is not pushing the second one yet.
     
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