1. Limited time only! Sign up for a free 30min personal 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!

A box slides on top of a cart as a force is applied to the cart

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data

    A box of mass = 20 kg rests on top of a cart mass = 30 kg. A force of 200 N is applied to the cart, and the box starts to slide. If the coefficient of kinetic friction is 0.25, what is the time it takes for the box to slide 1.5 m. Both the cart and the box start from rest.

    m1 = 20 kg
    m2 = 30 kg
    F = 200 N
    μk = 0.25
    Δs = 1.5 m
    t = ?

    2. Relevant equations

    Ffriction = (μk)(N)

    Fnet = ma

    aa/b = ab - aa

    Δs (from rest) = (1/2)(a)(t2)

    3. The attempt at a solution

    Here is my free body diagram of the box:

    -----------------------------[itex]\uparrow[/itex] N1
    --------------------|--------------------|
    --------------------|--------------------|
    --------------------|--------------------|
    --------------------|--------------------|
    --------------------|--------------------|
    ---------------------------[itex]\downarrow[/itex]m1g ----- [itex]\rightarrow[/itex]Ffriction (FF)

    y
    |__x

    Here is the kinetic diagram:

    --------------------|--------------------|
    --------------------|--------------------|
    --------------------|--------------------| [itex]\rightarrow[/itex] m1a1
    --------------------|--------------------|
    --------------------|--------------------|


    And the force analysis:

    [itex]\Sigma[/itex] Fy = m1a1y = 0; 0 = N1 - m1g ; N1 = m1g = (20)(9.81) = 196.20 N

    m1g = 196.20 N [itex]\downarrow[/itex]
    N1 = 196.20 N [itex]\uparrow[/itex]

    FF = (N1)(μk) = (196.2)(0.25) = 49.05 N [itex]\rightarrow[/itex]

    [itex]\Sigma[/itex] Fx = m1a1 = 20a1; FF= 20a1; 49.05 = 20a1

    a1 = (49.05)/(20) = 2.4525 m/s2 [itex]\rightarrow[/itex]


    Here is my FBD of the cart:

    -------------------- [itex]\uparrow[/itex] m1g --- [itex]\uparrow[/itex] N2 --- [itex]\leftarrow[/itex]FF
    --------------------|--------------------|
    --------------------|--------------------|
    --------------------|--------------------| [itex]\rightarrow[/itex] F (200 N)
    --------------------|--------------------|
    --------------------|--------------------|
    ---------------------------- [itex]\downarrow[/itex]m2g

    y
    |__x

    Here is the kinetic diagram:

    --------------------|--------------------|
    --------------------|--------------------|
    --------------------|--------------------| [itex]\rightarrow[/itex] ma2
    --------------------|--------------------|
    --------------------|--------------------|


    Here is the force analysis:

    ƩFx = ma2; 200 - FF = ma2

    200 - 49.05 = 150.95 = ma2

    And here, is the problem I am having troubles understanding:

    I am not sure what mass to insert here; 30 kg (the mass of the cart) or 50 kg (the mass of the cart and the box)

    The solutions book puts in 30 kg, and doesn't say why it neglects the mass of the box, which would make the total mass 50 kg. Is the solutions book correct or does it make an error here? If it is correct, can somebody explain what exactly is happening and why you can do this?

    putting in the 30 kg (the mass of the cart) gives an acceleration of 5.03166.
    subtracting 2.4525 (the acceleration of the box) gives a relative acceleration of 2.579166
    Using the kinetic formula the time for the box to move 1.5 meters gives a solution for time of about 1.08 seconds.

    on the other hand, putting in 50 kg (the mass of the cart and the box) gives an acceleration of 4 and subtracting 2.4525 gives a relative acceleration of 1.5475. the kinetic formula then spits out an answer of t = 1.39 seconds.
     
    Last edited: Feb 18, 2012
  2. jcsd
  3. Feb 18, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    When you drew the free body diagram you considered both object separately. The resultant of all forces acting on one object is the acceleration of that object multiplied by the mass of the same object.



    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A box slides on top of a cart as a force is applied to the cart
  1. Force on carts (Replies: 1)

  2. Box on Cart (Friction) (Replies: 5)

Loading...