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Truck/Box - Accel, distance, time

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    A box of mass 11.3 kg rests on the flat bed of a truck; the box is not tied down in any way. The coefficients of friction between the box and the flat bed are mu_s = 0.19 and mu_k = 0.14. The truck stops at a stop sign and then starts to move with a carelessly large acceleration of 2.20 m/s^2.

    I found the force of friction first --- 15.5 N

    then i found the accel of the box --- 1.37 m/s^2

    Now i am having trouble finding this next part.

    If the box is a distance 1.83 m from the rear of the truck when the truck starts, how much time elapses before the box falls off the truck?

    2. Relevant equations

    f=ma

    vf2=vi2+2ad

    v=d/t

    3. The attempt at a solution

    So for this last part this is what i did.

    vf22= 0 + 2(1.37)(1.83)

    vf= 2.24

    t = 1.83/2.24

    t= 0.817 seconds

    this was wrong so i thought oh well the truck is accelerating right and the box is accelerating in the opposite direction so i took a new approach.

    vf22= 0 + 2(2.2-1.37)(1.83)

    vf= 1.74

    t = 1.83/1.74

    t= 1.05 seconds

    but to no avail. this was wrong also so i thought again well the truck is accelerating right and the box is accelerating left so maybe i use total acceleration

    vf22= 0 + 2(2.2+1.37)(1.83)

    vf= 3.61

    t = 1.83/3.61

    t= 0.507 seconds

    wrong as well :(

    i dont know what else to try.

    what am i doing wrong?
     
  2. jcsd
  3. Feb 20, 2010 #2

    Draw a picture, with two points, on a cartesian x coordinate system. One point is the box, and the other point is the edge of the truck where the box is going to fall off. Each point is moving with constant acceleration, but different accelerations, so each is represented by some

    [itex]\rm x = x_0+v_{0_x} t + \frac{1}{2}a_x t^2[/itex]

    , an [itex]\rm x_1[/itex] and an [itex]\rm x_2[/itex]. You want the points to meet at some future time t' > 0, so you equate [itex]\rm x_1 (t') = x_2 (t') [/itex], and then solve for t'.

    Hint: one of the [itex]\rm x_0[/itex]'s will be zero, since you can place the origin of your frame on the box, or the back of the truckbed. Plus the [itex]\rm v_0[/itex]'s are all zero since the system starts from rest.
     
    Last edited: Feb 20, 2010
  4. Feb 20, 2010 #3
    If i understand you correctly this is what i came up with. Does it look right?

    1.83 + 0 + 1/2(1.37)t^2 = 0 + 0 + 1/2(2.2)t^2

    1.83 = 1.1t^2 - 0.685t^2

    1.83 = 0.415t^2

    t^2 = 4.410

    t = 2.10 seconds

    Thanks for the help :)
     
  5. Feb 20, 2010 #4
    Yep, that's it! Good job. A lot of the physics work, in this problem, was showing that the acceleration of the truck was greater than the maximum acceleration that the static friction force could provide, which implied that it was now a kinetic friction force, and the box would slip. The last step is really very similar to projectile motion, except in one-dimension. Projectile motion problems consist of one object in 2-d (x_1,y_1) being at another point in space (x_2,y_2), at some future time t'. For this problem we have the box, for example, at x_2 at t = 0, and we want it at x_1, the back of the truck, at t' > 0:

    http://img297.imageshack.us/img297/1973/trucki.jpg [Broken]

    [itex]\rm x_1 = \frac{1}{2} a_{truck} t^2[/itex]

    [itex]\rm x_2 = L + \frac{1}{2} a_{box} t^2[/itex]

    [itex]\rm x_1 (t') = x_2 (t')[/itex]

    [itex]\rm \Rightarrow t' = \sqrt{\frac{2L}{a_{truck}-a_{box}}}[/itex]

    Note: the place where it's very easy to make a mistake is the sign (+/-) of the acceleration components in our equations for x_1 and x_2. Even though the box is sliding to the left, relative to the truck, it's acceleration vector is pointing to the right, in the same direction as the acceleration of the truck.
     
    Last edited by a moderator: May 4, 2017
  6. Feb 20, 2010 #5

    PhanthomJay

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    This is a very good point: Mybro.. you made a lot of guesses on the direction of the accelerations, all of them wrong :frown:; it is sometimes difficult to see that the kinetic friction force in this case actually is providing an acceleration in the same direction of the truck;accelerations using newton's laws are with respect to the ground. You could have solved this problem using relative accelerations: the acc of the block with respect to the truck is 2.20 -1.37= 0.83m/s^2; thus using the displacement with respect to the truck of 1.83m., then 1.83 = 1/2(0.83)t^2, solve for t and get the same result. But often relative accelerations are confusing also, so, do it the way you feel is easier (if any!).
     
  7. Feb 20, 2010 #6
    Thank you both. That all helped greatly and i can really picture it now :)
     
  8. Feb 20, 2010 #7
    Definitely. I've always found that motion diagrams help to get the point across. In this problem, one could say the box is sliding to the left relative to the truck, and we always know that the kinetic frictional force points in the opposite direction of relative motion, but this is confusing me even saying it. :rofl: Drawing a motion diagram, for the box, is often times very helpful:

    http://img136.imageshack.us/img136/2343/truck1s.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Sep 29, 2011 #8
    Everything makes sense but when we look at the free body diagram of the box as seen by an external observer, the accelaration and kinetic friction are in the same direction.
    How is that possible?
    Am I correct to say that the box displaces to left as seen by an obeserver on the truck and is moving to the right as seen by on observer on the ground since a' =0.83 is < a = 2.20 m/s/s??
     
  10. Oct 5, 2011 #9
    Good Morning Samirgaliz,

    The box is accelerating to the left, for an observer at rest on the truck, but this observer's reference frame in non-inertial, i.e. an accelerated frame. The observer, at rest, on the ground is the inertial frame, and he/she sees both the box and the truck accelerating to the right, albeit with different accelerations. Although the problem can be solved in the non-inertial frame, this is usually beyond the scope of introductory Freshman Physics, and it's best to always choose a coordinate system that is inertial.


    [PLAIN]http://img136.imageshack.us/img136/2343/truck1s.jpg [Broken]

    If you look carefully at the motion diagram for the box, from the inertial reference frame, it is accelerating to the right. Newton's 2nd postulate says

    [tex]{\bf F}_{net} = m {\bf a}[/tex]

    so the net force,

    [tex]{\bf F}_{net} = {\bf f}_k[/tex]

    must point in the same direction as the acceleration, i.e to the right, since m > 0.
     
    Last edited by a moderator: May 5, 2017
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