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Box in a Car

  1. May 14, 2008 #1
    Hi, i'm new here. I have a problem.

    1. The problem statement, all variables and given/known data
    A box has the mass m, held by two horizontal ropes and two vertical ropes. The box is inside a still car. If the car accelerated suddenly with the accekeration of a then the box will remain still inside the car. What is :
    a. The acceleration of the box/car? (state it in t1, t2, t3, t4 and g)
    b. The separation after the time t? (state it in t1, t2, t3, t4, t and g)

    The illustration is here -> xttp://i31.tinypic.com/izvnev.jpg (change x to h)

    2. Relevant equations
    [tex]\Sigma F = ma[/tex]

    3. The attempt at a solution

    t3 + t1 = ma
    m = (t3 + t1)/a (1)

    t4 + t2 = mg
    m = (t4 + t2)/g (2)

    (t3 + t1)/a = (t4 + t2)/g
    a = g(t3 + t1)/ t4 + t2 (3)

    S = Vt
    S = a * t * t
    S = (g(t3 + t1)/ t4 + t2) * t^2
    S = g (t^2) ((t3 + t1)/(t4 + t2))

    I actually don't really know how the tensions work. I hope someone can explain it to me. And sorry, i'm not used to using latex yet.
    Last edited: May 14, 2008
  2. jcsd
  3. May 14, 2008 #2


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    Hi, Vermillion, and welcome to the Forums! A few points here to help you along in part (a), if I understand the problem correctly, and assume horizontal motion and massless, inextensible ropes:: First, Tension forces always pull away from the object they act on, so in your equations, check your plus and minus signs. Secondly, you have equated your 2 equations when solving each for m, which is unnecessary. The acceleration in the horizontal and vertical directions are independent of each other. Thirdly, with the box accelerating to the right (with respect to the ground), the box would appear to fare quite well without the left and bottom ropes present.

    For part (b), what separation is the problem talking about? Since the box remains still inside the car, there can be no separation between the box and car, so I guess the problem means to ask about the displacement of the box with respect to the ground, in which case you need to use your basic motion eqations.
  4. May 15, 2008 #3
    Oh wow thanks! I actually never knew that tension forces always pull away from the object they act on, duh.... As for the 2 equations of m, i needed them to solve the a(acceleration) in t1, t2, t3 ,t4, and g. My third equation is actually m = m. Sorry, i didn't type it.

    I think i have done the correct equations now.
    ma + t1 = t3
    ma = t3 - t1
    m = (t3 - t1)/a (1)

    mg + t2 = t4
    mg = t4 - t2
    m = (t4 - t2)/g (2)

    m = m
    (t3 - t1)/a = (t4 - t2)/g
    a = g(t3 - t1)/(t4- t2 (3)

    So the separation would be

    s = vt
    s = a*t^2
    s = (t^2*g(t3-t1)) / t4-t2

    Is that correct?
  5. May 15, 2008 #4


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    Part (a) is correct, since the problem asked you to solve for the acceleration in terms of those variables. Part (b), however, is not correct. In the equation s=vt, the 'v' refers to the average velocity (the average of the initial velocity and final velocity at time 't'), whereas in the equation V=at, the 'V' refers to the final velocity at the end of the time period of an object starting from rest. Average velocity and final velocity are not the same. How are they related?
  6. May 16, 2008 #5
    You're right, i forgot the velocity is not constant. Then i should use

    [tex]s = v0t + \frac{at^2}{2}[/tex]

    Because the car was resting then

    [tex]v0 = 0[/tex]

    [tex]s = \frac{at^2}{2}[/tex]

    [tex]s = \frac{gt^2(t3-t1)}{2(t4-t2)}[/tex]

    Is that correct?
    Last edited: May 16, 2008
  7. May 16, 2008 #6


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    Yes, where 's' represents the distance the box (and car) have moved together from their original starting position with respect to the ground. I'm not sure what the problem means by 'separation', we're assuming it means 'displacement'.
  8. May 16, 2008 #7
    Yeah i think that's what the question meant. Thanks for the help! :)
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