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Homework Help: Shopping Cart Problem

  1. May 29, 2008 #1
    1. The problem statement, all variables and given/known data
    A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of 22.0 m. She pushes in a direction 29° below the horizontal. A 48.0 N frictional force opposes the motion of the cart. What is the magnitude of the force that the shopper exerts?

    2. Relevant equations
    F = mg
    W = (F cosθ)s

    3. The attempt at a solution
    1. F = (16kg)(9.8m/s2) = 156.8 N
    2. FN= 156.8 N - 48.0 N = 108.8 N

    Correct answer: 54.9 N

    I know if you divide 108.8 by two you get 54.4 (which is really close to the answer), but I don't think you're supposed to divide by 2. Can someone help me out? :s
  2. jcsd
  3. May 29, 2008 #2
    You're told that the cart moves at a constant velocity, i.e. there's no NET force acting on the cart. In other words, the sum of the forces in the horizontal direction is zero. I would suggest drawing a free body diagram if you haven't already. The mass of the cart and the distance it moves is irrelevant here. Use Newton's Second Law.
  4. May 29, 2008 #3
    Cursed, in your solution you subracted a vertical force, the weight of the mass, by the friction, a horizontal. This is invalid. The weight is in the vertical direction, and the force of friction always points in the direction opposite of the motion, which is in this case horizontal. To subract or add forces, they must have the same direction.
  5. May 29, 2008 #4
    Thanks, Markus.

    I'm still not getting the answer, though. (I used F = ma, and I had the force be 156.8 N.)

    1. tan(29°) = 156.8 N/x; x = 282.8746 N
    2. cos(29°) = z/48; z = 41.9817 N
    3. 282.8746 - 41.9817 = 240.8929

    What am I doing wrong? :/
  6. May 29, 2008 #5
    How would you set up Newton's Second Law for the horizontal direction? You'll end up with a very simple equation.
  7. May 29, 2008 #6

    The weight of the mass is completely irrelevant for this question. The mass is on some horizontal surface, and by Newton's Third Law, it is canceled out by a normal force.

    The unknown here is what force the person exerts. We'll call this Fext. Only the horizontal component of this force contributes to motion. The vertical does not. Try pushing down on a brick and see if it moves! So using trig, the horizontal component of the force exerted is Fext x cos29. (DO NOT take my word for it, come to this conclusion yourself!)

    Since it is stated in the problem that the velocity is constant, then there is no acceleration. If there is no acceleration there is no net force on the mass. So, we can equation something to zero here... After you've figured out what, the answer is easily obtained.
  8. May 29, 2008 #7
    No matter what I do, I can't get the answer:

    1. cos29 * F = (F * cos29)s
    2. (cos29 * ma)s = mg
    3. (cos29 * F)s = 0
    4. (ma * cos29)s = 0.5mv2

    What am I supposed to set equal to zero?
  9. May 29, 2008 #8
    You need to read the above posts more carefully. You've been told that the mass of the object and distance it travels are of no consequence, so don't use mass or distance.. at all. The "key" to solving this is Newton's Second Law. Look at the free body diagram below:


    What two forces are acting in the x-direction? More specifically, what component of the pushing force are we concerned with? And what force opposes it such that the cart moves at a constant velocity? Call the opposing force Force2, and the component it works against Force1.

    Once you know this, you can use Newton's Second law [tex] \Sigma F_x = ma_x [/tex] to sum the forces in the x-direction:

    [tex] \Sigma F_x = Force1 + Force2 = ma_x [/tex]

    But what should this equation equal seeing as the velocity is constant?
    Last edited: May 29, 2008
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