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Homework Help: Forces in an Elevator

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data
    In the figure, a 25 kg block is pulled by a 100 N force and slides across the floor of the elevator. The coefficient of friction is µ = 0.35. If the elevator is accelerating downward at 2.0 m/s2, what is the resulting acceleration of the block?

    See Figure 1

    2. Relevant equations
    F = u * Fn
    where F = friction force, u = coefficient of friction force and Fn = normal force
    Fnet = ma
    where m= mass and a = acceleration

    3. The attempt at a solution
    So I decided the three forces acting on the block are Fn (normal force), Fg (force due to gravity), and F (horizontal force). I can find Fn of the block relatively easily, but I'm not sure what to assign the force Fg as. I don't think that it's as simple as Fg = mg because the block is not in free fall. Can I say Fg is 2.0 m/s2 * 25 kg? Then I would theoretically have all of the forces acting on the block and would be able to solve the equation Fnet = ma for a. I find this idea unsettling however, since I think that means the block would be slammed into the ceiling of elevator during this descent, since the downward force Fg is only 50 N and the upward force Fn is about 286 N.

    Also, since the block is moving in two dimensions, this means I should solve for ax and ay separately, correct? So when I give my answer should I give it in Cartesian coordinates like a= 5 xhat + 8 xhat or should I convert it to magnitude angle notation? I didn't know if there was a standard preference.

    Attached Files:

  2. jcsd
  3. Sep 18, 2010 #2


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    yes, good.
    please show your calculation for the normal force
    the block is not in free fall, but the force of gravity is always mg near the surface of the earth.
    Calculate the normal force
    either way is oK. Sometimes a problem asks specifically for an angle and magnitude of acceleration.
  4. Sep 18, 2010 #3
    F = u * Fn
    where F is the force causing the friction and u is the coefficient of friction and Fn is the normal force. So I said Fn = F/u = 600 N/.35 = 286 N.

    Then since Fg = mg, then Fg = 245 N. So the block still has an overall upwards acceleration, unless the acceleration of the elevator affects the normal force and so I calculated the normal force wrong.

    I'm having trouble seeing how the acceleration of the elevator has anything to do with this. Unless I just find the acceleration of the block in the elevator then subtract the elevator's acceleration from that to give the acceleration relative to the ground.
  5. Sep 18, 2010 #4


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    You are not calculating the Normal force corrrectly. In the vertical direction, the acceleration is downwards. Per Newton 2, the net Force in the vetical direction must be in the direction of the acceleration in the vertical direction. Therefore, the weight mg acting down must be greater than the Normal force acting up. Use newton 2 in the vertical directtion to calculate the normal force. Use free body diagrams.
  6. Sep 19, 2010 #5
    I did
    Fnet,y = ma
    FN - Fg = ma
    FN - 245 N = 25 kg * 2.0 m/s2
    FN = 295 N

    Fnet,x = ma
    Fk = Fn * u
    where Fk is the friction force pointing in the opposite direction of the force causing the friction (i.e, the force opposing the movement of the object)
    so Fk = 103.25 N
    Fnet, x = 100 N - 103.25 = -3.25 N

    which I don't think makes sense, because it's implying that the net force in the x direction is opposite that of the 100 N. Does that mean the block is moving to the left or does that mean the force of 100 N is not sufficient to move the block and thus the block has no horizontal acceleration? Then the block's acceleration is just 2 m/s2?
  7. Sep 19, 2010 #6


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    You are on the right track, and your answer would be correct IF the elevator was accelerating UP. But it is given that it is accelerating DOWN. So the block's weight must be greater than the normal force. Redo your calc for the Normal force, and watch your plus and minus signs, and you should be OK.
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