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Newton's Law/Mechanics Problems - Answers Found, Looking for Confirmation

  1. Oct 17, 2006 #1
    Hey, I'm working on some problems for physics, I've gotten answers for the first few, but I am not sure if my answers are right. It's important I get these problems right, so I'm hoping someone here can help me get through them. Thank you in advance to anyone who takes the time to read through this.

    Here's the first problem and my work:

    1. "A 10-kg crate is pulled with a force Fa at an angle x to accelerate the create at 0.9 m/sec^2. The coefficient of friction between the floor and the crate is 0.45. At what angle should the create be pulled so that the applied force is a minimum?"

    The forces I have acting on the crate in the diagram I drew - normal force going straight up, gravity going straight down, friction going to the left, Fa going up and to the right at angle x (from the horizontal) with the acceleration going towards the right as well. I then split Fa into its x and y components.

    Here's my work and my answer:

    Fa(cosx) - Ff = ma -> Fa(cosx) - (mu)(Fn) = ma
    and
    Fa(sinx) + Fn - Fg = 0 -> Fn = mg - Fa(sinx)

    Facosx - (mu)(mg - Fasinx) = ma
    Facosx - (0.45)(10*9.8 - Fasinx) = 10*0.9
    Facosx + 0.45Fasinx = 53.1
    Fa (cosx + 0.45sinx) = 53.1
    Fa = 53.1/ (cosx + 0.45sinx)

    Then since you're looking for the minimum value of Fa, I found the maximum value of (cosx + 0.45sinx), which I found to be 24.23 degrees. I wasn't sure how to calculate that value, so I took the lazy way out and graphed it in my calculator. If there's a more proper way of finding that answer (if it's correct), please let me know.

    Here's the first part of the second problem, which I got an answer for, and then the second part, which I don't really understand completely (there's more parts beyond that, but I should of course nail the first two before moving on).

    "A 300-kg box rests on a platform attached to a forklift, as shown above. Starting from rest at time t = 0, the box is lowered with a downward acceleration of 1.5 m/s^2."

    The "as shown above" refers to a simple diagram, pretty much the same thing as this only flipped - http://www.pbeinc.com/forklift.gif only the forklift is partially raised (although it labels the box as being 300 kg, that's also given in the problem).

    Anyway, the first part of the question is as such:

    "(a) Determine the upward force exerted by the horizontal platform on the box as it is lowered."

    And here is my work and answer:

    Forces acting on the box - Fg going down (with acceleration going down as well) with Fn going up.

    mg - Fn = ma
    (300*9.8) - (300*1.5) = Fn
    Fn = 2490 N

    That seems a bit too simple though, so I'm not sure if it's right.

    Whether it's correct or not, I'm not sure how to approach the 2nd part of the problem.

    "At time t = 0, the forklift also begins to move forward with an acceleration of 2 m/s^2 while lowering the box as described above. The box does not slip or tip over."

    "(b) Determine the frictional force on the box."

    I'm not sure where to go with this because I'm not entirely sure what forces are being applied to the box. We should still have Fg going down and Fn going up from the last problem, with the y component of acceleration going down. Then there's friction going to the left, with the box accelerating forward to the right at 2 m/s^2. However, I'm not sure if there is a force acting to the left and what it would be...is it just a "Fa" like in the first problem?
     
  2. jcsd
  3. Oct 17, 2006 #2

    Fermat

    User Avatar
    Homework Helper

    Your working out is fine and your answer is correct.
    To find a minimum value, use differential calculus - for finding maximum and minimum points.
    Differentiate Fa wrt x and set to zero. You will end up with -sinx + 0.45cosx = 0 giving tanx = 0.45 giving x = 24.23 deg.


    Yes. that's right.

    It is similar to the first problem, in the sense that the normal reaction is reduced by an external force.
    In the first case it's from the vertical component of the pulling rope.
    In the 2nd case, the normal reaction is reduced by the falling effect. The faster it falls, the less support it receives from the fork lift. If the box were falling freely, under gravity, then there would be zero support from the fork lift.
    In your 2nd problem, the only forces acting are a reduced reaction upwards (lke you guessed), a friction force to the left (the only force to the left) and an accelerating force to the right.
     
  4. Oct 17, 2006 #3
    Ah, thanks a lot Fermat, that's a big help.
     
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