1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mininum Work to Pull Crate

  1. Jan 7, 2008 #1
    [SOLVED] Mininum Work to Pull Crate

    1. The problem statement, all variables and given/known data

    A student could either pull or push, at an angle of 30 degrees from the horizontal, a 50 kilo crate on a horizontal surface, where the coefficient of kinetic friction between the crate and the surface is 0.2. The crate is to be moved a horizontal distance of 15m. (a) Compared with pushing, pulling requires the student to do (1) less, (2) the same, or (3) more work. (b) Calculate the minimum work required for both pulling and pushing.

    2. Relevant equations

    3. The attempt at a solution

    This is a problem that has been posted before but remained unsolved. I just need assistance with part b.

    For part a, I know that pulling requires less work because pushing creates a greater normal force and therefore, friction. Onto part B:


    uk = coefficient of kinetic friction, .20
    m = mass of crate, 50kg
    g = acceleration constant, 9.8m/s^2
    fk = force of kinetic friction
    N = normal force exerted by the crate
    F = unknown minimum force

    Calculate the minimum work required for pulling:

    1. The minimum work occurs when the horizontal force equals the force of kinetic friction. The force of kinetic friction is:

    fk = uk*N = uk*(mg-F*sin(30))

    Setting fk equal to the horizontal component of the applied force:

    uk*(mg - F*sin(30) ) = F*cos(30)
    ==> F = 101 N

    The minimum work in this case should be 101 N*15 m = 1521 J. But it turns out that this wasn't correct. What am I doing wrong?
    Last edited: Jan 7, 2008
  2. jcsd
  3. Jan 7, 2008 #2
    I had a helluva time checking the math on windows calculator(this can't count as slacking as work, I'm sharpening my mind!)but I think you need to check your units
  4. Jan 7, 2008 #3
    I'm stumped. I must be blind or something because I don't see where the units are inconsistent.
  5. Jan 7, 2008 #4
    That's my mistake, kilograms are the SI unit, durr etc.

    I think the mistake is just as unfortunate though. How precise do you need to be? It's like 101.44 or so Newtons. If you're sposed to round to the nearest Newton, then you have a problem because 101*15=/=1521

    If you're sposed to round to the nearest joule you have a problem 'cuz 101.44*15=1521.6ish. Make sure you kept track of significant figures and all that otherwise, and I'll assume you typed it inconsistently for brevity
  6. Jan 7, 2008 #5
    The solution in my textbook is ~1200 N, so that has me second-doubting myself.
  7. Jan 7, 2008 #6
    I'm hoping that you didn't typo and the solution really says 1200 N, since that's obviously the wrong unit for work and either the answer is muffed or you're looking at the wrong solution

    If it's 1200 joules, I dunno and patiently await a wiser man myself. Surely it's something stupid
  8. Jan 7, 2008 #7

    Well one thing, when you found F that's still the applied force's net magnitude. F*cos(30)=horizontal component that's actually responsible for work, using the result that you find from that(87ish N) you get a work of....1300ish joules.

    THEN I believe you need to find the actual force of friction(just use your formula with your 101 value for F)and subtract that off of the 1300ish to get 1200ish

    That may be a case of me working it 'till I got the expected answer, but I believe that does it. You solved for F correctly but misused it

    Final edit: Of course with friction you'd expect to do MORE work to move it, so I have no idea why I'm subtracting the work done by friction
    Last edited: Jan 7, 2008
  9. Jan 7, 2008 #8
    Following your directions:

    uk*(mg-101 N*sin(30))*d = 1318 J

    This is the same number as 101 N*cos(30)*d so the two cancel out don't they?

    Edit: I think you're absolutely right about taking horizontal component of the force I got since the problem is asking for the minimum work. Ie. it's assumed we are pulling directly horizontally, not a 30 degree angle. I can't check at the moment but now I believe 1318 J to be correct.
    Last edited: Jan 7, 2008
  10. Jan 7, 2008 #9
    I successfully failed to multiply by d when I was fiddling with it. As I pointed out it's wrong for obvious reasons anyways.

    But yah, obviously the work by friction will equal the work done pushing because we set the forces equal to each other earlier when we solved for F. My money's on 1312 joules, and your book gets my middle finger. At least until someone explains why we fail
  11. Jan 7, 2008 #10
    The more I reason it, the more I am convinced it's correct. The initial work that I found pertains to the minimum work required when the crate is pulled at a 30 degree angle. Of course, the absolute minimum work would involve applying a horizontal force. Hence it's F*cos(30).

    Thanks blochwave for pointing that out!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Mininum Work to Pull Crate
  1. Crate being pulled (Replies: 3)