1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Find acceleration? Are two solution possible?

  1. Jul 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Find acceleration-x ̈ of the load. Force F=10G, friction μ=1/3, mass of he body is G.

    I tried two ways of solving this problem but not sure if any is good. I think second is good. But wondering if first is also good?

    2. Relevant equations

    Ek=1/2 G/g x ̇^2

    Fμ=G μ

    3. The attempt at a solution

    Please look at the picture.There you can see my effort in solving this problem. Thanks for any help.

  2. jcsd
  3. Jul 5, 2011 #2


    User Avatar
    Science Advisor

    I don't recognize this formula. Is "g" the acceleration due to gravity? This problem does not involve falling under gravity. You have an object moving across a surface with friction, under a force. And, although you don't say it, I suspect [itex]\mu[/itex] is the coefficient of friction, not the friction force itself.

  4. Jul 5, 2011 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The way you're using it, G is the object's weight, not its mass.

    Both approaches will work.

    In the first approach, the initial equation, though correct, seems to come out of nowhere. It would be clearer if you used A.L. Bruce's suggestion and started with Newton's second law.

    In the second approach, you didn't calculate the work (A) correctly.
  5. Jul 5, 2011 #4
    Yes μ is the coefficient of friction. "g" is the acceleration due to gravity. Ek-would be kinetics energy of the system. G/g*X-Inertial force of the body.Fμ-friction force.G is the object's weight. Is any of the solution correct? If not how it should be done?

    Why work is not correctly calculated?

    Work shouldn't have coefficient of friction wright?

    I was thinking that pulling force is acting on one side and friction force+ inertial force on the other. That is how I come up with initial equation of the first approach. I don't see how to get same equation with Newton's second law?

    Thanks for all for your fast response
    Last edited: Jul 5, 2011
  6. Jul 5, 2011 #5
    Work would be dA=Fdx-Gμdx

    dA=29/3 Gdx

    from which acceleration is x=29/3 g

    I still don't see how to get G μ+G/g x ̈ =F from Newton's second law?
    Last edited: Jul 6, 2011
  7. Jul 5, 2011 #6


    User Avatar

    Staff: Mentor

    Fnet = M*A.

    M is the mass of the object. What is M in this case? What do you get for Fnet?
  8. Jul 6, 2011 #7
    Mass of the object G/g. For Fnet=F-Gμ. Is this correct ?

    G1 is weight on first load and 1 is just a subscript. That also stand for G2,G3. (Can't type subscript don't know why)

    Can the same principle be used for multiple loads. If there were 3 loads tied one to another with rope(light unstretchable) G1,G2,G3 than would it be correct to write (G1+G2+G3)μ+(G1+G2+G3)/g *x ̈ =F ?

    S- force in rope
    S1-force in first rope etc. Again 1 is just a subscript That also stand for S2,S3.

    And the force in the ropes S3 (G3 is the last in line)

    S3=G3 μ+G3/g x ̈

    S2=S3+G2 μ+G2/g x ̈

    S1=F-S2+G3 μ+G3/g x ̈

    Is this correct? If not any suggestion?
    Last edited: Jul 6, 2011
  9. Jul 6, 2011 #8


    User Avatar

    Staff: Mentor

    Yes. So you can plug these values into the Fnet = M*A expression and solve for A.
    Use the X2 and X2 buttons on the reply panel header.
    You have to make the assumption that the rope stays taught, of course. But sure, it seems okay to do that -- all the masses will behave as one larger mass as far as the pulling force is concerned.
    It looks okay.
  10. Jul 6, 2011 #9
    Thanks everybody for helping me, for taking time and effort.

    Best regard to all
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook