Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Weird problem

  1. Dec 17, 2003 #1
    what minimum horizontal force is needed to pull a wheel of radius R and mass M over a step of height H. force is supplied at center of wheel.
  2. jcsd
  3. Dec 17, 2003 #2


    User Avatar
    Science Advisor

    Draw yourself a force diagram. Just to get you started right, make the step height smaller than R, otherwise no force will be sufficient. The verticle component of the force exerted by the step upon the wheel must counteract gravity, so that it lifts off of the ground.

    You might also want to show that the process "runs away". As the wheel climbs over the step, less force is needed. This ensures that you don't have to worry about climbing partly up the step and rolling back down.

  4. Dec 17, 2003 #3
    Given force is horizontal
  5. Dec 17, 2003 #4
    We should look for conservation of Energy
  6. Dec 17, 2003 #5
    torque = F * R, but where do i factor in the H?
  7. Dec 17, 2003 #6
    nope torque is not FR here
  8. Dec 17, 2003 #7
    is torque even used in this problem?
  9. Dec 17, 2003 #8
    [tex]Fsin \theta R=I \alpha[/tex]

    Attached Files:

    Last edited: Dec 17, 2003
  10. Dec 17, 2003 #9


    User Avatar
    Science Advisor

    Yes, but reactive forces are not.

    The applied forces are a horizontal force upon the axis, and gravity. The reactive forces are from the ground, and the step. The wheel will roll over the step when the reactive force from the step has a verticle component equal and opposite to the gravitaional effect upon the wheel. This will mean the ground supplies no force, because the wheel will stop touching the ground.

  11. Dec 17, 2003 #10
    How that is going to help to solve the prob,
  12. Dec 17, 2003 #11
    im stuck at F*sintheta = m * a now.
    how do i factor in the H?
  13. Dec 17, 2003 #12
    [tex]Fsin\theta r = i \frac{d\omega}{dt}[/tex]
    [tex]FRsin\theta d\theta=\omega I d\omega[/tex]

    solving this from fig uget

  14. Dec 17, 2003 #13
    Apply Conservation of Energy

    [tex]F=\frac{mg \sqrt{R^2+(R-H)^2}}{R}[/tex]
  15. Dec 17, 2003 #14


    User Avatar
    Science Advisor
    Homework Helper

    Do the torque calculations about the edge of the step. Torque due to the normal force is zero. That leaves torque due to gravity, and torque due to the applied force which must be larger.
    I get
    (negative because it is clockwise)
    [tex]F > \frac{mg \sqrt{2RH-H^2}}{R-H}}[/tex]
  16. Dec 17, 2003 #15
    Sorry i lost NateTG is correct
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook