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: Moving a box with a force that is less than gravity

  1. Oct 3, 2012 #1
    1. The problem statement, all variables and given/known data
    The object is a box with a given mass m. Our person has the choice between pushing the box with a horizontal force, or pulling the box with a wire with an angle of θ=30o. The magnitude of Fo is the force vector he affects the box with in both cases.

    Is it possible (If you can freely choose a value for θ) to keep the box moving without using a force fo that is greater than the gravity on the box

    2. Relevant equations
    A drawing of the situation:

    I believe I need an equation that describes the force Fo of θ, which I've found is given as:
    [itex]F_o(θ)=\frac{\mu_k mg}{cos(θ)-μ_k sin(θ)}[/itex]

    3. The attempt at a solution

    I could choose to solve F'o(θ)=0 for theta which gives θ=arctan(μk)

    This would describe the minimum force required but I don't know how to relate that to Fg
  2. jcsd
  3. Oct 3, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    What is the relationship between friction and the weight of the box?

    But what would the turning point tell you?
    Don't you want to know where F0 < Fg
  4. Oct 4, 2012 #3
    So if I use that Fnet(x)=Fo*cos(θ)-μn

    And that Fnet(x)<0, then the value of θ<arctan(μ), and therefor I'm applying a force Fo that is less than gravity? Wouldn't that technically mean the box is not moving and is being held back by the friction?
  5. Oct 5, 2012 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Why would that be? Gravity is mg.... and force less than mg would be less than gravity. To overcome friction it just has to overcome μmg.cos(θ) ...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook