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!

Work & Inclided Plane

  1. Jan 13, 2007 #1
    1. The problem statement, all variables and given/known data
    What is the minimum work needed to push a car of mass [tex]m[/tex] kilograms, an absolute distance of [tex]d[/tex] meters (or a height of [tex]h[/tex] meters) up a [tex]\theta[/tex] degree incline plane. a) Ignore Friction. b) Assume that the effective coefficient of friction is [tex]\mu[/tex]

    2. Relevant equations
    [tex]W = F_{\parallel}d[/tex]
    [tex]F_{fr} = \mu F_N[/tex]

    3. The attempt at a solution
    For part a), if you want to push the car up the inclined plane, then you would need to apply a vertical force with a magnitude equal to that of gravity, right? So you would want

    [tex]F_y = F_g[/tex]

    [tex]\Rightarrow F\sin(\theta) = F_g \Rightarrow F = \frac{F_g}{\sin(\theta)}[/tex]
    where [tex]F[/tex] is a force in the direction of motion of the car. So the work needed to push the car would be
    [tex] W = Fd = \frac{F_g}{\sin(\theta)} d[/tex]

    and since [tex]h = d\sin(\theta)[/tex],

    [tex] W = \frac{F_g h}{{\sin}^2(\theta)} = \frac{mgh}{{\sin}^2(\theta)}[/tex]

    I think the answer should be [tex]mgh[/tex]? What did I do wrong?
     
    Last edited: Jan 13, 2007
  2. jcsd
  3. Jan 13, 2007 #2
    what about the horizontal component?
     
  4. Jan 13, 2007 #3
    Why would I need the horizontal component? I already calculated the force [tex]F[/tex] which is being exerted in the direction of motion of the car.
     
  5. Jan 13, 2007 #4

    PhanthomJay

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You have mistakenly assumed that F_y = F_g. Rather than over complicate the problem, try using work -energy methods and note the the minimum work required with or without friction occurs when the block is moved at constant velocity.
     
  6. Jan 13, 2007 #5
    Why is that assumption wrong?
     
  7. Jan 13, 2007 #6

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    First draw a free body diagram. In order to push the box up the slope, apply a force F parallel to the slope. Then resolve the weight into parallel and perpendicular components. You will find a familiar answer will drop out easily using this method.
     
  8. Jan 13, 2007 #7
    Assuming you are pushing parallel to the incline, the force is just [itex] F [/itex], the force of gravity that you are pushing against is the component of gravity along the incline, or [itex] F_g(\sin \theta)[/itex].

    Dorothy
     
    Last edited: Jan 14, 2007
  9. Jan 13, 2007 #8
    You're pushing a car up an inclinced plane =).

    Remember, your normal force is perpendicular to the object that is at rest. So your normal force isn't at a 90 degree angle =). So think about that.

    Also, as said draw a free body force diagram. It will help get you with how the forces are working and balancing out in the entire picture.
     
  10. Jan 14, 2007 #9
    Thank you very much you guys. Thanks to you guys, I figured out what was wrong with my analysis.

    First of all, my choice of the coordinate system wasn't a very smart one -- I chose the y-axis to point directly upwards and the x-axis to point directly to the right.

    Second of all, I was thinking that there was only one vertical force that was acting on the car -- gravity. But I forgot to take into account the vertical component of the normal force (I hadn't taken into account this vertical component previously because I thought it canceled out with one of the components of gravity perpendicular to the inclined plane. But in my coordinate system there was no such component! -- there was only one component of gravity which was acting in the -y direction.).
     
  11. Oct 4, 2009 #10
    I have an identical problem. I'm not sure where to proceed from where I left off, though. Could anyone help?
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Work & Inclided Plane
  1. Inclined plane work (Replies: 4)

Loading...