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!

How big is the friction force on the truck?

  1. Mar 5, 2013 #1
    Calculating friction force on a truck on an incline?

    1. The problem statement, all variables and given/known data
    A 4200kg truck is parked on a 20 degree slope.
    How big is the friction force on the truck? The coefficient of static friction between the tires and the road is 0.90.


    2. Relevant equations
    F=ma
    Ff=mu*m*a

    3. The attempt at a solution
    Friction Force on the truck is (4200kg*9.81m/s^2 * .90) = 37081.8N
    since it's on an incline I am supposed to do 37081.8 * sin20?
    so the answer would be 12682.7N but that is wrong
    I also tried doing 37081.8 * cos20 which is 34845.5N which is also wrong
    Please help me! Idk what I am doing wrong
     
    Last edited: Mar 5, 2013
  2. jcsd
  3. Mar 5, 2013 #2
    Please post the whole problem statement.
     
  4. Mar 5, 2013 #3
    sorry about that. I posted the whole problem now.
     
  5. Mar 5, 2013 #4
    Draw a free body diagram and the forces acting on the truck.
     
  6. Mar 5, 2013 #5
    What you found is the max static friction meaning that you would need a force greater than that friction to move the truck. If you were to draw out a FBD and put the friction you had found in you'll see that it's greater than the gravity force acting on the truck down the incline. But you know that this can't be true since the truck is in equilibrium.
     
  7. Mar 5, 2013 #6
    I did draw a FBD, but I really suck at physics. Wouldn't there just be a force going down on the truck which would be (4200*9.81)cos20? where would the friction point?
    can you point me in the right direction?
     
  8. Mar 5, 2013 #7
    No it would be sin20. Frictions always point against motion so it will be pointing opposite of the gravity force down the incline. Don't forget that a=0 either.
     
  9. Mar 5, 2013 #8
    alright thanks a bunch Maiq. What was the point of the static coefficient in that problem?
    The answer was just (4200*9.81)sin20
     
  10. Mar 5, 2013 #9

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The static friction does not mean a definite magnitude. It is the force necessary to keep a body in rest, but it can not exceed (normal force) times (coefficient of static friction).

    ehild
     
  11. Mar 5, 2013 #10
    They probably put it in there to throw you off or maybe just to show what ehild said.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted