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Box Sliding Wagon problem

  1. Nov 27, 2006 #1
    a 50 kg wagon is being pushed up a 15 degree incline. The coefficient of friction between the wagon and the slope is 0.15. A 15 kg box sits on top of the wagon and the coefficient between the wagon and the box is 0.30. What is the maximum force that you can push the wagon up the slope with before the box starts to slide?
  2. jcsd
  3. Nov 27, 2006 #2
    Start by posting your current work on the problem.
  4. Nov 27, 2006 #3
    well first i drew it and came up with:
    F-127-71+42.6=50a for the wagon and concluded that an extra 4.6 N must be directed down the slope in order to move the box. I am clueless now.
    Last edited: Nov 27, 2006
  5. Nov 27, 2006 #4
    How'd you get those numbers? Can you be a little more clear?
  6. Nov 27, 2006 #5
    lol too much typing ok here I go
    the 127 is the force of gravity acting on the big box down the slope, the 71 is the force of friction between the big box and the slope the 42.6 is the force of friction acting up on the wagon.
  7. Nov 27, 2006 #6
    Sum the forces on the block in one free-body diagram, then sum the forces on the cart in another.

    There should be no force of friction acting up the ramp on the wagon. The wagon's motion is upwards, and friction resists the direction of motion.

    Once we have four equations from summing our forces, find a way to relate the maximum static friction between the block and cart to the force being applied to the cart. Plug in the maximum friction and get the maximum force.
  8. Nov 27, 2006 #7
    I put the force of friction acting down
  9. Nov 27, 2006 #8
    If friction is acting down, we should be subtracting 42.6.
  10. Nov 27, 2006 #9
    i am only able to come up with 2 equations: F-155.4=50a for the wagon and F=4.6 for the box. What else is there?
  11. Nov 27, 2006 #10
    isnt the 42.6 N acting upwards tho? bcuz the box isnt moving and the friction has to be acting upwards to keep it from moving.
  12. Nov 27, 2006 #11
    The friction between the box and wagon acts in different directions on each object. It acts downwards on the wagon and upwards on the box.

    We can sum the forces up/down the ramp for each object, then perpendicular to the ramp for each object. Maybe you already took that into account and found numerical answers for everything. I haven't plugged the numbers and compared them to what you've gotten. My carpel tunnel is acting up and I'm going to sign off for the night. Good luck!
  13. Nov 28, 2006 #12
    What I don't get is how can you supply extra newtons downwards without actually touch the box? The force of friction stays the same, this has to do with inertia but I have no idea how to solve for this? Maybe I missed something in Physics last year, could any one help?
  14. Nov 29, 2006 #13
    I'm not sure what extra newtons you're referring to. Lay out to me what you have and I will go from there. Try to be as specific as possible. Use a format such as:

    Fx(on wagon) = Fa - Ff(ramp) ... etc.
  15. Nov 30, 2006 #14
    Sorry I don't understand that format, is Fx sposed to be parallel to the slope? and what is Fa?

    For the wagon:
    Ff = 70.95 N
    Fg = 127 N
    F = ?

    For the box:
    Ff = 42.6 N
    Fg = 38 N
  16. Nov 30, 2006 #15
    If the x direction is parallel to the slope, Fx means the forces parallel to the slope.

    I mean force applied by Fa.
    Last edited: Nov 30, 2006
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