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!

How much mass is needed to prevent an object from sliding down a ramp

  1. Jul 8, 2014 #1
    1. The problem statement, all variables and given/known data
    An 80 kg object is on top of a 40°ramp and held by a box that is filled with sand bags. Each sand bag has a mass of 12 kg, the static friction coefficient between the box and the ramp is .507. What is the least amount of sand bags needed to prevent the object from sliding.


    2. Relevant equations
    F,net,,x,object = 0 = F(box) - F(grav,x)
    F(box) = F(grav,x,obj)

    F,net,box = friction - F(object) - F(grav,x,box)
    friction = μ(s)mgcos(40°)
    F(grav,x,box) = mgsin(40°)
    F(object) = mg(sin°40)


    3. The attempt at a solution

    F(box) = F,grav,x,obj = 504 N

    F,net,box = friction - F(obj) - F,grav,x,box
    = > μ(s)mg(cos40) - 504 - mg(sin40) = 504
    => m[μ(s)gcos40 - gsin40] = 1008
    => m = 1008/(-2.49) = -404 kg ??

    so lost
     
  2. jcsd
  3. Jul 8, 2014 #2

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Is there no friction between the 80kg object and the ramp?
     
  4. Jul 8, 2014 #3

    Bandersnatch

    User Avatar
    Science Advisor

    This looks like a trick question(providing all the data is correct).

    Perhaps if you calculate the net force acting on the box parallel to the ramp, completely disregarding the 80kg body, you'll see why.
     
  5. Jul 8, 2014 #4
    There is no information regarding static friction between the object and ramp..

    If i disregard the object and focus on the box that is on the ramp, wouldnt the net force be 0 with friction and F,g,x cancelling each other out if it does not move?
     
  6. Jul 8, 2014 #5

    BiGyElLoWhAt

    User Avatar
    Gold Member

    As bandersnatch pointed out, just do a N2L for the box with the sand bags and mass M, ignore the 80kg object, and see what happens.
     
  7. Jul 8, 2014 #6
    F box = friction - F,gx --> u(s)mgcos40 - mgsin40 = 0 --> factor out m --> m = 0??
     
  8. Jul 8, 2014 #7

    BiGyElLoWhAt

    User Avatar
    Gold Member

    If you factor out the m (and g) from your final eqation you get
    ##mg(\mu_{s}cos(40) - sin(40)) = ma = 0##
    If you plug in your known values, what does that tell you? (you're on the right track, just think about the implications)

    Try dividing.
     
  9. Jul 8, 2014 #8
    divide out m and your left with a negative acceleration meaning that box is moving meaning friction would not keep the box up on the ramp no matter what?
     
  10. Jul 8, 2014 #9

    Bandersnatch

    User Avatar
    Science Advisor

    Looks that way.
     
  11. Jul 8, 2014 #10

    BiGyElLoWhAt

    User Avatar
    Gold Member

    Basically yea.

    ##\mu_{s}\text{cos}(40)## is always less than ##\text{sin}(40)## for the given value of mu s.
     
  12. Jul 9, 2014 #11

    verty

    User Avatar
    Homework Helper

    I calculate that there is an answer only when Θ < 26.885°.
     
  13. Jul 9, 2014 #12

    BiGyElLoWhAt

    User Avatar
    Gold Member

    That could be, I didn't go that far, but you need mu s cos theta >= sin theta which doesn't happen at 40 degrees.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted