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I need the normal force, the force of friction, and her acceleration

  1. Nov 27, 2006 #1
    I've got 2 problems:
    Number 1: Greg is pushing a 32 kg crate across the floor. He is making like difficult by applying his 800 N force down at an angle of 37 degrees below the horizontal. The coefficient of kinetic friction between the crate and the floor is .50.
    I need to find the normal force, and the acceleration.
    Number 2: After completing her jump, a ski jumper is going back up a hill on the other side, which is designed to slow her down. The hill is angled at 37 degrees to the horizontal. The coefficient of kinetic friction between her bargain basement skis and the snow is .20. Her mass, including all equipment, is 80kg
    I need the normal force, the force of friction, and her acceleration:cry:
     
    Last edited by a moderator: Jan 7, 2014
  2. jcsd
  3. Nov 27, 2006 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF, Jenny. One of the rules here is that you have to show us some of your work in order for us to help you. What can you tell us about frictional forces? What direction do they act in? What is the equation relating a force F to an object's mass and its resulting acceleration?
     
  4. Nov 27, 2006 #3
    here is what I know force of friction=coefficient of friction*Normal force. To find normal force you use SOH CAH TOA depending. They act opposing the force of in the first problem 800N. g*sin(theda)-acceleration/g*cos(theda) acceleration= 2*distance/time squared. I'm really having a difficult time understanding the whole tension concept with friction thrown in. By themselves I can somewhat understand, but when they are combined I got lost.
     
  5. Nov 27, 2006 #4
    I may have figured it our #1 normal force sin37=320/N N=531.7
    acceleration .5=10sin37-a/10cos37 a=2m/s*s with major rounding
    #2 normal force= 800N? force of friction=160N? accleration .2=10sin37-a/10cos37 a=-4.4m/s*s
     
  6. Nov 27, 2006 #5
    Hi Sheebajenny,

    I find drawing a freebody diagram makes life much simpler. Have you done that?

    Once you've done that, you can find the net forces in the horizontal and vertical direction.

    One hint: the equation you gave, a = 2d/t^2, probably won't be that useful in this problem. Consider using Newton's second law:

    Net Force = ma

    Where Net force is the sum of all the forces in a certain direction.

    Try this, and let us know if you run up against more problems.

    Dorothy
     
  7. Nov 27, 2006 #6
    I did draw free body diagrams. For the second problem my only force I have is weight it seems like I must be missing something.
     
  8. Nov 27, 2006 #7
    correction then I first found acceleration and then I used that to find normal force and then force of friction for the second problem one so I got normal force=325N and force of friction as 70.4N
     
  9. Nov 27, 2006 #8
    Are you using g=10 m/s^2?

    This is a good try, but I don't think it's quite right yet.

    The normal force would be a combination of the downward part of the push and the pull of gravity on the crate.

    If you draw a free body diagram, you'll see the normal force pushes up, from the floor, and the other forces (or the components) are straight down. Since there is no motion, they add up to zero:

    N - ForceFromPush - ForceFromGravity = 0

    Try that, and see what you get.

    Dorothy
     
  10. Nov 27, 2006 #9
    Yes I'm using 10m/s*s for gravity
     
  11. Nov 27, 2006 #10
    I truely understand now, thank you very much for your help I had trouble with this one since the day before Thanksgiving. I really appreciate the help.
    Sadie
     
  12. Nov 27, 2006 #11
    Could we focus on one problem at a time :confused:

    This normal force is much too low for an 80 kg weight on a 37 degree slope.

    I'll help you with both, but just one at a time, please.
     
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