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!

Homework Help: Incline Friction Problem?

  1. Sep 8, 2007 #1
    1. The problem statement, all variables and given/known data

    A 2.00 kg block is held in equilibrium on an incline of angle = 75° by a horizontal force applied in the direction shown in Figure P4.50. If the coefficient of static friction between block and incline is µs = 0.300, determine the following.
    (a) the minimum value of F
    (b) the normal force exerted by the incline on the block

    Illustration of Problem: http://www.webassign.net/sercp/p4-50.gif...

    2. Relevant equations

    F=ma

    3. The attempt at a solution

    For Y Component:
    n-mgcos75=0
    n=mgcos75
    n=2.00*9.8cos75
    n=5.07 N
    It says this is wrong because of: Your answer differs from the correct answer by orders of magnitude.

    For X
    F-fs=0
    F=usn=0
    F=usn
    F=0.300 x 5.07 N = 1.52 N

    Does this look like I did it correctly?
     
    Last edited: Sep 8, 2007
  2. jcsd
  3. Sep 8, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    No. You have missing components. The F force has a component in the x direction (parallel to the incline) and y direction (perpendicular to the incline).

    Also the mgsintheta component is missing (component of gravity parallel to the incline)...

    And friction acts upward along the incline... it's preventing the block from sliding down. It works together with F to prevent it from sliding down.
     
  4. Sep 8, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    You neglected the Y component of the applied force F.
     
  5. Sep 8, 2007 #4
    Ok, well now I really don't even almost have a clue what to do.

    Of course the Force has x and y components, but if I don't know the force how would I solve to find those?
     
  6. Sep 8, 2007 #5

    learningphysics

    User Avatar
    Homework Helper

    Resolve F into its components using the angle.
     
  7. Sep 8, 2007 #6
    Would the normal force be equal to mg?

    That is, 19.6 N?
     
  8. Sep 8, 2007 #7
    I am so confused, I don't even know what angle goes with F.

    I can figure out the components of mg, but haven't a clue how to go about finding it for F.
     
  9. Sep 8, 2007 #8

    learningphysics

    User Avatar
    Homework Helper

    It's the same theta. Just have to be careful about sin vs. cos.

    From the tail of the vector F (the part without the arrow) draw a line perpendicular to the incline. And from where that line intersects the incline draw another line to the head of the vector F (the part with the arrow). See the two components? One of the angles of this triangle is theta.
     
  10. Sep 8, 2007 #9

    learningphysics

    User Avatar
    Homework Helper

    No. that's not the answer.
     
  11. Sep 8, 2007 #10
    I think I'm just screwed.

    Thanks for the help anyways.
     
  12. Sep 8, 2007 #11

    learningphysics

    User Avatar
    Homework Helper

    Draw an x - axis parallel to the incline... y -axis perpendicular to the incline... now rotate your picture so that the x-axis is horizontal, and y -axis is vertical... now try to resolve the components of F along the x-axis and y-axis.
     
  13. Sep 8, 2007 #12
  14. Sep 8, 2007 #13

    learningphysics

    User Avatar
    Homework Helper

  15. Sep 8, 2007 #14
    So should the component parallel to N be pointing up (above F)?
     
  16. Sep 8, 2007 #15

    learningphysics

    User Avatar
    Homework Helper

    No sorry... it's opposite to N...

    And fs should be upward along the plane... and the other component of F is parallel to fs upward along the plane.
     
  17. Sep 8, 2007 #16
  18. Sep 8, 2007 #17

    learningphysics

    User Avatar
    Homework Helper

  19. Sep 8, 2007 #18
    Ok, so I drew it and there are 3 angles: 90, 45, and 45.

    Which angle should I be working with?
     
  20. Sep 8, 2007 #19
    Should I get Fsin45 for my x component and Fcos45 for my y-component?
     
  21. Sep 8, 2007 #20
    Ok, I got it. It's n=35.64 N and F=31.66, correct?
     
  22. Sep 8, 2007 #21

    learningphysics

    User Avatar
    Homework Helper

    I uploaded a picture...

    http://server4.pictiger.com/img/1317497/picture-hosting/force.jpg [Broken]

    Fx = Fcos75
    Fy = Fsin75
     
    Last edited by a moderator: May 3, 2017
  23. Sep 8, 2007 #22

    learningphysics

    User Avatar
    Homework Helper

    No. I was wrong it can't be any right triangle... look at the picture I uploaded...
     
  24. Sep 8, 2007 #23

    learningphysics

    User Avatar
    Homework Helper

    Yeah, I think that's right.
     
  25. Sep 9, 2007 #24
    Here you will have to resolve two components. 1) mg 2) F.
    For mg one of its componets[mgsin(theta)]is parallel to the plane which will be trying to bring the block downwards.Another would be perpendicular to the plane[mgcos(theta)](towards the plane).

    For F one of its components would be perpendicular to the plane(towards the plane)[fsin(theta)] and another would be parallel to the plane which will be trying to pull the block upwards[fcos(theta)].

    hence fcos(theta)-mgsin(theta)-mu{fsin(theta)+mgcos(theta)}=0
    Now as it is on the verge of moving we consider a=o. so ma=o
    Now you solve the above equation and you get the answer.

    Sorry doc al, i had to give the equation.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook