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!

Torque to find forces

  1. Nov 24, 2008 #1
    1. The problem statement, all variables and given/known data

    A 6 meter long ladder rests against a wall which is 4 m tall. The base of the ladder is 1.5 m away from the base of the wall (in the picture you can see the wall touches the ladder about 3/4 way up the ladder). The wall is frictionless, but the ground is not.
    a- Find the normal force on the ladder due to the wall (remember it is perpendicular to the ladder)
    b- Find the normal force on the ladder due to the ground.
    c- If the ladder is just about to slide, find the coefficient of static friction between the wall and the ground.

    2. Relevant equations

    torque = r * force * sin (angle between them)

    3. The attempt at a solution
    a- I put the pivot at the point where the floor meets the ladder. I found the angle to be 69.5
    -torque of normal wall - torque of weight = 0
    r*normal wall * sin 90 = r*weight * sin 200.5
    normal wall = 17.2 N

    b- I put the pivot at the point where the wall meets the ladder.
    torque of normal floor - torque static friction - torque weight = 0

    I dont know normal floor or friction force, so how do I solve this?
     
  2. jcsd
  3. Nov 24, 2008 #2
    How are the forces acting on the ladder related? Remember, the ladder is at rest.
     
  4. Nov 24, 2008 #3
    Can I just say the normal force of the floor = the weight - y component of normal of wall?
    Is the y component just sin 45 * normal wall?
     
  5. Nov 24, 2008 #4
    Yes, this is right.

    Hmm, Don't know where you get 45 from. The direction of the normal force due to the wall is given to you. You're even asked to remember it :wink:
     
  6. Nov 24, 2008 #5
    Oops! I meant to say 90. It's been a long day...

    So the normal force from the ground is mg-normal wall *sin 90 =

    10kg*9.8-17.2*sin 90 = 80.8 N?
     
  7. Nov 24, 2008 #6
    Well , 90 isn't right either.

    "a- Find the normal force on the ladder due to the wall (remember it is perpendicular to the ladder)"

    Along which axis are you equating the forces?
     
  8. Nov 24, 2008 #7
    God, I was looking at the normal force for the wall again. Sorry. So it should be sin 69.5.
     
  9. Nov 24, 2008 #8
    Good :smile:

    Edit: I don't know which of the angles is 69.5, so I don't know if you have the sine term right. Looks like you may be off -- I'd check it again
     
  10. Nov 24, 2008 #9
    Oh! This is the part that always messes me up. Looking again, I think the angle should be 90 + (90-69.5) = 110.5.

    So I get normal = 10kg*9.8m/s^2 - 17.2*sin 110.5 = 81.9 N. Does this seem right?
     
  11. Nov 24, 2008 #10
    I can't really reproduce the 17.2N number, but there is nothing wrong with your approach.
     
  12. Nov 24, 2008 #11
    1." r*normal wall * sin 90 = r*weight * sin 200.5"

    Are the r's the same? What values do they take?

    2. "I think the angle should be 90 + (90-69.5) = 110.5. "

    Hmm, sin 69.5 = sin 110.5, therefore this is not correct either. You need to look at it a little more.
     
  13. Nov 24, 2008 #12
    1) I used 4.27m*normal wall*sin90 = (4.27m/2)*weight*sin200.5

    2) For this, I really don't know. I just drew a diagram, and now it looks t me like the y component of the normal wall should use sin 20.5 because the original theta (69.5) would make a 90 degree angle with it. I'm not sure though...
     
  14. Nov 24, 2008 #13
    The ladder is 6 m long :smile: I also assume you mean 20.5 degrees

    Yes, now it is correct. You should probably practise these diagrams a few times, so that you don't get confused.
     
  15. Nov 24, 2008 #14
    The r from the normal wall to the pivot on the ground should be 4.27, but the r for the weight should be 6/2 = 3, right? Using all this, I get normal wall = 24.1 N. And yes, I did mean 20.5. Sorry for being so sloppy and confused! This homework makes me flustered.
     
    Last edited: Nov 24, 2008
  16. Nov 24, 2008 #15
    That's fine. I think you know enough to solve this problem now.
     
  17. Nov 24, 2008 #16
    Great. Thank you so much for your help (and your patience!).
     
  18. Nov 25, 2008 #17
    I thought I had this figured out, but now I am having trouble trying to solve for the coefficient of static friction. I am using friction = coeffiient*normal ground

    looking at the diagram, I said -friction = normal of wall * cos 69.5 which gave me friction = -8.44

    solving for the coefficient with friction / normal ground is -8.44 / 89.6, which is negative, so I am doing something wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?