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Friction/tension at an angle

  1. Dec 11, 2011 #1
    So I tried working this problem out. I'm fairly certain it will be on my final tomorrow morning, but I can't find anywhere to verify my answers. So if someone can please just check over my math for me, it would help immensely! I have been confused on the normal force when tension at an angle is involved, so thats why I'm so desperate to have this verified. Thank you in advance!

    A 10 kg box is sitting on a rough floor. A worker has attached a cord to the box so that he can pull at an angle of 30 degrees above the horizontal. If μk=.1 and μs=.2

    a. How hard must the worker pull if the box is to start moving?

    F=uN

    F= .2 * (mg-Fsin30)

    Fsmax= 17.82 N

    Fcos30=17.82

    F=20.57 N

    b.If he continues to pull with the force you found in part a, what will the acceleration be?

    Fy=0
    Fx= Pull-μkN
    Fx-(μK*(mg-Fsin30))=ma
    19.6-(.1*(98-20.57sin30))=10a

    a=1.083


    c. He slowly increases the magnitude of the pulling force. What is the value of the acceleration along the floor just before the box is lifted off the floor?

    I know the normal force=0

    Fy=mg
    Fy=98N

    tan30=98/Fx
    Fx=169.74N

    Fx=max

    169.74=10a
    a=16.9
     
  2. jcsd
  3. Dec 11, 2011 #2
    I didn't get the same answer and you have left out steps so let's start from the beginning summing the forces in the x and y direction

    I think it comes out to be the same equation you have so it may just be that I ran it all at once rather than in two parts.

    Fx max = Ff = mu Fn

    Fy +Fn = Fw

    If you solve equation 2 for Fn and put in your sin and cos to get the components you can then solve the first equation for F.

    Run it one more time and check your answer as I didn't get quite the same thing. It could have been rounding but just double check. I don't see any errors in part a.

    for part B how did you get the value of 19.6 for the x component of the force which is 20.5 (.866)? I get 17.7 which changes the answer.


    For c I get the same thing.
     
    Last edited: Dec 11, 2011
  4. Dec 11, 2011 #3
    Thank you for your response!

    I got a slightly different answer for part a after changing some things. Here is my work:

    Fx=uN
    N=mg-Fsin30
    Fx=Fcos30

    Fcos30=u(mg-Fsin30)
    Fcos30=umg-uFsin30
    F(cos30-usin30)=umg
    F=umg/(cos30-usin30)
    F=(.2*10*9.8)/(cos30-.2sin30)

    F=25.58N

    Like I said, finding the Fn at an angle has been tripping me up.
     
  5. Dec 11, 2011 #4
    I think you have a sign error

    Isn't it F cos 30 = mumg - muFsin30
    so it would be F(cos30 + mu sin30) = mu mg

    You did it correctly the first time. I just got 20.3 rather than 20.5

    If you always draw your forces and add the ups and set them equal to the sum of the downs you won't usually have a problem as long as they are in equilibrium.
     
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