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Another Forces Problem

  1. Oct 19, 2009 #1
    1. The problem statement, all variables and given/known data
    They're two boxes connected on a pulley, each weighing 2 kg. Box 1 is on a 30 degree incline and box 2 is on a 60 degree incline.
    a. Find the acceleration ignoring friction
    b. Find the acceleration if the coefficient of kinetic friction is 0.1.

    2. Relevant equations
    F=ma
    Ffr=ukFN

    3. The attempt at a solution
    I drew the free-body diagram. We are missing the pulling of pushing force (Fp), the tension force and the acceleration (which we are trying to find).

    So this what I did for part a (ignoring friction):
    BOX 1-
    - ∑Fx1=Fp+mgsinΘ-FT=m1a1
    - Ft=Fp + mgsinΘ - m1a1

    I solved for tension force since we don't have that.

    BOX 2-
    - ∑Fx2=Fp-mgsinΘ+FT=m2a2
    I substituted the equation from Box 1 for Ft so I could eliminate it.
    - Fp+Fp+mgsinΘ-mgsinΘ - m1a1=m2a2
    - 2Fp/m1m2=a

    So I am left with Fp still, which we don't know. I am confused.

    For part b, with coefficient of kinetic friction of .1, I did the same thing but I added Ffr for box #2 because I made it so that the pulling force and the tension force were in the same direction while the frictional force would obviously be opposing that. For box #1 I thought that Ft would be like the frictional force since it's opposing force (I made the box going down in #1) so I didn't even add it for that part.

    Can you tell me what I am doing wrong.
    physicspicture.jpg
     
  2. jcsd
  3. Oct 19, 2009 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    There is no external applied force (Fp) unless it is explicitly stated that there is. You seem to be under the impression that somebody needs to "do something" to get this system moving. Gravity is perfectly capable of doing that just fine. The system is exactly as described, nothing more, nothing less.
     
  4. Oct 19, 2009 #3
    Alright, I'll go see if I can solve it without having an applied force.
     
  5. Oct 19, 2009 #4
    So ignoring Fp would it go something like this...

    Box 1:
    Ft-mgsin30=m1a1
    Ft=m1a1 + mgsin30

    Box 2:
    Ft-mgsin60=m2a2
    m1a1+mgsin30-mgsin60 = m2a2
    mgsin30-mgsin60/m1+m2 = a
    a= -1.8 m/s^2

    -------------------------
    b.
    Box 1
    Ft-mgsin30-ukFn=ma
    Ft=ma + mgsin30+ukFn
    *Fn=mgcos30=16.97*

    Box 2
    Ft-mgsin60-ukFn=ma
    m1a1 + mgsin30 - uk1Fn1 - mgsin60 - uk2Fn2 = m2a2
    mgsin30 - uk1Fn1 - mgsin60 - uk2Fn2/m1+m2 = a
    9.8 - (.1)(16.97) - 16.97 - (9.8) (.1)/4 = a
    a = -1.61 m/s^2

    correct?
     
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