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Tension Along Rope with Masses

  1. Feb 4, 2012 #1
    I've worked through this problem a few different ways now and each one gives me different answers and I am looking for some guidance as to which is correct:

    Four blocks of mass m1=1kg, m2=2kg, m3=3kg (very original right?), and unknown mass m4 are on a frictionless surface (pic attached). Force Fl pulls in the left direction at 30N and Fr to the right at 50N. The tension between m2 and m3 is 36N. What is m4?

    As I worked through the problem I labeled all tensions as shown in the picture. I know T1=30N and T5=50N. My biggest problem arises from whether T3=36N or if 2*T3=36N.

    The first time I did this problem I used 2*T3-36N or T3=18N. Then using Newton's 2nd law I set up the equations T2-T1=m1*a, T3-T2=m2*a, and so forth. Using this method I got that a=4m/s2 to the right and m4=5kg.

    The second time I did this I used T3=36N and using the same equation setup as before I got a=2m/s2 to the left which seemed plausible, however using T5-T4=m4*a I ended up with m4=-10kg which seems not so plausible.

    Any suggestions?
     

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  3. Feb 4, 2012 #2
    T3 = 36N not 2x36N.... it is pulling with 36N to the right on the combination (m1+m2) and pulling with 36N to the left on the combination (m3+m4)
    You are given the tension between m2 and m3 to be 36N so (m1+m2) experience a resultant force to the right of what?... can you see that and calculate the acceleration of (m1 and m2).
    Once you know the acceleration you can apply resultant force on (m3+m4) to calculate m4
     
  4. Feb 4, 2012 #3
    So since m1+m2=3kg and the experience a force of 6N then a=2.

    Then m3 and m4 experience 14N so

    14=(m3+m4)*2 => 14=6+2*m4 => 8=2*m4 => 4kg=m4?

    Is that right?
     
  5. Feb 4, 2012 #4
    that is exactly what I got
     
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