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Frictionless pulley- incline

  1. Sep 25, 2006 #1
    Two packing crates of masses m1 = 10.0 kg and m2 = 7.00 kg are connected by a light string that passes over a frictionless pulley. The 7.00 kg crate lies on a smooth incline of angle 43.0°. Find the acceleration of the 7.00 kg crate up the incline. Find the tension in the string.

    I found the acceleration (correctly) to be 3.0126 but I can't seem to get the tension part of the question. I tried the formula T=(m2)(g)(sinTHETA)+(m2)(a) using 7(3.0126)sin43+7(3.0126) and 10(3.0126)sin43+10(3.0126) but neither gave the correct answer.
     
  2. jcsd
  3. Sep 25, 2006 #2
    This might be oversimplifying it a bit but shouldnt f=ma work?
     
  4. Sep 25, 2006 #3
    What would you use as the mass? 98 isn't the right answer.
     
  5. Sep 25, 2006 #4
    Well, I'm just wondering on this one...but isnt the T of the string that is holding the block on the incline equal to the T of the string holding the hanging block?

    So how would you figure out T on the hanging block?
     
  6. Sep 25, 2006 #5
    well no because theyre moving theyre not in equilibrium, i assume?

    just write out how you got the accel so we can see how ur goin about this as i agree with ur workin for T so maybe accel is wrong? ( i know you said you got it right but just check)
     
    Last edited: Sep 25, 2006
  7. Sep 25, 2006 #6
    Wouldn't they be moving in equilibrium? The string is not growing shrining, they are just shifting arent they?
     
  8. Sep 26, 2006 #7
    I got the acceleration by a= (m2-m1sinTHETA/m1+m2)g so it was (10-7sin43/17)(9.8) = 3.0126

    We can check to see if we got them right on the internet so that's definately the right answer, i just can't figure out the second part.
     
  9. Sep 27, 2006 #8

    try using the equation T= m1(g-a) = (10.0kg)[(9.80 - 3.0126) m/s^2] = 67.9N

    i'm sure that's right cause i just did the same problem for hw.
    good luck :cool:
     
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