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A system of pulleys

  • Thread starter Chiara
  • Start date
  • #1
Chiara
You have a puley on which a light string passes. At one end ofthe string a mass of 5 Kg is attached and on the other end another pulley of mass 1 Kg is attached. A second light string passes over this second pulley. At the extremities of this string two masses of 2 and 1 Kg are attached. Calculate the accelerations of the masses and the tension in the two strings.

I attempted this problem, solving a system of 4 equations:
pulley: 1(9.8)-(T1-T2)= a
5Kg mass: 5(9.8)-T1=-5a
1Kg mass: 1(9.8)-T2=-(f-a)
2Kg mass: 2(9.8)-T2=(f+a) where a=acceleration of 5 Kg mass
F=acceleration of the 1 and 2 Kg masses relative to the second pulley
T1= tension in the first string
T2 Tension in the second string
Where is my mistake?
 

Answers and Replies

  • #2
508
0
Why do you think you made a mistake?
 
  • #3
Chiara
because i don't get the same reult given in the Book
 
  • #5
Chiara
Thank you gnome, but in my problem the second pulley has a mass of 1 Kg.this should complicate the problem . . . I think
 
  • #6
508
0
Chiara,
from what you wrote, I think I can deduce the following:
You assume a to be positive if 5kg mass goes up.
You assume f to be positive if 1kg mass goes up.
Right?

OK, if so, I have the following changes (let g = 9.8):
pulley: 1g -(T1 - 2T2) = a
I inserted the factor 2 because each of the two branches of the string exerts a downward force of T2 on the pulley.
5kg mass: 5g - T1 = -5a (as you said.)
1kg mass: 1g - T2 = -(f-a) (as you said).
2kg mass: 2g - T2 = 2(f+a)
I inserted the factor 2 because of Newton's law F = ma, so we must have the mass on the RHS.

Is this in better accordance with the book?
 
  • #7
508
0
I get a=-2/13 g, f=+5/13 g. Is that correct?
 
  • #8
Chiara
yes your answers are correct, Thank you!
 

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