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Pulley and cords tension physics

  • #1
The pulley in (Figure 1) is suspended by a cord C. Suppose that m1 = 1.0kg and m2 = 3.9kg .
Determine the tension in the cord that supports the pulley C after the masses are released and before one hits the ground. Ignore the mass of the pulley and cords.

I'm having trouble getting started please help.

i know the tension connecting the masses is the same i just dont know how to solve for it.
 

Answers and Replies

  • #2
24
0
To solve these types of problems, I would always recommend starting by drawing force diagrams. This will help you answer such questions as: What upward force is acting on m1? What downward force is acting on m1? What is the net force on m1? What about m2?
 
  • #3
so weight would be the downward force for both m1 and m2 and tension would be the upward force. correct?
 
  • #4
24
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Correct. And, as you mentioned earlier, tension (T) is the same for both.

Now set up your equations.
 
  • #5
To solve these types of problems, I would always recommend starting by drawing force diagrams. This will help you answer such questions as: What upward force is acting on m1? What downward force is acting on m1? What is the net force on m1? What about m2?
would the tension for m1 and m2 be 5.8?
 
  • #6
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Where are you getting 5.8? That's not the correct answer. How did you set up your equations?
 
  • #7
thats the problem i have is getting the equations.

would the first be Fnet=T-mg=ma?
 
  • #8
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Yes. Do that for both masses.
 
  • #9
would the second equation be mg-T=ma or would it be the same as the first?
 
  • #10
24
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Technically yes, it would be mg - T = ma. Let "a1" be the acceleration of the first mass and "a2" be the acceleration of the second mass.

We have the equations:
(1) T - m1*g = m1*a1
(2) T - m2*g = m2*a2

Both m1 and m2 undergo the same magnitude of acceleration, but in opposite directions, therefore a2 = -a1.

Thus, equation 2 becomes T - m2*g = -m2*a1
Or, as you put it: m2*g - T = m2*a1
 
  • #11
so would T-9.8=a1? correct
 
  • #12
would T=15.6N
 
  • #13
24
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Yep, that's the answer.
 
  • #14
my mastering physics tells me its wrong? is there more to the problem?
 
  • #15
24
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Ah, I see. I just re-read the original question: it's "the tension in the cord that supports the pulley". I was confused slightly by the wording of the question, and didn't have access to the problem's diagram.

The net force on the pulley is 0. Draw a force diagram for the pulley; it is acted upon by an upward force (which you are trying to find) and the tensions below it (which you have already solved for).
 
  • #16
so tension would be the downward and would upward force be the norma force?
 
  • #17
24
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The upward force should be the tension in the cord that holds the pulley up. The downward force is the tensions in the ropes (technically, 2*15.6N since there are two ropes hanging from the pulley). And the net force is 0.
 
  • #18
oh okay i get it. the tension in the cord holding the pulleys has to be equal to the tensions of both the cords below to bring the net force = to 0.
 
  • #19
24
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Yes. Exactly.
 

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