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Man hanging on a ropeby khuko
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#1
Aug2511, 01:29 PM

P: 7

1. The problem statement, all variables and given/known data
Calculate the force that the man (weight 70 kg) should exert on the rope to stay in balance. (answer = 137,2 N) 2. Relevant equations Like I see, there are two ropes who are in equilibrium. In each rope the tension is the same but different from each other. I suppose the sum of the tensions of both ropes are equally to the weight of the person 3. The attempt at a solution The netforce is zero cause there's no movement or acceleration Rope 1 = the rope who holds the chair. Rope 2 = the rope the man holds. in positive ydirection : T(rope1)mg/2=0 (/2 because the weight from the two ropes are from the same person) so T(rope1)=mg/2 the other one T(rope2)=mg/2 like you see my equations doesn't match the answer. I know that the rope 2 depends on rope 1 cause they are attached to each other. But i can't write that mathematical equation. Can you help me? 


#2
Aug2511, 01:33 PM

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How does the tension in rope 2 relate to the tension in rope 1? (Look at the middle pulley.)



#3
Aug2511, 01:44 PM

P: 7

If he pulls on rope 2 , then will the pulley follow and the chair go up. So it's related to each other or else the two ropes where indepented and the weight (and tension) where the same for each rope. ?



#4
Aug2511, 02:01 PM

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P: 41,443

Man hanging on a rope
I'm looking for the mathematical relationship between the two tensions. Analyze the force on the middle pulley. (The net force on the pulley must be zero.) If the tension in rope 2 is "T", what's the tension in rope 1?



#5
Aug2511, 02:09 PM

P: 7

i must be "T". So the equation for the rope 2 => T=mg/2
and for rope 1 => T=mg/2. (The opposite ). But a negative force doesn't exist :s But i don't understand why the answer is 137,2 N. Cause my answer would be for rope 2 = 343,35 


#6
Aug2511, 02:16 PM

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P: 41,443




#7
Aug2511, 02:36 PM

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P: 41,443

Maybe it will be easier if you call the tension in rope 1 T_{1} and the tension in rope 2, T_{2}. Then write a force equation for the middle pulley.



#8
Aug2511, 04:20 PM

P: 7

Allright. the middle pulley is the "center" point of all the tensions from the ropes.
everythings comes to this point (correct ?). T_{1} is the tension in rope 1, T_{2} is the tension in rope 2 There's no force in the xdirection. ydirection : The net force of the pulley must be zero cause there is no movement ([itex]\sum[/itex] F_{y} =0 T_{1} is coming to this pulley but also goes to rope 2 so we have two components of T_{1}. T_{2} is also coming to this pulley but also goes to rope 1 so we have positive T_{2} and a negative T_{2}. so : T_{1} + T_{1} + T_{2}  T_{2}  mgmg = 0. Am i close to it ? 


#9
Aug2511, 04:29 PM

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P: 41,443

There are two rope segments pulling upward: What force do they exert on the pulley? There is one rope segment pulling down: What force does it exert on the pulley. Note that we assume that the pulleys have no mass. 


#10
Aug2511, 04:55 PM

P: 7

Hmm
The Two forces that are pulling the pulley upwards are T_{2}. The first T_{2} goes to the ceiling and haves a normalforce called T_{2}' who's going back to the rope 2 (and pulls the pulley upwards). The force that pulls the pulley downwards is the normalforce of T_{1} called T_{1}' . We have also the mg from the first pulley who's pulling downwards. so T_{2} + T_{2}'T_{1}' mg would be zero ? 


#11
Aug2511, 05:05 PM

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P: 41,443

The ropes pulling upward have a tension of T_{2}, thus the force they exert is 2*T_{2} upward. The rope pulling down has a tension of T_{1}, thus the force it exerts on the pulley is T_{1} downward. Note that since the pulley is massless, there's no 'mg' term. The net force must be zero, thus: 2*T_{2}  T_{1} = 0 Or: 2*T_{2} = T_{1} Now you have the relationship between T_{1} and T_{2}. You'll need this in a bit. Next step: Analyze the forces on the man+chair+bottom pulley (treat them as one 'object'). 


#12
Aug2511, 05:24 PM

P: 7

There are two forces that are pulling the pulley up, both are T_{1} and one goes down. That is the weight of the object (man+chair).
therefore the equation is 2*T_{1}mg =>T_{1}=mg/2 And if I substitute this in 2*T_{2}=T_{1} We get for T_{2}= mg/4 ow no. just saw that the bottom pulley haves two tensions going up with the same size of Tension 2 ( same object like the object on rope 2. Hmm, i'm confused :) 


#13
Aug2511, 05:52 PM

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#14
Aug2511, 06:08 PM

P: 7

Thanks, i now see that you must do mg/5 for the answer. Thank you very much!. How do i must learn to solve some exercises like this? caus i don't understand the essential bout this exercise 


#15
Aug2511, 06:23 PM

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P: 41,443

In this case, one 'trick' is to realize that the net force on a massless pulley is always zero (even if it accelerates)that will give you at least one equation to use. 


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