Ek 1001 pully 466/471 I made pictures!

GRod18 · Aug 2, 2010

** the empty box on 471 is also 1 Kg.

466. What minimum force F is required to lift the mass?
Answer: 30 N

EK says to ignore the angle 30 degrees for this problem because tension is same everywhere, but that doesn't seem to be very intuitive, why am I not just using the straight vertical force?

471. What is the tension T in the rope.
　Answer: 6 N

EK says that mass 1 (empty box, left one) will accelerates at twice the rate of mass 2, is it because that mass 2 has two Tension forces holding it up?

Thanks all.

wolverinepwns · Aug 5, 2010

GRod18 said:
** the empty box on 471 is also 1 Kg.

466. What minimum force F is required to lift the mass?
Answer: 30 N

EK says to ignore the angle 30 degrees for this problem because tension is same everywhere, but that doesn't seem to be very intuitive, why am I not just using the straight vertical force?

471. What is the tension T in the rope.
　Answer: 6 N

EK says that mass 1 (empty box, left one) will accelerates at twice the rate of mass 2, is it because that mass 2 has two Tension forces holding it up?

Thanks all.

for Question 466: you are using the vertical force which is why you can ignore the angle! the weight is 6 x 10 = 60, and since the down force involves 2T and the upward pull (you pulling on the rope) is T, you get 60/2 = 30N. You have to pull twice as long but with 1/2 the amount of force.

477: yes, it is due to the 2T compared to T, (exact same senario at question 466.

divajazzerina · Jul 12, 2011

GRod18 said:
** the empty box on 471 is also 1 Kg.

471. What is the tension T in the rope.
　Answer: 6 N

Can someone explain how T=6N in 471 ? Thanks!

MT Headed · Jul 12, 2011

divajazzerina said:
Can someone explain how T=6N in 471 ? Thanks!

If the big pulley is fixed in space and the pulleys and rope are massless, the total force down on the big pulley (as transmitted through the rope) is 2kg x g = 20N.

Assuming the pulleys are also frictionless (in addition to being massless), the tension will be a constant throughout the rope. The rope pulls down on the big pulley at three places, so the tension in the rope is 20N/3 = 6.66N.

Anything else, and the big pulley would fly away or go crashing into the ground.

divajazzerina · Jul 12, 2011

Awesome, thanks! I was making it wayyyyyy to complicated than it need be!

premed1001 · Mar 12, 2014

MT Headed said:
If the big pulley is fixed in space and the pulleys and rope are massless, the total force down on the big pulley (as transmitted through the rope) is 2kg x g = 20N.

Assuming the pulleys are also frictionless (in addition to being massless), the tension will be a constant throughout the rope. The rope pulls down on the big pulley at three places, so the tension in the rope is 20N/3 = 6.66N.

Anything else, and the big pulley would fly away or go crashing into the ground.

this is incredibly amazing explanation and i hope it is the correct way to deal with this problem. can anyone else verify this as the answer is 6 not 6.666?

chenzt · Mar 12, 2014

premed1001 said:
this is incredibly amazing explanation and i hope it is the correct way to deal with this problem. can anyone else verify this as the answer is 6 not 6.666?

6.666 would be true if it is in equilibrium. But since the OP said the blank box is also 1kg, or 10N in force, and based on answer in 466, you only need 5N to lift the box, which means there's also an acceleration upward with the small pulley. I know theres some annoying calculation required to chug out the answer for tension, so I won't do it, but it explains the discrepancy.

Ebio · Mar 13, 2014

Here is my answer
a=2a'
(m1g-F)/m1=2(2F-m2g)/m2

premed1001 · Mar 14, 2014

Ebio said:
Here is my answer
a=2a'
(m1g-F)/m1=2(2F-m2g)/m2

That is incorrect

Maxxxx · Mar 17, 2014

premed1001 said:
That is incorrect

care to explain? I also did it that way

butters221 · Mar 17, 2014

I actually just explained this problem in another thread. The above poster is not correct with 6.66.
http://forums.studentdoctor.net/threads/tension-question.1060109/

Ek 1001 pully 466/471 I made pictures!

GRod18

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wolverinepwns

Si vis pacem, para bellum

divajazzerina

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MT Headed

snow, PBR, and bears

divajazzerina

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premed1001

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chenzt

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Ebio

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premed1001

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Maxxxx

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butters221

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