Pulley Problem

Classical Mechanics Level 3

A ball mass of mass 30 kg rests on a pan attached to a rope that goes around a pulley and is attached to another pan.

Anand, who has mass 60 kg, wants to send this ball up to a platform.

Anand has 3 options:

1) Anand can himself sit on the other pan.

2) Anand can pull the other pan down by rope attached below it using force 600 N.

3) Anand can place another ball of mass 30 kg on the other pan and then pull the rope attached below it down with 300 N.

Which method will provide the fastest acceleration to send the ball up to the platform?

Rank the methods from fastest to slowest acceleration as your answer.

Note: The gravitational acceleration is $g=10\text{ m/s}^2.$

2, 1, 3 3, 2, 1 2, 3, 1 3, 1, 2

4 solutions

Satyen Nabar
Aug 24, 2014

Method 2, 3, 1

The net force in all 3 systems is the same 300N but the mass in the system is different which produces different acceleration F=ma.

Method 2) Anand pull s with 600 N, 300 N will neutralize balls weight and the 300 N will produce acceleration of 300/30 = 10 m/s^2

Method 3) The other ball neutralizes the balls weight, but now Anands 300 N has to accelerate both masses together, 60 Kg so acceleration of 300/60 = 5 m/s^2

Method 1) When Anand sits himself, there is net force of 300 N acting downward after neutralizing the balls weight. This 300 N has to pull both Anand 60 kg and the ball 30 kg= 90 kg mass down. So acceleration is 300/90 = 3.33 m/s^2.

oh i see .. i feel like a fool lol, i just considered the net force being the same and thought it was a trick question, so selected any answer. forgot to think about what the net force is acting on, thanks for solution.

Jord W - 6 years, 9 months ago

Tx guys. I am glad you enjoyed it! :)

Satyen Nabar - 6 years, 9 months ago

very nice question clears lots of concepts

swapnil rajawat - 6 years, 9 months ago

It was a nice question.Thank you sir for posting it.

satvik pandey - 6 years, 9 months ago

Very nice question.

Fauzan Yulianto - 6 years, 9 months ago

tricky question

Piyush Bharti - 6 years, 9 months ago

Yogesh Ghadge
Nov 30, 2014

In first condition

m1g - T = m1a ...................................................................... 1

T - m2g = m2a ..........................................................................2

taking eq 1 and 2

m1g - m2g = (m1+m2)a

600 - 300 = 90 a

300 = 90 a

10/3 = a

In second condition the rope has been pulled with 600N so the tension is 600N

substituting in eq 2

T - m2g = m2a

600 - 300 = 30 a

300 = 30 a

10 = a

In third condition one ball of mass 30 kg is kept on it the 300 N force is applied to the rope so the eq becomes

F+ m1g - m2g = (m1 + m2)a

300 + 300 - 300 = 60 a

600 -300 = 60 a

300 = 60 a

5 = a

so from all the conditions we get 2nd con>3rd con>1st con

Jaivir Singh
Sep 17, 2014

IN SECOND CASE a IS 10 IN THIRD CASE IT IS 5 AND IN FIRST CASE IT IS 10/3

Lakshay Sethi
Sep 8, 2014

This problem can be solved using F=ma. Case 1: We have anand sitting on pan thus the total mass of system is 90 kg now whereas total pulling force is 300N in the side of Anand. There acceleration comes out to be $\frac { 10 }{ 3 } m{ /s }^{ 2 }$ . Case 2: Here total mass of system is 30 kg whereas net pulling force is 300N therefore acceleration= $10 m{ /s }^{ 2 }$ . Case 3: Here total mass is 60 kg and net pulling force still remains 300N there acceleration= $5 m{ /s }^{ 2 }$

Pulley Problem

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