Pulley concept question

[frog301]

What is the difference between these 2 problems? (taken from EK 1001 physics)

I don't understand why they solved 465 like that, but then solved for 471 differently. Based on how they solved 465, for 471 I set F=2T=mg as well. T=(1)(10)/2=5N

465. Does not account for 'ma' in this problem. F=2T=mg. F = 30

471. This problem accounts for 'ma' and sets T+m2a=mg for the mass on left, and 2T=mg+ma on right.

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[Flufpot12]

The two questions aren’t the same. In the first question the acceleration is constant throughout the rope while in the second question you know that the acceleration isn’t constant because there appears to be a greater net force on the lower pulley (they both have the same mass yet this ones lower). Greater net force=greater acceleration

 

[shouldvestudied]

The two questions aren’t the same. In the first question the acceleration is constant throughout the rope while in the second question you know that the acceleration isn’t constant because there appears to be a greater net force on the lower pulley (they both have the same mass yet this ones lower). Greater net force=greater acceleration

You're mistaken here. The first problem has zero acceleration. The second problem has constant acceleration.

The first problem asking for "minimum force" means how much force will be enough to just counter act the gravitational force on the object.

The second problem is a slightly more complicated Atwood machine which has a constant acceleration.

 

[Flufpot12]

You're mistaken here. The first problem has zero acceleration. The second problem has constant acceleration.

The first problem asking for "minimum force" means how much force will be enough to just counter act the gravitational force on the object.

The second problem is a slightly more complicated Atwood machine which has a constant acceleration.

If the acceleration is zero in the first problem, then how is the mass supposed to move from rest?
In the second problem, the acceleration on the two masses can’t be the same because of string constraint

 

[shouldvestudied]

The trick to the first question is understanding the meaning of "minimum force". If we are going to be super precise, the minimum force needed to move the weight is an infinitessimal amount more than what is needed to counter balance the weight. But, to avoid being pedantic, simply solving for a Force that cancels out mg is enough to come to an answer. And that problem is one with zero acceleration:

1) F = T since that's how tension works
2) FBD on weight yields: 2T - mg = ma = 0 (because a = 0)
3) Substituting for T yields: F = mg/2 = (6*10)/2 = 30 N

Notice that acceleration was set to zero when solving this.

The second problem involves acceleration, but it is constant. Yes, the weights speed up over time, but that is not inconsistent with a constant acceleration:

1) FBD of left weight yields (make downward the positive direction): mg - T = m2a
2) FBD of right weight yields: 2T - mg = ma ---> 4T - 2mg = m2a
3) Combine the equations to get: 4T - 2mg = mg - T ---> 5T = 3mg ---> T = 3mg/5 = 6 N

But notice that acceleration was constant here. It was not changing with time.

 

[Flufpot12]

The trick to the first question is understanding the meaning of "minimum force". If we are going to be super precise, the minimum force needed to move the weight is an infinitessimal amount more than what is needed to counter balance the weight. But, to avoid being pedantic, simply solving for a Force that cancels out mg is enough to come to an answer. And that problem is one with zero acceleration:

1) F = T since that's how tension works
2) FBD on weight yields: 2T - mg = ma = 0 (because a = 0)
3) Substituting for T yields: F = mg/2 = (6*10)/2 = 30 N

Notice that acceleration was set to zero when solving this.

The second problem involves acceleration, but it is constant. Yes, the weights speed up over time, but that is not inconsistent with a constant acceleration:

1) FBD of left weight yields (make downward the positive direction): mg - T = m2a
2) FBD of right weight yields: 2T - mg = ma ---> 4T - 2mg = m2a
3) Combine the equations to get: 4T - 2mg = mg - T ---> 5T = 3mg ---> T = 3mg/5 = 6 N

But notice that acceleration was constant here. It was not changing with time.

Yeah i guess it makes sense to assume that the acceleration is zero. I was saying the same thing for the second problem though. By not constant I meant that the accelerations of the two masses are different because one of the ropes on the upper mass is fixed. Hence it moves half the distance that the lower mass moves and the lower mass has to move with twice the acceleration