Max F & Tension in String for m2 to Remain at Rest on Floor

[atavistic]

Someone exerts a force F directly up on the axle of an ideal pulley. Two objects, m1 of mass 1.2kg and m2 of mass 1.9kg, are attached to the opposite ends of the string, which passes over the pulley. The object m2 is in contact with the floor.
a) What is the largest value the force F may have so that m2 will remain at rest on the floor?
b) What is the tension in the string if the upward force F is 110N?
c) With the tension determined in part b, what is the acceleration of m1?

For a) T in the string should be equal.So for the body(or bodies considering both have to be at rest so that the larger one remains at rest) to be at rest,its acceleration shud be 0.

This implies F-2t=0 => F=2t.Now we need to take t=1.2g as if we take t=1.9g the smaller body will get acceleration .So t=12N and F=24N.Am i correct in this explanation??

I can only think something of the first part.i am blank for the next part.Plz help

 

[learningphysics]

Yes, a is essentially correct... Though I usually use g=9.8 instead of g=10. For your explanations, I would do it more so by using the freebody diagrams of the pulley and m2... since it is an ideal pulley, the pulley is massless...

So for the pulley:
F - 2T = m*a_pulley
F-2T = 0*a_pulley
F = 2T

Note that this relationship is valid throughout this problem... it is independent of the accleration of the pulley.

In the same way use the freebody diagram of m2... since m2 must remain at rest its acceleration is 0. So

T - m2g = m2*a2
T - m2g = 0
T = m2g

For part b), you already have a relationship between F and T from part a... this relationship is still valid... so just solve for T

For part c), use the freebody diagram of m1

 

[atavistic]

But sir,we know that 110N far exceeds min force for acceleration so this implies that m1 and m2 will get acceleration upwards.And m1 will further get another acceleration relative to m2 as it is lighter than m2.Isnt it?

 

[learningphysics]

But sir,we know that 110N far exceeds min force for acceleration so this implies that m1 and m2 will get acceleration upwards.And m1 will further get another acceleration relative to m2 as it is lighter than m2.Isnt it?

Yes, this is all true.

 

[learningphysics]

I think I made a mistake in the explanation for part a... when you consider the freebody diagram of m2... you should consider the normal force...

T + N - m2*g = m2*a2
T + N - m2*g = 0
T = m2*g - N

The maximum tension occurs when the normal force becomes 0 (normal force can't be negative)

then substitute N = 0 and finally get,

T = m2*g

 

[atavistic]

So it doesn't matter in the equations we are making?I simply use the one i.e F=2t for finding this tension in string?

EDIT: On ur preceding post:But then with this T the smaller mass will start accelerating upwards isn't it?

 

[learningphysics]

So it doesn't matter in the equations we are making?I simply use the one i.e F=2t for finding this tension in string?

Yes. Use the freebody diagram of the pulley... the pulley is massless so the net force on the pulley is 0, so F-2T = 0... this doesn't change...

 

[learningphysics]

So it doesn't matter in the equations we are making?I simply use the one i.e F=2t for finding this tension in string?

EDIT: On ur preceding post:But then with this T the smaller mass will start accelerating upwards isn't it?

In part c) ? Yes, the smaller mass accelerates upwards. so does the bigger mass...

Remember the entire pulley is being lifted upwards... if you look in the frame of reference where the pulley is still... you'll see one mass go up, and the other come down... But in the rest frame you'll see both masses going upwards... is that what is confusing you?

 

[FedEx]

In part c) ? Yes, the smaller mass accelerates upwards. so does the bigger mass...

Remember the entire pulley is being lifted upwards... if you look in the frame of reference where the pulley is still... you'll see one mass go up, and the other come down... But in the rest frame you'll see both masses going upwards... is that what is confusing you?

I think that we should apply pseudo force in this case. The whole system is moving upwards and hence we will consider the pseudo force downwards.
So if the whole thing is moving upward with an accelertion a than the gravity on the block is m1(g+a) and for other block m2(g+a).
Now for the two blocks in the normal case one sholud go up and other should go down. but in this case a is so huge that both the blocks are going up the heavier block with an acc of a-a' and the second one with an acc a+a'. Here a' is the acc of the blocks when the force is not being applied. Now with this things we may get our answer.; Is anything wrong in this application , then please let me know i will try again

 

[learningphysics]

Hi Fedex. Yes, I think that would be another method to do it if you are careful... But I'm finding a little confusing... one thing to remember is that in the normal pulley operation situation, F isn't 0...

 

[atavistic]

WOw fedex that's exactly what i needed to know since my very smart friend also said about the application of pseudo force(which i haven't studied yet).I think pseudo force will solve it.

learningphysics:yes that was exactly what was confusing me. :)

 

[learningphysics]

WOw fedex that's exactly what i needed to know since my very smart friend also said about the application of pseudo force(which i haven't studied yet).I think pseudo force will solve it.

learningphysics:yes that was exactly what was confusing me. :)

That's cool. But you can also solve it the way I mentioned... compare the two methods afterwards to check the answers are the same.