Calculating Tension and Acceleration in a Pulley System with Applied Force

[Panphobia]

Homework Statement

A force of 25 N is applied on the pulley. M1 = 1.5 kg, M2 = 2.5 kg, light frictionless strings and pulley.

a) What is the tension in the strings?
b) What is the acceleration of the masses?
c) What is the minimum Force to apply on the pulley so that M2 comes off the ground?

Homework Equations

ƩF = ma

The Attempt at a Solution

I am not totally sure how to do this question because of the extra force applied. I mean since there is another acceleration, I am thinking in my head that it won't be just mg but it will be something like m(g+ 25/9.8) but I am not sure. Also it seems like there won't be an acceleration in mass 2, but will be in mass 1. Can anyone point me in the right direction?

 

[STEMucator]

Homework Statement

A force of 25 N is applied on the pulley. M1 = 1.5 kg, M2 = 2.5 kg, light frictionless strings and pulley.

a) What is the tension in the strings?
b) What is the acceleration of the masses?
c) What is the minimum Force to apply on the pulley so that M2 comes off the ground?

Homework Equations

ƩF = ma

The Attempt at a Solution

I am not totally sure how to do this question because of the extra force applied. I mean since there is another acceleration, I am thinking in my head that it won't be just mg but it will be something like m(g+ 25/9.8) but I am not sure. Also it seems like there won't be an acceleration in mass 2, but will be in mass 1. Can anyone point me in the right direction?

Treat the two masses separately and draw free body diagrams for each. Don't forget to define a reference frame for each one separately.

The tension in the string will be uniform.

 

[Panphobia]

What do you mean defining a reference frame for each one separately?

 

[STEMucator]

What do you mean defining a reference frame for each one separately?

Notice mass 2 is heavier than mass 1. This means gravity will pull mass 2 downwards.

It could be safe to say then that you should define down as positive in the case of mass 2.

Does it make sense to define it as such for mass 1?

 

[Panphobia]

Yea I know that, Like for the fbd for mass 2 is m2g - T = 0, and for mass 1 its T - m1g = m1a, that is what I wasn't sure about, because the second mass doesn't accelerate but the first mass does...because of the force

 

[STEMucator]

Yea I know that, Like for the fbd for mass 2 is m2g - T = 0, and for mass 1 its T - m1g = m1a, that is what I wasn't sure about, because the second mass doesn't accelerate but the first mass does...because of the force

What? Both masses are accelerating at the same rate.

If you broke things down into the FBDs you would get two equations:

$F_{net_1} =F_T - F_G$

$F_{net_2} =F_G - F_T$

Now you have some information to work with. Namely gravity and the actual values of mass.

 

[Panphobia]

I could have solved that part on my own, its just then T - m1g = m1a and m2g - T = m2a. But my real problem is incorporating the force on the pulley.

 

[STEMucator]

I could have solved that part on my own, its just then T - m1g = m1a and m2g - T = m2a. But my real problem is incorporating the force on the pulley.

Okay very good. You have two linear equations now.

What happens if you.. per say sub one into the other?

 

[Panphobia]

Yea I know two equations two unknowns, and I can solve for both T and a, a = g(m2 - m1)/(m2+m1) = 9.8/4
T = m1g +m1a = (1.5*9.8 + 1.5*9.8/4) N

 

[STEMucator]

Yea I know two equations two unknowns, and I can solve for both T and a, a = g(m2 - m1)/(m2+m1) = 9.8/4
T = m1g +m1a = (1.5*9.8 + 1.5*9.8/4) N

Yup looks good.

 

[Panphobia]

But that doesn't solve the whole problem...the pulley is still accelerating up

 

[STEMucator]

For $m_2$ to come off the ground, there's two options. Suppose we choose the option to introduce an applied force $F_A$, which pulls up on the second mass (imagine pulling on the string with your hand).

What happens now? Think in terms of an FBD.

 

[Panphobia]

you mean like pulling down on m1? Then its just an extra F in the m1 in the m1 FBD, and a -F in the m2 FBD.

 

[STEMucator]

you mean like pulling down on m1? Then its just an extra F in the m1 in the m1 FBD, and a -F in the m2 FBD.

From part a) and b) you have lots of information at your disposal. You know $F_T ≈ 18.4N$ and $F_G ≈ 24.5$ for $m_2$.

Keep it simple, apply the force to the second mass. The difference between the tension force and the force of gravity should give you the force you would need to apply to get the box to stop accelerating downwards.

Then applying even 0.1N of extra force afterwards would cause the box to accelerate upwards.

 

[Panphobia]

But that tension and acceleration was taken as if the pulley had 0 acceleration

 

[STEMucator]

But that tension and acceleration was taken as if the pulley had 0 acceleration

Yes exactly. You have to assume the limiting case of $F_{net_2} = 0$ so that the applied force is at a minimum. If you find the point where the mass stops accelerating, then you can conjecture how much force you would need to get it to accelerate upwards (based on the significant figures, which If I've been following there should only be 2).

 

[Panphobia]

Maybe you didnt understand the question. The pulley is being pulled up with 25 N of force..wouldn't the tension and acceleration be different because of this fact?

 

[STEMucator]

Maybe you didnt understand the question. The pulley is being pulled up with 25 N of force..wouldn't the tension and acceleration be different because of this fact?

Ohh I didn't see that. I'm sorry about that. That would change the answer numerically a bit yes.

I'm not sure exactly what they're pulling on the pulley for, but it's simply another force to consider in the FBD and equation.

The wording of the question isn't the best IMO.

 

[Panphobia]

But where would that force go? The pulley is massless

 

[STEMucator]

But where would that force go? The pulley is massless

If the pulley is being pulled on, and the strings are attached to the pulley, then the strings will also be pulled on by the same force.

Since the masses are attached to the strings, they will follow suit.

 

[Panphobia]

So F and T are in the same direction?

 

[STEMucator]

What I don't understand from your question is why the 25N force is being applied or when it is being applied. From what I can gather they want you to exert a force on the pulley in question c), but they've given you a 25N force already. This doesn't seem to make sense.

If you consider this 25N force for part a) and b), you would get the acceleration is up and the tension force is down. This obviously doesn't make sense to either of us because $m_2$ is going to the ground for part c).

Could you perhaps provide the original question?

 

[Panphobia]

Well the original question was on my final exam.. this was a bonus question

 

[Panphobia]

For a and b you use thr 25 N force and for c) its the minimum force to get m2 off the ground

 

[STEMucator]

For a and b you use thr 25 N force and for c) its the minimum force to get m2 off the ground

I think I understand the context of the question now. For part a and b use the 25N force they've given you.

For part c), treat it completely independently of part a) and b). You need to find that applied force yourself in this case.

The methods to do this are no different than what has been done on the first page.

 

[Panphobia]

The only thing I am currently having trouble with is the FBD of the pulley because it has no mass. If I don't have to Havre an FBD for the pulley then which direction is the force on the masses? In the direction of tension or mg?

 

[STEMucator]

The only thing I am currently having trouble with is the FBD of the pulley because it has no mass. If I don't have to Havre an FBD for the pulley then which direction is the force on the masses? In the direction of tension or mg?

Why does the pulley matter? You don't need a FBD of the pulley here. You know the magnitude and direction of the force on the pulley already (from your diagram), which is all you need.

 

[Panphobia]

For c I would just make a=0?

 

[STEMucator]

For c I would just make a=0?

Indeed that's what I was trying to say about the limiting case. The mass is on the ground so the net force on it is zero at the time.

 

[ehild]

Draw the free-body diagram for both masses and the pulley. The pulley is said massless. What should be the net force acting on?.

The tension in the string is the same in both sides. How is the tension and the applied force related?

What forces act on m2 when it rests on the ground? What should be the applied force so as mass m2 comes off ground?

Supposing m2 is not on the ground, what forces act on it? What forces act on m1?

The length of the string does not change. What does it mean for the accelerations - that of the masses and the pulley?

ehild

 

[ehild]

Homework Statement

A force of 25 N is applied on the pulley. M1 = 1.5 kg, M2 = 2.5 kg, light frictionless strings and pulley.

a) What is the tension in the strings?
b) What is the acceleration of the masses?
c) What is the minimum Force to apply on the pulley so that M2 comes off the ground?

Homework Equations

ƩF = ma

The Attempt at a Solution

I am not totally sure how to do this question because of the extra force applied. I mean since there is another acceleration, I am thinking in my head that it won't be just mg but it will be something like m(g+ 25/9.8) but I am not sure. Also it seems like there won't be an acceleration in mass 2, but will be in mass 1. Can anyone point me in the right direction?

Yes, the pulley also accelerates. First you have to figure out the tension. The pulley is massless, so its mass times acceleration is zero: the net force acting on the pulley has to be zero. The net force is F-2T=0. Now you have T. What is it?

What forces act on the heavier mass? It is on the ground. There is an upward force T and the downward force _m2g. If it is negative (is it?) the block can not accelerate upward. But it can not move downward, because of the support. Its acceleration is zero. The net force includes also the normal force and the sum of all forces is zero. The block stays on the ground. . How much should be F so it can rise?

You know T, so it is easy to find the acceleration of m1 from the equation you have shown : m₁a₁=T-m₁g.

You see that the accelerations are not the same!

ehild

 

[Kishlay]

how did you put that image on the question??

 

[Panphobia]

URL not relevant to the question, but there it is.

 

[haruspex]

So F and T are in the same direction?

The direction of T depends on your standpoint. For the pulley, the Ts act downwards; for the masses, upwards.
Just to point out something I didn't notice mentioned elsewhere in the thread: the two tensions are the same because the pulley is massless (and frictionless). No mass means no moment of inertia, so no torque required to accelerate it on its axis.