Finding the acceleration of two masses on a pulley system

[jbriggs444]

So what do we do about friction?

We are concerned about the force of friction on m₁ you would agree? The acceleration of the surface is already given and friction does not act directly on m₂.

What is the magnitude of this frictional force? In what direction does it act? Do we have enough information to determine either or both?

 

[Jpyhsics]

We are concerned about the force of friction on m₁ you would agree? The acceleration of the surface is already given and friction does not act directly on m₂.

What is the magnitude of this frictional force? In what direction does it act? Do we have enough information to determine either or both?

SO the frictional force acts on mass 1 toward the right and it is supposed to be normal force times the coefficient of friction, but we are not given the coefficient, so would we have to find that?

 

[Jpyhsics]

SO the frictional force acts on mass 1 toward the right and it is supposed to be normal force times the coefficient of friction, but we are not given the coefficient, so would we have to find that?

So since it is in equilibrium can we assume forces in the x and y direction equal zero? So is the tension force equal to m2g?

 

[FriedChicken885]

Remember that the surface, including the pulley, is accelerating.

So where I’m at so far is:

1. I found μ=0.5 based off of the fact that the system began in equilibrium, and therefore
μm(1)g = m(2)g

2. ΣF(m2) = F(tension) - (m2)*g = (m2)(a2)

ΣF(m1)(x) = F(friction) - F(tension)
= μ(m1)*g - F(tension) = (m1)(a1x)

ΣF(m1)(y) = F(normal) - (m1)*g = 0

At this point I decided to mess around with the numbers, and found that the magnitude of
a1 = (1/2)(a2)

I don’t know if I’m doing this right because I always get stuck at this point because I don’t know how to use the acceleration of the surface and the pulley...

 

[Jpyhsics]

So where I’m at so far is:

1. I found μ=0.5 based off of the fact that the system began in equilibrium, and therefore
μm(1)g = m(2)g

2. ΣF(m2) = F(tension) - (m2)*g = (m2)(a2)

ΣF(m1)(x) = F(friction) - F(tension)
= μ(m1)*g - F(tension) = (m1)(a1x)

ΣF(m1)(y) = F(normal) - (m1)*g = 0

At this point I decided to mess around with the numbers, and found that the magnitude of
a1 = (1/2)(a2)

I don’t know if I’m doing this right because I always get stuck at this point because I don’t know how to use the acceleration of the surface and the pulley...

That is exactly where I am too!

 

[Jpyhsics]

That is exactly where I am too!

But i think that the positive x direction has been switched in your x direction equations.

 

[haruspex]

So since it is in equilibrium can we assume forces in the x and y direction equal zero? So is the tension force equal to m2g?

It is in equilibrium just before the surface starts to accelerate, so yes, at that time the tension equals the weight of m2. But not once the acceleration starts.

 

[haruspex]

based off of the fact that the system began in equilibrium

Not just that it was in equilibrium, but that it was on the point of not being so.

I don’t know how to use the acceleration of the surface and the pulley...

I refer you to posts #2, #6 and #27, which @Jpyhsics is studiously ignoring.

 

[Jpyhsics]

Not just that it was in equilibrium, but that it was on the point of not being so.

I refer you to posts #2, #6 and #27, which @Jpyhsics is studiously ignoring.

I am not ignoring it, I just don't know. As I was taught that the accelerations of both the objects would be the same just in opposite directions, but you have said that is not the case, so I don't know what to assume.

 

[FriedChicken885]

I refer you to posts #2, #6 and #27

Is it something along the line that the difference in the accelerations of m1 and the surface would be the maximum possible acceleration of m2?

Since the string is fixed length, the mass 2 can't fall any faster than the length of the vertical part of the string is growing? Am I on the right track here?

 

[haruspex]

I was taught that the accelerations of both the objects would be the same just in opposite directions,

That must have been for some specific context that does not occur here. What context was that in?

I am not ignoring it,

You are in that you have not responded to it. If you don't understand a hint say so.
Think about the two lengths of string, the vertical part and the horizontal part. Each of those has has an "acceleration", i.e. a second derivative of length wrt time.
What is the relationship between them?
How does the "acceleration" of the horizontal part relate to the acceleration of the platform and the acceleration of m1?

 

[haruspex]

Is it something along the line that the difference in the accelerations of m1 and the surface would be the maximum possible acceleration of m2?

Since the string is fixed length, the mass 2 can't fall any faster than the length of the vertical part of the string is growing? Am I on the right track here?

Yes, but you can assume the string remains taut.

 

[FriedChicken885]

Yes, but you can assume the string remains taut.

Okay, from this point I get to a(2) = a(surface) + a(1) if a(1) is towards the left and a(2) = a(surface) - a(1) is a(1) is towards the right

How would I find the direction of a(1) and go about solving it from here? Also is the previously found equation a1 = (1/2)(a2) valid?

 

[bluejay27]

Homework Statement

Two blocks of the masses m1=7.40 kg and m2=m1/2 are connected via a massless pulley and massless string. The system is currently in equilibrium but is about to start sliding, if m2 would increase even by a bit. For the friction between the surface and m1 assume that µs=µk. The block m2 is suspended from the pulley and is free to swing. What will be the acceleration of m1 if the surface starts to move in a horizontal direction to the right with an acceleration a=5.43 m/s2? Take +x direction to be toward the left. Give your answer in m/s2.

Homework Equations

F=ma

The Attempt at a Solution

For m1
ΣF_x=F_T
m₁a₁=F_T

For m2
ΣF_y=F_T-mg
m₂a₂=F_T-mg

I am not sure how to use the given acceleration in the problem. Also, does µs=µk mean there is no friction?If anyone could help me with this, I would be grateful.

Were you given a diagram for this question?

 

[jbriggs444]

Were you given a diagram for this question?

See post #3 for a diagram.

 

[haruspex]

How would I find the direction of a(1)

You do not need to guess the direction. Just define a particular direction as positive and write the equation accordingly. If you get a negative answer then the acceleration is in the other direction.

 

[haruspex]

Also is the previously found equation a1 = (1/2)(a2) valid?

Looks right.

 

[Jpyhsics]

Looks right.

So does that mean that the acceleration of 1 is half the acceleration of 2? Why would that be so?

 

[Jpyhsics]

Okay, from this point I get to a(2) = a(surface) + a(1) if a(1) is towards the left and a(2) = a(surface) - a(1) is a(1) is towards the right

How would I find the direction of a(1) and go about solving it from here? Also is the previously found equation a1 = (1/2)(a2) valid?

You do not need to guess the direction. Just define a particular direction as positive and write the equation accordingly. If you get a negative answer then the acceleration is in the other direction.

So if you were to define the right as a negative direction, as indicated in the question would the formulae become:

a₂= -a_surface + a₁ ? Does that seem valid?

 

[Jpyhsics]

So if you were to define the right as a negative direction, as indicated in the question would the formulae become (defining left as a positive direction):

a₂= -a_surface + a₁ ? Does that seem valid?

So based on the previous post #43 would that mean that
a₂= -2a₁

If that is the case then it seems as though the acceleration of the surface is not needed as
T₂ = m₂(a₂ +g)
T₁ = m₁(a₁+μg)

Which I would equate the two equations and just for a₁

Based on intuition, I don't think we would have been given the acceleration of the surface if we were not required to use it...
So where am I going wrong?

Woul

 

[Jpyhsics]

So I guess my question is which of these two equations do we use in the tension force equations (if either is correct...and if my tension force equations are correct...):
a₂= -a_surface + a₁
a₂= -2a₁

 

[haruspex]

a2= -a_surface + a1 ?

Which way are you defining as positive for a_surface?

 

[haruspex]

it seems as though the acceleration of the surface is not needed

You are to find the accelerations of the masses in terms of the acceleration of the surface.

 

[FriedChicken885]

So I went through the entire question with +x as left and +y as up, I got to the two equations a(2) = 2a(1) and -a(2) = a - a(1) and solved for a(1) = -a. Yet I got the incorrect answer, what did I do wrong?

 

[haruspex]

-a(2) = a - a(1)

If you are taking a₂ as positive up and a₁ positive left then you have a sign error there.

 

[FriedChicken885]

If you are taking a₂ as positive up and a₁ positive left then you have a sign error there.

I also tried it for:
a₂ = 2a₁ and a₂ = a - a₁ and solved for a₁ = ⅓a, but that's still the incorrect answer. Also I wonder if I'm missing something since I don't use the given value of m₁.

 

[haruspex]

a₂ = a - a₁

still the wrong combination of signs.
The easiest way to get this right is to imagine holding one acceleration at zero and thinking how the other two interact.
Since a₁ and a₂ are both positive towards the pulley, if a were zero their sum would be constant. So we have a₁+a₂= f(a).
If we imagine a₂ held at zero but the string remaining taut then m₁ must be at constant distance from the pulley. Since a is positive right but (bizarrely) a₁ is positive left then, again, their sum must be constant. This leads to a₁+a₂= -a.

 

[Jpyhsics]

So does this solution make sense to you?

 

[FriedChicken885]

a₁+a₂= -a.

So does this solution make sense to you?

I again tried solving that solution with the system of equations:

a₁+a₂= -a and a₂ = 2a₁

and I got a₁ = (-1/3)a which is still incorrect despite every step seeming to be correct.

Also do we need to use the numerical value for m₁ at all?

 

[haruspex]

and I got a₁ = (-1/3)a which is still incorrect despite every step seeming to be correct.

hmm.. that is the answer I get.

Also do we need to use the numerical value for m₁ at all?

No. Only the ratio of the masses can be relevant.