Two masses, two pulleys and an inclined plane

[Valerion]

Homework Statement

A block of mass m₁ is at rest on an inclined plane that makes an angle (theta) with the horizontal. The coefficient of static friction between the block and the inclined planed is µ_s. A massless, inextensible string is attached to one end of the block, passes over a fixed pulley, pulley 1, around a second freely suspended pulley, pulley 2, and is finally attached to a fixed support. The pulleys are massless, but a second block of mass m₂ is hung from the suspended pulley. Gravity acts downwards.

Now assume that the block on the incline plane is sliding up the plane. The coefficient of kinetic friction is µ_k. Find the magnitude of the acceleration of the block on the inclined plane a__x1 Express your answer in terms of some or all of the variables m₁, m₂, theta, µ_k and g

Homework Equations

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F = ma

The Attempt at a Solution

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I choose coordinate systems. For block 1 I chose the positive X axis downward along the surface of the inclined plane and the positive Y axis up, perpendicular to the surface of the inclined plane with the x origin at the top of the incline. For block 2 I chose the positive Y-direction downwards with the origin at the ceiling.

I made free-body diagrams of both block 1, 2 and the suspended pulley which gave me the following equations:

I can treat the suspended pulley and block 2 as one system because the length between them is constant so they move with the same acceleration a_2

-T + µ_k m_1 gcos(theta) + m_1 gsin(theta) = m_1 a_x1
N = m_1 gcos(theta)
-2T + m_2 g = m_2 a_2

The length of the cord connecting block 1 with the ceiling is constant. So after taking two derivatives of the length. I get the relation between a_x1 and a_2 which is a_1 = -2a_2

T = ((m_2*g- m_2*a_2)/2 )

T = µ_k*m_1*g*cos(theta) + m_1*g*sin(theta) - m_1*a_1

((m_2*g-m_2*a_2 /2 ) = µ_k*m_1*g*cos(theta) + m_1*g*sin(theta) - m_1*a_1

2*(µ_k*m_1*g*cos(theta) + m_1*gsin(theta) - m_1*a_1 = (m_2*g-m_2*a_2)

2*(µ_k*m_1*g*cos(theta) + m_1*g*sin(theta)) - m_2*g = -m_2*a_2 + 2*m_1*a_1

substitute a_1 / -2 for a_2 and factor a_1 out

2*(µ_k*m_1*g*cos(theta) + m_1*g*sin(theta)) -m_2*g = (m_2/2 + 2*m_1) * a_1

a_1 = (2*(µ_k*m_1*g*cos(theta) + m_1*g*sin(theta))-m_2*g ) / (m_2/2 + 2*m_1)

For some reason this answer is not correct and I have been looking for that reason for several hours but I can't seem to find it. I think its a math error rather than a physics error. Can someone help me out?

 

[PhanthomJay]

Looks like you have signage errors. You have one block moving down the plane and the other hanging block moving downward. If the hanging block moves down the other moves up the plane, as stated. Acceleration of each block is in the direction of the net force.

 

[Valerion]

So the acceleration of block 1 has a negative sign and the acceleration of block 2 has a positive sign. Then,

-T + µ_k m_1 gcos(theta) + m_1 gsin(theta) = m_1 a_x1 should be -T + µ_k m_1 gcos(theta) + m_1 gsin(theta) = - m_1*a_x1 ?

 

[PhanthomJay]

Yes , and a2 is one half of a1.

 

[Valerion]

Ok, I think I got it. a_1 = (2*(µ_k*m_1*g*cos(theta) + m_1*g*sin(theta))-m_2*g ) / (-m_2/2 - 2*m_1)

 

[PhanthomJay]

Yes that looks correct. I find it easier to assume the positive direction in the direction of the acceleration, positive up incline for m1 and positive down for m2. Helps a bit with signage errors

 

[Valerion]

Thank you for your help!

 

[PhanthomJay]

Thank you for your help!

You're welcome.

 

[ThEmptyTree]

Hi. Has anyone made parts (a) and (b)?

(a) What is m2,min, the minimum value of m2 for which block 1 just barely slides up
the incline? Express your answers in terms of some or all of the variables m1, θ,
µs and g.
(b) What is m2,max, the maximum value of m2 for which block 1 just barely slides
down the incline? Express your answers in terms of some or all of the variables
m1, θ, µs and g.

I got m_2_max/min = 2*m_1*(sin(θ) +- µ_s*cos(θ)).

 

[haruspex]

Hi. Has anyone made parts (a) and (b)?

(a) What is m2,min, the minimum value of m2 for which block 1 just barely slides up
the incline? Express your answers in terms of some or all of the variables m1, θ,
µs and g.
(b) What is m2,max, the maximum value of m2 for which block 1 just barely slides
down the incline? Express your answers in terms of some or all of the variables
m1, θ, µs and g.

I got m_2_max/min = 2*m_1*(sin(θ) +- µ_s*cos(θ)).

Good.

 

[vcsharp2003]

Yes , and a2 is one half of a1.

Is ##a_2= \dfrac {a_1} {2}## because mass ##m_2## moves half the distance that mass ##m_1## moves in the same interval?

 

[ThEmptyTree]

Is ##a_2= \dfrac {a_1} {2}## because mass ##m_2## moves half the distance that mass ##m_1## moves in the same interval?

@vcsharp2003 that can be explained using "constrained motion", making use of the fact that the string is inextensible, so its length is constant. Writing the string's length in terms of motion functions of the objects and other constants, we get the identity. Would you like me to post the actual proof so you can understand better?

 

[vcsharp2003]

Would you like me to post the actual proof so you can understand better?

Yes, please.

 

[vcsharp2003]

@vcsharp2003 that can be explained using "constrained motion", making use of the fact that the string is inextensible, so its length is constant. Writing the string's length in terms of motion functions of the objects and other constants, we get the identity. Would you like me to post the actual proof so you can understand better?

I think I got it as shown below.

In analysis below, ##l_1##, ##l_2##, ##l_3## are instantaneous lengths of respective sections of the string, while
##a_1##, ##a_2## are the instantaneous accelerations of masses ##m_1## and
##m_2##.

I am not very clear why I wrote ##\dfrac {d^2l_1} {dt^2} = -a_1##. I know that ##l_1## is decreasing with time and so ##\dfrac {dl_1} {dt}## is negative.

 

[haruspex]

I am not very clear why I wrote ##\dfrac {d^2l_1} {dt^2} = -a_1##.

Because you are measuring a1 as positive up the slope but the distance of that mass from the pulley, l1, as positive down the slope.

 

[vcsharp2003]

Because you are measuring a1 as positive up the slope but the distance of that mass from the pulley, l1, as positive down the slope.

You mean the origin of the axis for the ##l_1## axis is at top pulley and I have assumed ##l_1## to be positive so the positive direction of axis is pointing down along the plane?

 

[haruspex]

You mean the origin of the axis for the ##l_1## axis is at top pulley and I have assumed ##l_1## to be positive so the positive direction of axis is pointing down along the plane?https://www.physicsforums.com/attachments/288104

I mean because an increasing value of l1 would mean the mass is sliding down, so implicitly you are measuring the displacement of the mass from the pulley as positive down the plane.

 

[vcsharp2003]

I mean because an increasing value of l1 would mean the mass is sliding down, so implicitly you are measuring the displacement of the mass from the pulley as positive down the plane.

I see.

I think it's very important to define the axis with origin and direction when analysing kinematic situations else signs can get messed up.

 

[vcsharp2003]

I mean because an increasing value of l1 would mean the mass is sliding down, so implicitly you are measuring the displacement of the mass from the pulley as positive down the plane.

From the differential analysis we get the relationship between instantaneous accelerations ##a_1## and ##a_2##, but why would we assume that these instantaneous accelerations are constant with time i.e. uniform?

 

[haruspex]

From the differential analysis we get the relationship between instantaneous accelerations ##a_1## and ##a_2##, but why would we assume that these instantaneous accelerations are constant with time i.e. uniform?

In which equation are you making that assumption?

 

[vcsharp2003]

In which equation are you making that assumption?

When applying Newton's second law to each of the masses.

We are treating acceleration of each mass as constant else we would end with acceleration as a function of time.

Or maybe we apply Newton's second law to instantaneous values of forces and acceleration and after we solve the simultaneous equations it turns out that tensions and accelerations are pure numbers and not a function of time.

 

[ThEmptyTree]

I think I got it as shown below.

In analysis below, ##l_1##, ##l_2##, ##l_3## are instantaneous lengths of respective sections of the string, while
##a_1##, ##a_2## are the instantaneous accelerations of masses ##m_1## and
##m_2##.

I am not very clear why I wrote ##\dfrac {d^2l_1} {dt^2} = -a_1##. I know that ##l_1## is decreasing with time and so ##\dfrac {dl_1} {dt}## is negative.

View attachment 288103

You are on the right track, to some extent. When tackling constrained motion problems, it is very important to clearly define coordinate axis for each object, as things can easily overcomplicate. Take a look, for example, at the system of pulleys I have attached, and think of how you would solve it.

In my approach I have chosen for object 1 an axis with origin at distance d away from pulley 1, and for object 2 an axis that is at height h below the ceiling. I denoted the small string between pulley 2 and object 2 with s. When considering the length of the string I have not neglected the part wrapped around the pulleys, which is also constant.

These steps are not necessary here, however in the future treating everything carefully will bring you benefits.

Good luck.

 

[haruspex]

When applying Newton's second law to each of the masses.

No, that law does not assume constant acceleration. It relates the net force at some instant to the resulting acceleration at that instant.
It will turn out that the accelerations are constant here because all the masses are constant and the forces will be proportional to the weight of the suspended mass.

 

[ThEmptyTree]

When applying Newton's second law to each of the masses.

We are treating acceleration of each mass as constant else we would end with acceleration as a function of time.

Or maybe we apply Newton's second law to instantaneous values of forces and acceleration and after we solve the simultaneous equations it turns out that tensions and accelerations are pure numbers and not a function of time.

When we write F=ma, we do not "assume" constant acceleration, but why we answer the question "what is the acceleration?". Here, because the forces are not time dependent, the acceleration also turns out to be constant. However there are applications where the forces are for example velocity dependent (drag forces) or x dependent (mass spring system) and will require solving a differential equation.

 

[vcsharp2003]

Here, because the forces are not time dependent, the acceleration also turns out to be constant

We can say that gravity forces are time independent but how can we say that tesion T is time independent?

 

[vcsharp2003]

No, that law does not assume constant acceleration. It relates the net force at some instant to the resulting acceleration at that instant.
It will turn out that the accelerations are constant here because all the masses are constant and the forces will be proportional to the weight of the suspended mass.

Since pulley mass problems always end up with constant accelerations, can we safely assume before diving into solving it that accelerations are constant?

 

[ThEmptyTree]

We can say that gravity forces are time independent but how can we say that tesion T is time independent?

The question that you are asking is a bit complicated.

First of all, in this problem the strings are considered massless and firctionless when wrapping around a pulley. Assume a piece of the string, denote its ends with A (left) and B (right). The forces acting on it are Ta (tension in point A oriented to the left) and Tb (tension in point B oriented to the right). Applying F=ma : Tb - Ta = ma. But m ~ 0 so Tb - Ta = 0, yielding Tb = Ta, for any piece of the string. The conclusion is that the tension is uniform in the string.

For a massive spring, the tension can be computed distance dependent, using differential analysis.

Now we only have to worry for the tension at the extreme points / where the string links to another object. Tension can be "extracted" from writing F=ma for the motion of each moving object: mass 1, mass 2 and pulley 2. You will end up with a system with 4 equations and 4 unknowns: T (tension in main string), T' (tension in little string), a1, a2. Solving for T will prove that it is constant, and it is a good exercise to try out.

 

[vcsharp2003]

Take a look, for example, at the system of pulleys I have attached, and think of how you would solve it.

Is the pulleys masses diagram a part of some question? It looks like a challenging problem and worth trying/learning from it.

We assume the positive direction of vertical axis to point downwards and origin to be at center of pulley P when looking at the masses 2 and 3.

If the mass 1 on left side moves down by x units then the pulley P will also move up by x units. Therefore, mass 1 and pulley P have same accelerations.

Regarding masses 2 and 3, it gets challenging. Let's say the mass 3 moves up by z units relative to pulley P, then mass 2 will move down by z units relative to pulley P. So relative to ground, mass 3 will move up by x + z units and mass 2 will move down or up by z- x or x- z units. So masses 2 and masses 3 will also not same accelerations as distances traveled by them are different in a certain time interval. We would therefore end up with 3 different accelerations as unknowns in this situation.

The solution is as shown below. Once we have ##a_1##, then by substitution in other equations we can determine ##a_2## and ##a_3##.

 

[ThEmptyTree]

Is the pulleys masses diagram a part of some question? It looks like a challenging problem and worth trying/learning from it.

We assume the positive direction of vertical axis to point downwards and origin to be at center of pulley P when looking at the masses 2 and 3.

If the mass 1 on left side moves down by x units then the pulley P will also move up by x units. Therefore, mass 1 and pulley P have same accelerations.

Regarding masses 2 and 3, it gets challenging. Let's say the mass 3 moves up by z units relative to pulley P, then mass 2 will move down by z units relative to pulley P. So relative to ground, mass 3 will move up by x + z units and mass 2 will move down or up by z- x or x- z units. So masses 2 and masses 3 will also not same accelerations as distances traveled by them are different in a certain time interval. We would therefore end up with 3 different accelerations as unknowns in this situation.

The solution is as shown below. Once we have ##a_1##, then by substitution in other equations we can determine ##a_2## and ##a_3##.

View attachment 288113

Yes, the system is from an MIT problem. If you are interested, I have a very good resource of problems and explanations from an MIT OCW.

The first 4 equations are OK, but the 5th one is RIP. We cannot consider the acceleration of a system of objects, only if we consider that to be a non-inertial frame and consider the fictitious force, which messes up the problem in this case.

I have attached a PDF with the full solution.

 

[vcsharp2003]

The first 4 equations are OK, but the 5th one is RIP.

I think that the equation (5) is ok because we have the system of pulley and the hanging masses in black box move up with acceleration ##a_1## relative to ground. Is that correct? The acceleration of internal parts of this system is not needed but just the acceleration of whole system.

 

[vcsharp2003]

If you are interested, I have a very good resource of problems and explanations from an MIT OCW.

Yes, I am interested.

 

[ThEmptyTree]

I think that the equation (5) is ok because we have the system of pulley and the hanging masses in black box move up with acceleration ##a_1## relative to ground. Is that correct? The acceleration of internal parts of this system is not needed but just the acceleration of whole system.

I understand your intuitive logic, but here's the thing.

Whenever you have objects that are considered massless (pulleys / strings), avoid grouping them into systems of objects, because you risk getting such contradictions. In this case it's the fault of pulley P which you have to treat separately, because it's massless. When you group it with other objects, you jump over using this fact. You can compare it to dividing by x in an inequality, but you don't know if x is positive or not, and you dilute that information and you get to something wrong.

As you can see, according to equation (8.6.69) (which I also arrived to when solving the problem), the final answer is not conclusive with yours.

This is mainly the problem with "ideal physics" where you have to make some assumptions, because otherwise it will be insanely hard to treat everything (ignore mass and friction for rope/pulley, ignore drag force in air falling, consider uniform density, sin x = x for the pendulum, also known as fundamental theorem of engineering). And when you make such assumptions sometimes they turn against you. Hope you understand it.

 

[ThEmptyTree]

Yes, I am interested.

https://ocw.mit.edu/courses/physics/8-01sc-classical-mechanics-fall-2016/

 

[vcsharp2003]

https://ocw.mit.edu/courses/physics/8-01sc-classical-mechanics-fall-2016/

Thankyou.

 

[vcsharp2003]

Whenever you have objects that are considered massless (pulleys / strings), avoid grouping them into systems of objects, because you risk getting such contradictions

I was comparing it to a plane that is moving on the runway. Passengers inside could be walking relative to the plane, but Newton's second law could be applied to the plane using the total mass of passengers and plane as the m term in F=ma equation.