Finding Acceleration in a Pulley and Slope System

In summary, the problem is trying to find the forces on a massless pulley. The equation of torque is used to solve for the torque on the pulley. The two unknowns are the height of the pulley above the ramp and the angle at which the string is pulled.
  • #1
Jazz
103
5

Homework Statement



Moment_of_Inertia_1.jpg


NOTE: This is a problem I found and the question is actually the part B of it. That of ''massless pulley'' I suppose was dealing with the part A (which I don't know what question was), so please don't consider that (:.

Moment_of_Inertia_3.jpg


Homework Equations



##F_{net} = ma##
##\tau = I\alpha##

The Attempt at a Solution



I followed this reasoning:

I started focusing on the pulley.

- The force acting on it are the tensions due to ##m_1## and that due to ##m_2##, so ##F_{net} = T_2 - T_1##.

- Since the pulley is massive, there is a torque there, so ##net\ \tau = I\alpha##. This torque is caused by the forces of tension acting on the pulley, then: ##T_2r - T_1r = I\alpha##.

- In this case ##T_1## is the pulling force exerted on ##m_1## to accelerate it upward: ##m_1g + m_1a##.
- ##T_2## is the pulling force exerted on ##m_2## to prevent it from accelerating at ##g##: ##m_2g - m_2a##.

- Replace this into the equation of torque:

##(m_2g - m_2a)r - (m_1g + m_1a)r = I\alpha##

- We know that ##\alpha = \frac{a}{r}##, so:

##(m_2g - m_2a)r - (m_1g + m_1a)r = I\frac{a}{r}##

And here I'm stuck. I have two unknowns: ##a## and ##r##. I suppose the two heights and the angle are given to be used to solve the problem but I don't know how.

The other thing I've been thinking is that, at ##B##, the vertical component of ##F_{net}## has two components: one perpendicular to the slope and the other parallel to it, but I don't know how to figure out the angle that the string makes with the ground in order to find the vertical component of ##F_{net}## and on the other hand, if I'm able to find it, the torque wouldn't be taken into account I guess.
 
Last edited:
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  • #2
Jazz said:
- ##T_2## is the pulling force exerted on ##m_2## to prevent it from accelerating at ##g##: ##m_2g - m_2a##.
How do you get that for T2? m1 and m2 do not have the same acceleration, m2's acceleration is not in the vertical direction, and you've omitted the normal force from the slope.
 
  • #3
haruspex said:
How do you get that for T2? m1 and m2 do not have the same acceleration, m2's acceleration is not in the vertical direction, and you've omitted the normal force from the slope.

Sorry for my late reply.

Here is my diagram:

Moment.jpg


True. From the fact that the pulley needs a net torque for it to rotate and hence that both tensions must not be in balance, the acceleration is not the same.

Well, I must start again. This is what I think:

- Forces on ##m_1##:

##F_1 = T_1 - w_1##

- Forces on ##m_2##:

Here I'm a little confused. It seems that I need to figure out the forces along the axes in order to find ##F_2##:

##F_{2(y)} = w_2 - (N\cdot \cos(\theta) + T_2\cdot \cos(\beta))##

##F_{2(x)} = N\cdot \sin(\theta) - T_2\cdot \sin(\beta)##

But I got that angle ##\beta## there |:

- Forces on the pulley:

##T_2r - T_1r = I\alpha##

I think there will be a ##F_{net}## on ##m_2## making it accelerate at ##\frac{F_{net}\cdot \sin(\theta)}{m_2}##, but I still don't know what I'm missing.
 
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  • #4
Seems to me there's not enough information. There's nothing that fixes the height of the pulley above the ramp. You can easily work out the answer for two extreme case (1. rope parallel to ramp; 2. with the pulley at great height, the rope effectively vertical at all times) and see that the results are different.
 
  • #5
I would like to try this case:

Moment4.jpg


In my new diagram the forces are:

- On ##m_1##:

##F_1 = T_1 - w_1##

- On ##m_2##:

##F_{2(y)} = w_1 - (N\cdot \cos(\theta) + T_2\cdot \sin(\theta))##

##F_{2(x)} = N_2\cdot \sin(\theta) - T_2\cdot \cos(\theta)##

- On the pulley:

##I\alpha = (T_2-T_1)r\hspace{35pt}(1)##

Trying to find the tensions:

##T_1 = w_1 + F_1##

##T_2 = w_2\cdot \sin(\theta) - T_1##

##T_2 = w_2\cdot \sin(\theta) - (w_1 + F_1)##

Replace into ##(1)##:

##I\alpha = (w_2\cdot \sin(\theta) - (w_1 + F_1)-(w_1 + F_1))r##

##I\frac{a}{r} = (w_2\cdot \sin(\theta) - 2(w_1 + F_1))r##

I have another ##a## within ##F_1##, but I still have that ##r## there that don't know how to get rid of.
 
  • #6
You need to bring the accelerations of the masses into it to get rid of F1, F2. The relationship between the angular and linear accelerations will get rid of the r.
You have a couple of typos above. Your first m2 equation references w1. Your last equation has an r on the right which should not be there.
 
  • #7
Jazz said:
I would like to try this case:

Moment4.jpg


In my new diagram the forces are:

- On ##m_1##:

##F_1 = T_1 - w_1##

- On ##m_2##:

##F_{2(y)} = w_1 - (N\cdot \cos(\theta) + T_2\cdot \sin(\theta))##......(1)

##F_{2(x)} = N_2\cdot \sin(\theta) - T_2\cdot \cos(\theta)##.........(2)

Hi, Jazz. I would like modify your eq(1) and eq(2). I will write ##F_{2(y)}## as ##m_{2}a_{2y}## and ##F_{2(x)}## as ##m_{2}a_{2x}##.

So ##m_{2}a_{2y}= w_1 - (N\cdot \cos(\theta) + T_2\cdot \sin(\theta))##

and ##m_{2}a_{2x}=N_2\cdot \sin(\theta) - T_2\cdot \cos(\theta)##.
I did it just to include the expression of accelerations.

Now we have to relation between ##a_{1}, ##a_{2y}## and ##a_{2x}## and ##\alpha##.
Moment_of_Inertia_3.jpg

Let ##L## be the distance between pulley and ##m_{1}## when the ##m_{2}## is at B. Let ##l_{1}## be the distance between ##m_{2}## and pulley. Now, can you find the horizontal and vertical displacement of ##m_{2}## from initial position in terms of ##l_{1}##? Differentiate the expressions twice wrt time.

Now in this type pf problem we assume that length of the string is constant. Let ##L_{0}## be the total length of the string.
So ##l_{1}+L=L_{0}##. What will you get if you differentiate this equation twice wrt time?

Use these two equation and find relation between ##a_{1}##, ##a_{2y}## and ##a_{2x}##.

As no information is given so I assume that the strings do not slip on the pulley. So from this assumption can you find relation between ##\alpha## and ##a_{1}##.
 

Related to Finding Acceleration in a Pulley and Slope System

What is a massive pulley and slope?

A massive pulley and slope is a simple machine that consists of a large wheel (pulley) and a sloping surface. It is used to change the direction of a force and can also be used to multiply the force applied.

How does a massive pulley and slope work?

When a force is applied to the rope or string wrapped around the pulley, the direction of the force changes as it is redirected along the slope. The larger the diameter of the pulley, the more effective it is at changing the direction of the force. The slope also helps to multiply the force applied.

What are the advantages of using a massive pulley and slope?

One advantage is that it can change the direction of a force, making it easier to lift or move heavy objects. It can also multiply the force applied, making it possible to lift objects that would otherwise be too heavy to lift. Another advantage is that it is a simple machine, meaning it is easy to understand and use.

What are some examples of how a massive pulley and slope is used?

A common example is in a construction site, where a pulley and slope system is used to lift heavy materials to higher levels. Another example is in rock climbing, where a pulley and slope system can assist in pulling a climber up a steep face. In physics experiments, a massive pulley and slope can be used to demonstrate concepts such as work, energy, and momentum.

What factors affect the efficiency of a massive pulley and slope?

The efficiency of a massive pulley and slope depends on factors such as the angle of the slope, the friction between the rope and the pulley, and the weight of the object being lifted. The larger the angle of the slope, the less efficient the system is. Additionally, the greater the friction, the less efficient the system will be. Lastly, the heavier the object being lifted, the more force will be required to lift it, making the system less efficient.

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