# Particle dynamics, incline

• Strawer
In summary: F=maThe relevant forces acting on m2 is friction, f=\muN = \mu mgTension T=m2gGravitational component, Fx=sin(37)mgAnd so \mu mg - sin(37)mg + m2g = mag( \mu - sin(37) + m2 ) = aa = g( \mu - sin(37) + m2 ) In summary, we have two blocks with equal masses connected by a pulley. With a coefficient of friction of 0.25, we can find the acceleration of the system when block m1 is moving down and when block m1 is moving up

## Homework Statement

Two blocks with equal masses m1=m2=5kg are connected via a pulley, as shown. Take $\mu$ = 0.25. Find the acceleration given that a) m1 is moving down, b)m1 is moving up. If m2=6 kg for what values of m1 will the pair move att constant speed?

## Homework Equations

F=ma, f=$\mu$N

## The Attempt at a Solution

I've managed to solve a) and c) but not b).
I don't understand how m1 can be moving up, since m1=m2 and the force m2 affects m1 with is sin(37)*m2g < m1g.
If m1 is moving up doesn't that mean that it will not affect the other mass? I mean shouldn't the tension on m2 be zero? If so then we have $\Sigma$F=5*9.8*sin(37)-0.25*9.8*cos(37)*5=19.7
Which gives a=19.7/5=3.94 m/s^2 However it should be 2.94 m/s^2 according to the answer sheet.

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Strawer said:
I don't understand how m1 can be moving up, since m1=m2 and the force m2 affects m1 with is sin(37)*m2g < m1g.

No. I highly recommend that you draw a free body diagram for m2, and then another free body diagram for m1. Apply Newton's 2nd law, namely that ƩF = ma. You should be able to solve for a. It doesn't matter that it is impossible for m2 to pull m1 up by gravity alone. All that matters is that you're told that right now m2 happens to be moving down the incline, while m1 is moving upward. These are the initial conditions you've been given. (Maybe somebody applied a force to make this happen. Who cares? It doesn't matter, just accept this as the initial condition).

Indeed, if you choose the "down the incline" direction to be positive in your force balance equations for m2, you'll find that you get a negative value for a, meaning that although m2 is presently moving down the incline, its acceleration points up the incline, meaning that it is slowing down and will eventually reverse direction.

Strawer said:
If m1 is moving up doesn't that mean that it will not affect the other mass? I mean shouldn't the tension on m2 be zero?

No, the fact that m1 is moving up doesn't mean that the rope is going slack. As I said before, it means that m2 is moving down the incline while m1 moves upward (and the rope remains taut).

cepheid said:
No. I highly recommend that you draw a free body diagram for m2, and then another free body diagram for m1. Apply Newton's 2nd law, namely that ƩF = ma. You should be able to solve for a. It doesn't matter that it is impossible for m2 to pull m1 up by gravity alone. All that matters is that you're told that right now m2 happens to be moving down the incline, while m1 is moving upward. These are the initial conditions you've been given. (Maybe somebody applied a force to make this happen. Who cares? It doesn't matter, just accept this as the initial condition).

Indeed, if you choose the "down the incline" direction to be positive in your force balance equations for m2, you'll find that you get a negative value for a, meaning that although m2 is presently moving down the incline, its acceleration points up the incline, meaning that it is slowing down and will eventually reverse direction.

No, the fact that m1 is moving up doesn't mean that the rope is going slack. As I said before, it means that m2 is moving down the incline while m1 moves upward (and the rope remains taut).

Experience tells me that if I would give m1 a push the rope would go slack. Or is that because i accelerate it? Also wouldn't the rope go slack if v1 > v2 because s=vt and if the rope is taut then s1=s2?

Anyways, new try
ƩF=ma

The relevant forces acting on m2 is

friction, f=0.25*cos(37)*5*9.8
Tension T=5*9.8
Gravitational component, Fx=sin(37)*5*9.8

And so
f-Fx+T=ma
a=(0.25*cos(37)*5*9.8-sin(37)*5*9.8+5*9.8)/10=2.93 m/s^2

Strawer said:
Experience tells me that if I would give m1 a push the rope would go slack. Or is that because i accelerate it?

That's true. An example of the adage, "you can't push on a rope." However, nobody is saying that you (or anyone else) is pushing on m1. All that is being said is that, presently, m1 is moving upwards, whilst m2 is simultaneously moving down the ramp. That's it. Do you understand?

Yeah, that's right. I busted out a "whilst." :rofl:

Strawer said:
Also wouldn't the rope go slack if v1 > v2 because s=vt and if the rope is taut then s1=s2?

Once again, you're not wrong. Any upward advancement of m1 must be accompanied by an advancement of m2 down the ramp by an equal distance. That, in fact, is precisely what is happening (i.e. this is the situation you are asked to consider). It's just the exact reverse of what happens when m1 falls down, and m2 consequently gets pulled up the ramp. The difference is that the latter situation can happen spontaneously, while the former cannot.

Strawer said:
Anyways, new try
ƩF=ma

The relevant forces acting on m2 is

friction, f=0.25*cos(37)*5*9.8
Tension T=5*9.8

This part in red is not correct. If the tension were just equal to the weight, then the net force on m1 would be zero, and it would have zero acceleration. However, if m1 had 0 acceleration, then the whole system would have zero acceleration. It doesn't.

To get T, you need to do the force balance for m1, which tells you that T - mg = ma.

Strawer said:
Gravitational component, Fx=sin(37)*5*9.8

And so
f-Fx+T=ma
a=(0.25*cos(37)*5*9.8-sin(37)*5*9.8+5*9.8)/10=2.93 m/s^2

It's a fluke that you got the right answer here. You made a second mistake that negated the first one that I mentioned above (in red). Your second mistake was to assume that the 'm' in the ma term was the total system mass. However, since the force balance equation you wrote above was for m2 only, it should be that m = 5 kg, not 10 kg.

Without this second mistake, you see that your answer is too large by a factor of 2.

Also, I'd really recommend keeping everything algebraic, until the last possible step, and not plug in numbers until you've simplified things as much as possible. If you'd done that, you'd have seen that 'm' canceled from both sides of the equation, making it unnecessary to plug in the 5 kg. Furthermore, you'd see that g factors out of every term on the left-hand side.

Anyway, if you correct your first mistake (the one in red) then you'll have this problem nailed.

I would like to clarify a few things about the problem and your solution attempt. First, it is important to note that in this system, m1 and m2 are not directly connected to each other. They are connected via a pulley, which means that they are indirectly connected through the tension in the string. This tension is what allows the blocks to move together.

Now, in part b) of the problem, m1 is indeed moving up. This means that the tension in the string will be pulling m2 upwards, causing it to accelerate in the same direction as m1. This also means that m2 will be exerting a downward force on the pulley, which will in turn affect m1.

Your attempt at solving the problem is on the right track, but there are a few errors. First, the force of friction should be calculated using the normal force on m1, not m2. This normal force is equal to m1g, not m2g. Second, the tension in the string should not be considered as a separate force acting on m1. Instead, it should be included in the net force acting on m1, along with the force of friction and the weight of m1.

With these corrections, the correct solution for part b) should be:

\sumF = m1a = T - \mu(m1g) - m1g = 0

T = (1 + \mu)m1g = (1 + 0.25)(5kg)(9.8m/s^2) = 61.25 N

Therefore, the acceleration of the system in part b) is:

a = T/m1 = 61.25N/5kg = 12.25 m/s^2

As for part c) of the problem, the blocks will move at constant speed when the net force on the system is equal to zero. In other words, when the tension in the string is equal to the weight of m1. This means that:

T = m1g

So, when m2 = 6kg, the value of m1 that will result in the system moving at constant speed is:

m1 = T/g = (6kg)(9.8m/s^2) = 58.8 kg

In conclusion, as a scientist, I would recommend double-checking your calculations and taking into account all the relevant forces acting on the system

## 1. What is particle dynamics?

Particle dynamics is the study of the motion and behavior of particles in a system, such as their position, velocity, and acceleration, under the influence of various forces. This field is used to understand and predict the movement of particles in different environments, from microscopic particles in biological systems to large-scale particles in engineering applications.

## 2. What is an incline in particle dynamics?

An incline, also known as a ramp or slope, refers to a surface that is at an angle or inclined in relation to the horizontal plane. In particle dynamics, an incline is an important factor to consider when studying the motion of particles as it affects their acceleration and velocity, especially in systems where gravity is the dominant force.

## 3. How does an incline affect particle dynamics?

The presence of an incline can change the direction and magnitude of the force acting on a particle, which in turn affects its motion. The steeper the incline, the greater the force of gravity pulling the particle down, resulting in a faster acceleration and increased speed. Additionally, the angle of the incline can also change the direction of the force, causing the particle to move in a curved path instead of a straight line.

## 4. What is the role of friction in incline particle dynamics?

Friction is a force that opposes the motion of particles on an incline. It is caused by the contact between the surface of the incline and the particles. Friction can either slow down or speed up the motion of particles, depending on the direction of the force. Inclines with a rough surface tend to have more friction, which can significantly affect the motion of particles.

## 5. How are inclines used in real-world applications?

Inclines are commonly used in various real-world applications, such as ramps in wheelchair accessibility, ski slopes, and conveyor belts in factories. Inclines are also used in physics experiments to study the effects of gravity and other forces on the motion of particles. Understanding the dynamics of particles on an incline is crucial in designing and optimizing these systems for efficient and safe performance.