Mechanics- connected particles

In summary: I was able to solve this problem thanks to the link you provided.The pulleys only act to change the direction of the tension in the connecting string.It should be obvious that the mass on the incline slides down the inline, making the mass on the horizontal floor move left … no way the system moves otherwise.Set up two net force equations like the ones I’ve set up previously, one for each mass.You need to start taking ownership of these problems.
  • #1
Shah 72
MHB
274
0
20210531_213908.jpg

I have no clue how to do this. Pls help
 
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  • #2
The pulleys only act to change the direction of the tension in the connecting string.

It should be obvious that the mass on the incline slides down the inline, making the mass on the horizontal floor move left … no way the system moves otherwise.

Set up two net force equations like the ones I’ve set up previously, one for each mass.

You need to start taking ownership of these problems.
 
  • #3
skeeter said:
The pulleys only act to change the direction of the tension in the connecting string.

It should be obvious that the mass on the incline slides down the inline, making the mass on the horizontal floor move left … no way the system moves otherwise.

Set up two net force equations like the ones I’ve set up previously, one for each mass.

You need to start taking ownership of these problems.
Thank you! The pulley ones are a bit confusing. I need to practice more on these kind of problems.
 
  • #4
skeeter said:
The pulleys only act to change the direction of the tension in the connecting string.

It should be obvious that the mass on the incline slides down the inline, making the mass on the horizontal floor move left … no way the system moves otherwise.

Set up two net force equations like the ones I’ve set up previously, one for each mass.

You need to start taking ownership of these problems.
 
Last edited:
  • #5
skeeter said:
The pulleys only act to change the direction of the tension in the connecting string.

It should be obvious that the mass on the incline slides down the inline, making the mass on the horizontal floor move left … no way the system moves otherwise.

Set up two net force equations like the ones I’ve set up previously, one for each mass.

You need to start taking ownership of these problems.
In this
F= m×a
T- 0.2× 20cos 30-10= 2a
T- mu R= 2a.
I still don't understand how to calculate a
 
  • #6
You are given that it takes 1 second for the floor mass to move 0.6m from rest. You should be able to determine the magnitude of acceleration of the floor mass with that info using a kinematics equation.

Both masses undergo the same magnitude of acceleration.

also,

forces for the mass on the incline ...

$m_1g\sin{\theta} - \mu_1 m_1 g\cos{\theta} - T = m_1a$

mass on the floor ...

$T - \mu_2 m_2g = m_2a$

Tension in the string is the same for both masses.
 
  • #7
skeeter said:
You are given that it takes 1 second for the floor mass to move 0.6m from rest. You should be able to determine the magnitude of acceleration of the floor mass with that info using a kinematics equation.

Both masses undergo the same magnitude of acceleration.

also,

forces for the mass on the incline ...

$m_1g\sin{\theta} - \mu_1 m_1 g\cos{\theta} - T = m_1a$

mass on the floor ...

$T - \mu_2 m_2g = m_2a$

Tension in the string is the same for both masses.
Thanks a lot. So I get a= 1.2m/s^2 and speed of box B = 1.2m/s
 
  • #8
skeeter said:
You are given that it takes 1 second for the floor mass to move 0.6m from rest. You should be able to determine the magnitude of acceleration of the floor mass with that info using a kinematics equation.

Both masses undergo the same magnitude of acceleration.

also,

forces for the mass on the incline ...

$m_1g\sin{\theta} - \mu_1 m_1 g\cos{\theta} - T = m_1a$

mass on the floor ...

$T - \mu_2 m_2g = m_2a$

Tension in the string is the same for both masses.
For q(d) tension is zero as the string breaks. So I need to calculate the acceleration for A
2a=10- 0.2× 20 cos 30 , a=3.27 m/s^2, s = 1-0.6= 0.4, v^2=u^2+2as, I get v =2.01m/s
 
  • #10

FAQ: Mechanics- connected particles

1. What is the definition of "connected particles" in mechanics?

Connected particles in mechanics refer to a system of two or more particles that are linked together by a rigid or flexible connection, such as a string, rod, or spring. The motion of one particle affects the motion of the other particles in the system.

2. What is the difference between a rigid and a flexible connection in connected particles?

A rigid connection is one where the distance between the particles remains constant, while a flexible connection allows for changes in distance between the particles. This means that a rigid connection restricts the possible motions of the particles, while a flexible connection allows for more freedom of movement.

3. How do you calculate the motion of connected particles in mechanics?

The motion of connected particles can be calculated using Newton's laws of motion and the principles of conservation of energy and momentum. The forces acting on the particles, such as tension, gravity, and friction, must also be taken into account.

4. Can connected particles have different masses and still be considered a system?

Yes, connected particles can have different masses and still be considered a system as long as they are connected by a rigid or flexible connection and their motions are influenced by each other. The masses of the particles will affect the overall motion of the system.

5. What are some real-world examples of connected particles in mechanics?

Some common examples of connected particles in mechanics include a pendulum, a swinging door, a car's suspension system, and a pulley system. In each of these examples, the motion of one particle is affected by the motion of the other connected particles in the system.

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