# PhysicsMechanics- connected particles

#### skeeter

##### Well-known member
MHB Math Helper
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.

• topsquark and Shah 72

#### Shah 72

##### Member
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.

#### Shah 72

##### Member
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:

#### Shah 72

##### Member
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

#### skeeter

##### Well-known member
MHB Math Helper
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.

• Shah 72

#### Shah 72

##### Member
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

#### Shah 72

##### Member
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

• skeeter

#### skeeter

##### Well-known member
MHB Math Helper
have a look at the link ...

• Shah 72