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- #1

- Mar 1, 2012

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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.

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- #3

Thank you! The pulley ones are a bit confusing. I need to practice more on these kind of problems.

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.

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- #4

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:

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- #5

In this

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.

F= m×a

T- 0.2× 20cos 30-10= 2a

T- mu R= 2a.

I still don't understand how to calculate a

- Mar 1, 2012

- 935

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 ...

$T - \mu_2 m_2g = m_2a$

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- #7

Thanks a lot. So I get a= 1.2m/s^2 and speed of box B = 1.2m/s

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.

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- #8

For q(d) tension is zero as the string breaks. So I need to calculate the acceleration for A

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.

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

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- #10

Thank you so much!