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Hi everyone here is the picture of the this plane/friction problem.

http://phasedma.com/uploaded/Physics problem.JPG

The question asks what are the minimum and maximum values of m1 in the figure to keep the system from accelerating...

Take µk = µs = 0.50

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Ok when drawing my free body diagrams I have come up with this method of solving the problem... tell me if you agree.

The force of friction can either be up or down the slope, if m2 = 0 or sufficiently small, then m1 would tend to slide down the plane. so Ffr would be directed Up the incline.

We know that newtons second law for the y direction (i chose my xy coordinate axes along the plane-- IE horizontal x being the plane) shows that

Fnormal - m1*g*cos(30) = m1*ay = 0

since theres no y motion Fnormal = m1*g*cos(30)

Now for the x motion... For the first case (smallest m1) f = ma shows that

m1*g*sin(30) - Ftension - Ffr = m1*ax <---- x direction

since we want ax to be 0, we can solve Ftension since thats related to m2.

Since Ffr can be AT MOST µs * Fnormal= µs*m1*g*cos 30 the minumum value m2 can have to prevent motion (ax = 0) is (after dividing by g)

m2 = m1 * sin(30) - µs*m1*cos(theta).

And then finding the max value wouldnt be much more difficult from there since we already set up our equations..

Am I correct here? If you are willing can someone work it --- what range do you get for the mass?

Thanks for you help. Mechanics gets soooo tough!

http://phasedma.com/uploaded/Physics problem.JPG

The question asks what are the minimum and maximum values of m1 in the figure to keep the system from accelerating...

Take µk = µs = 0.50

------------------------

Ok when drawing my free body diagrams I have come up with this method of solving the problem... tell me if you agree.

The force of friction can either be up or down the slope, if m2 = 0 or sufficiently small, then m1 would tend to slide down the plane. so Ffr would be directed Up the incline.

We know that newtons second law for the y direction (i chose my xy coordinate axes along the plane-- IE horizontal x being the plane) shows that

Fnormal - m1*g*cos(30) = m1*ay = 0

since theres no y motion Fnormal = m1*g*cos(30)

Now for the x motion... For the first case (smallest m1) f = ma shows that

m1*g*sin(30) - Ftension - Ffr = m1*ax <---- x direction

since we want ax to be 0, we can solve Ftension since thats related to m2.

Since Ffr can be AT MOST µs * Fnormal= µs*m1*g*cos 30 the minumum value m2 can have to prevent motion (ax = 0) is (after dividing by g)

m2 = m1 * sin(30) - µs*m1*cos(theta).

And then finding the max value wouldnt be much more difficult from there since we already set up our equations..

Am I correct here? If you are willing can someone work it --- what range do you get for the mass?

Thanks for you help. Mechanics gets soooo tough!

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