What Forces Affect the Motion in an Inclined Pulley System?

[azn_oohwee]

I am having a problem finding the last part of this problem, and I'm not really sure if I did the first part right.

The 1.0 kg physics book in Figure P8.38 is connected by a string to a 600 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are Us = 0.50 and uk = 0.20.

http://www.webassign.net/knight/p8-38.gif

(a) How far does the book slide?

(b) At the highest point, does the book stick to the slope, or does it slide back down?

If the book sticks, what magnitude of force along the incline is required to make it slide down? If the book slides, what magnitude of force along the incline is required to make the book stick?

So this is what I did.

For the cup
∑(Fc)y = T - mg = ma

For the book.
∑(Fb)y = N - mgcos(@) = 0
∑(Fb)x = -T - fk - mgsin(@) = ma
∑(Fb)x = -T - uk(m(b)*a + (m(b)*g) - m(b)sin(@) = m(b)a

I add the cup to the book.

-m(a)g - uk(m(b)*a + (m(b)*g) - m(b)sin(@) = m(b)a + m(c)a = a(m(b) + m(a))

-{ g(m(a) + uk(m(b)*a + (m(b)*g) - m(b)sin(@) } / { m(b) + m(a) = a

after solving for acceleration I use a to find delta x

Vf = 0, Vi = 3 so

0 = (3)^2 + 2a(deltaX)

deltaX = -9/2a

so that gives me (a)

For (b) I reason the book slides back down.

And for C I'm having trouble visualizing how to set up the problem

I'm setting it up like this

Components to the left:

Tension
Ma
Mgsin@

Components to the right:

static friction
and the force?


I just need a little help on this one. Anything will be greatly appreciated. Thanks.

 

[ehild]

The 1.0 kg physics book in Figure P8.38 is connected by a string to a 600 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are Us = 0.50 and uk = 0.20.

...

So this is what I did.

For the cup
∑(Fc)y = T - mg = ma

For the book.
∑(Fb)y = N - mgcos(@) = 0
∑(Fb)x = -T - fk - mgsin(@) = ma
∑(Fb)x = -T - uk(m(b)*a + (m(b)*g) - m(b)sin(@) = m(b)a

I do not understand your last equation. Why did you replace friction by uk(ma +mg)? Friction is N*uk.

ehild

 

[azn_oohwee]

oops sorry..

I meant to write..

∑(Fb)x = -T - fk - mgsin(@) = ma

∑(Fb)x = -T - ukN - mgsin(@) = ma

N = mgcos@ so

∑(Fb)x = -T - ukmgcos@ = ma

-m(a)g - uk(m(b)gcos@ - m(b)gsin@ = m(b)a + m(c)a

a = - { g(m(c) - uk(m(b)cos@) - m(b)sin@ } / {m(b) + m(c) }

That should be more correct.

I have the answers to the problems now. The last question was a bit vague on the wording, i'll have to ask my professor about that one.

(a) 0.65m
(b) slide back down
(C) 4.63N