Question with slope, masses, pulley

[mystmyst]

[PLAIN]http://img413.imageshack.us/img413/4452/mechanicsdiagram2.png

Homework Statement

There is no friction between mass c and the slope. The pulley and rope have no mass. The coefficient of friction between the two masses on the slope is \mu_1

Get an expression for the acceleration of the system of masses when there is no relative motion between the two masses on the slope.

The Attempt at a Solution

There is something I don't get: If there is not relative motion between the two masses on the slope, then each mass has a different acceleration. The question makes it seem that I should still find a total acceleration of the entire system but I don't think that is possible.

Also, even though b and c have different accelerations, c and d still have the same acceleration, correct?

So, the way I would do it is find the acceleration for b, and find the acceleration for c and d. Is that correct or is there really one unifying acceleration?

 

[Delzac]

I think you are very confused here.

Firstly, b and c, if there is no relative motion between them, they must have the same velocity and acceleration.

Now, they can have the same acceleration but different force acting on them.
F = ma, they have different masses, hence they have different magnitude of force acting on them. This is because you need more force to accelerate something that is heavier (to be precise, more inertia).

Also, the acceleration of d is the same as acceleration of b and c. d cannot accelerate more than b and c since the rope cannot stretch, or break for all that matter.
Acceleration of d cannot be less that b and c, if not the string will not be taut.

delzac

 

[mystmyst]

I think you are very confused here.

Firstly, b and c, if there is no relative motion between them, they must have the same velocity and acceleration.

ooo. I thought relative motion means they are moving together at the same velocity and acceleration. So if there is no relative motion they are moving at different velocities and different accelerations.

Thanks for clearing that up.

 

[mystmyst]

so basically the only force acting on the system is the weight from mass d. so:

m_dg = (b+c+d)a

correct?

and if there was friction between the mass and the slope it would be:

m_dg - f_k = (b+c+d)a

 

[Delzac]

You have to take into account that the the force is not one sided, meaing, b and c also provides a force in the opposite direction.

The difference in the 2 forces will determine if the mass will move in either direction.

 

[mystmyst]

o you i forgot about the mg\cos \alpha and mg\sin \alpha stuff...
thanks

 

[mystmyst]

When I am calculating the net force, do I also include the force from the small mass (with mass b):

bg\sin \alpha

or I don't since the friction cancels it out?

as in either:

dg - cg\sin \alpha = (b+c+d)a

or

dg - cg\sin \alpha - bg\sin \alpha= (b+c+d)a

 

[Delzac]

Well, you will have to take into account both mass, since both b and c is tugging on the rope. You can imagine that there are glued together, so their mass will have to add up.

 

[mystmyst]

Well, you will have to take into account both mass, since both b and c is tugging on the rope. You can imagine that there are glued together, so their mass will have to add up.

I know I take their mass into account. that's why i have (b+c+d)a

but my question is referring to the net force. Do I take into account the force from the smaller mass or since the friction cancels it out (which is why it is not moving) I only take into account the large mass.

 

[Delzac]

Like i said, b and c is as good as glued together, meaning you should treat them as one object.

 

[mystmyst]

gotcha!

Thanks.