# Two masses sliding down a slope-help with the different forces.

1. Aug 29, 2011

[PLAIN]http://img221.imageshack.us/img221/541/slope.png [Broken]
1. The problem statement, all variables and given/known data
two masses sliding down a slope[slope cant move] like in the attached picture.
there is a static friction 'c' between the two masses. and between mass a and the slope there isnt any friction.
so im trying to figure out the equation of each mass , please tell me whether Im right or wrong.

Ma=mass of body a f=static friction force= Nb*c
Mb=mass of body b
Na=normal force of body a
Nb=normal for of body b
a= mass A and mass B acceleraiton

2. Relevant equations

forces on mass b :
f+Mb*g*sin(alpha)=Mb*a
Mb*g*cos(alpha)=Nb
forces on mass a:
Ma*g*sin(alpha) - f =Ma*a
(Ma+Mb)*cos(alpha)*g=Na

3. The attempt at a solution

thanks for the help

Last edited by a moderator: May 5, 2017
2. Aug 29, 2011

### Yuqing

Consider the situation where there is no friction between mass a and b. Is there any relative motion between the two?

3. Aug 29, 2011

no, they'll both have the same accelaration of g*sina(alpha),
but here u do have friction..
so my equations arent correct?

4. Aug 29, 2011

### Yuqing

Friction only acts when there will be relative motion between the two objects.

5. Aug 30, 2011

i dont that true with static friction...
static friction is when there arent relative motion between the two objects..
thanks

6. Aug 30, 2011

### PeterO

OK, I think the distinction trying to be made is that when you a frictionless surface, you will get no friction force, no matter what is happening.

When we have a surface where static friction is possible, we need to consider if it is needed or not.

If you place a block on a table while tilting the table a few degrees, you will soon see if there was the possibility of friction.
If yes, a friction force will probably stop the block sliding down the slope.
If no, there will be no friction force holding the block back, and the block will accelerate down the slope.

If we now adjust the table to be level:

No matter what the surface, there will be no friction force, as there is no friction force needed - the block is going to just sit there anyway.

With the two blocks in your problem, although there is the potential for a friction force between the two blocks, there is no necessity for a frictional force, so there will be no frictional force

7. Aug 30, 2011

ok , lets say we add an additional force on mass A with the same directions as its movment[someone pushes it..],
so what would be the equations now on each masses? would it be the same equations that I written plus the additional componants of the new force?

8. Aug 30, 2011

### PeterO

I couldn't actually follow your original equations nor comment on their validity.

In you original post you said "im trying to figure out the equation of each mass"

I wondered exactly what equations you may have meant??

Peter

9. Aug 30, 2011

the newton force equations ...
the two masses will slide down the slope because of forces [ gravity , friction ...]
so did i get the equations right?

10. Aug 30, 2011

### PeterO

Clearly body A won't be moving due to friction, as the surface is frictionless.

Which force(s) does "..." stand for?

List the force, and the reason for it this time.

eg: gravity - the Earth pulls the mass vertically down.

now the rest.

I am not sure what you mean by newton force equations.

All force equations involve newtons as that is the unit of force. Or did you mean Newton force equations - meaning equations based on Newtons laws of motion?

11. Aug 31, 2011

umm nevermind this question. I'll try to ask about something more simple .
[which help me understand my original question]
lets say there are two boxes with masses m1 and m2 , m2 sits on tops of m1.
and m1 lets say sits on a table . they arent moving . no friction between m1 and m2 and between m1 and the table.
and newtwos laws of motion equations are :
m2 : N2 - m2*g=0
m1 : N1 - m1*g -N2 =0
this is from physics text book, and what I really dont get is why they put N2 into consideration in m1 equation? isnt the equation suppose to be :
N1- m1*g-m2*g = 0?
i really dont get it.

12. Aug 31, 2011

### PeterO

m2 : N2 - m2*g=0 so N2 = m2*g

So presumably either could be listed in the expression for m1??