# 2 Masses, A pulley, Friction and an incline problem

1. May 24, 2004

### undertow2005

i'm trying to teach myself how to do this for hours and have come up with very little. here's the problem:

An inclined plan making an angle of 45 degress with the horizontal has a frictionless pulley on its top. A 3-kg block(call it B) on the plane is connected to a freely hanging 4-kg block(call it A) by means of a cord passing over the pulley(bascially the 4-kg block is hanging and the 3-kg is on the inclined plane). Find the magnitude and direction of acceleration of the moving system of the two blocks. Coefficient of friction between the block B and incline plane is .6. assume the cord and pulley are massless.

any help with this problem would be greatly, greatly appreciated.

2. May 24, 2004

### AKG

Draw free body diagrams, and figure out the forces on each block. The hanging block is subject to the force of gravity, and the tension in the cord. The block on the incline is subject to the normal force, the force of gravity, the tension in the cord, and the force of kinetic friction. Note, both masses must accelerate at the same rate, so if the acceleration is X, then the net force on the hanging block must be 4X [down] and the net force on the other block must be 3X [up the plane]. You should be able to easily find the force of gravity on both blocks, the normal force (from the force of gravity) and the force of friction (from the normal force). All that is left unknown would be the tension, which is the same on both objects (although in different directions). So, you'll have a few equations and a few unknowns, and should be able to solve this easily.

3. May 24, 2004

### Gokul43201

Staff Emeritus
Here are the equations. Draw the free body diagrams and you'll see why they are right.

For B :

Normal reaction, N = mgcos45 = 0.707mg
T("tension") - (mu)*N - mgsin45 = ma
Therefore, T = ma + 0.707mg + (0.6*0.707mg) |||| --(1)

For A :
Mg - T = Ma
So, T = M(g-a) |||| --(2)

From (1) & (2) : m[a + 0.707g + (0.6*0.707g)] = M(g-a)

Substitute numbers for m, M, and g to find 'a'

IMPORTANT NOTE : If you don't draw the force diagrams, you won't see why this works, and hence will not gain anything from all this.

4. May 25, 2004

### undertow2005

i dont quite get where u got this equation, T("tension") - (mu)*N - mgsin45 = ma. is it somethig that i should just memorize. this is so frustioning

5. May 25, 2004

### speg

Those are the forces acting on the block on the plane.
F=ma
The forces are: ( with + up the incline, - down the inclline ):

+Tension from the rope
-(mu)*N this is the force of friction
-mgsin45 this is the force of gravity

I've been reviewing these kind of problems this weekend and I know how they can be frustrating, making nice neat, thourough free body diagrams helps a lot! Hope this helped.

6. May 25, 2004

### arildno

undertow:
It seems to me that you are still struggling a bit to master the concepts in this exercise (i.e., your last reply)
It has been emphasized how important it is to draw a free body diagram here, and that remains fundamental in order to solve the problem.
However, it is also very important that you understand the following 2 points:

a) A massless cord implies that the tension in the cord is constant throughout the cord

b) Constant cord length implies that object B's acceleration up the inclined plane must equal object A's acceleration downwards.

I would like to argue in some detail on these subjects:

a)
(We will neglect the effect of friction here)
Let us call the tensile force acting on object B $$\mathcal{T}_{B}$$

Clearly, due to Newton's 3.law, the object B imparts a reaction force onto the rope, $$-\mathcal{T}_{B}$$
where I have included the opposite direction of the force by the minus sign.

Consider now a cord segment, s; for simplicity, let s lie completely on the inclined plane.
In reality, of course, no cord is massless, so let s' mass be denoted $$m_{s}$$
The component of the force of gravity along the inclined plane is
$$-m_{s}g_{p}$$

Now, upon the segment s, the rest of the cord imparts a tensile force (dragging it along); we call that $$\mathcal{T}_{c}$$

Hence, Newton's 2.law for the segment s reads:
$$\mathcal{T}_{c}-\mathcal{T}_{B}-m_{s}g_{p}=m_{s}a_{s}$$
Clearly, if we make the approximation $$m_{s}=0$$,
and require that the acceleration remains finite, we get:
$$\mathcal{T}_{c}=\mathcal{T}_{B}$$

This shows, that for a massless cord, we must have equal tension throughout the cord.
In particular, the tensile forces acting upon objects A and B must be equal in magnitude.

b)
Clearly, in order that the cord length is to remain constant,the part of the cord lying on the inclined plane must shorten with the same amount as the part of the cord in falling lengthens.
But this implies equality in B's upwards acceleration along the plane and A's downwards acceleration.

Last edited: May 25, 2004
7. May 25, 2004

### undertow2005

is my thought process correct here

for Mass A: Fnet= Ma*-T=Ma*A

For Mass B: Fnet=T-Fparallel-Ffriction
now, the Fparallel and the Ffriction are both negitive because they are going in the oppsite direction of the moving system, correct? i do know that T is positive because it's going in the same direction as the moving system.

8. May 25, 2004

### arildno

Do you with Fparallell mean the component of the force of gravity along the inclined plane?

I don't quite understand the terms in your 1.equation; could you explain them?

9. May 25, 2004

### undertow2005

For Mass B: Fnet=T-Fparallel-Ffriction
if i'm correct Fparallel means massB*g*SinX

the whole equation i think changes to this,
Fnet=T-massB*g*sinX-M*massB*g*cosX

Last edited by a moderator: May 25, 2004
10. May 25, 2004

### arildno

Yes, that's what I thought, and it is the correct expression for the forces acting on massB. (if M is the coefficient of friction)

However, I am still uncertain as to what you mean by your equation regarding object A.

11. May 25, 2004

### Staff: Mentor

Don't you mean: Ma*g - T=Ma*A,
where A is acceleration, Ma the mass of object "A"? (a bit confusing)