# Homework Help: Why is the acceleration zero?

1. Apr 22, 2015

### Brynhildur

I have an inclined pulley with inclination 30°. There is mass A on top of it equals 2 kg. And over a frictionless pulley there is a mass B with weight 0,83 kg.
The text says that the coefficient of friction is 0,18 between mass A and the incline.
I need to know the acceleration. and I also have to find the tension? in the string between the masses over the pulley.
I know the answer is zero, but I'm having trouble understanding why? Is it because the mass of object B is too small to move the system as in over come the friction?
The equations I'm using are
Fþverkraftur*=m*g*sin(θ)
Ffriction = μ*m*g*cos(θ)
Ftension - mB*g=mb*a

*This is icelandic, I dont know the english term for it, I´m sorry :)

2. Apr 22, 2015

### PhanthomJay

You have to crank out the numbers to see if the block A on the incline moves down the incline , up the incline, or remains stationary. Assume motion one way or the other or assume stationary and draw free body diagrams of each block to see what happens when applying newtons laws. Note that tension forces cannot be pushing forces. I also wish I could translate Icelandic for you.

3. Apr 23, 2015

### Staff: Mentor

First thing to determine: Does it move? Imagine the blocks starting from rest. How much of a force must the tension overcome to accelerate block A? What's the maximum value of tension in the string?

4. Apr 23, 2015

### Brynhildur

So I've done a calculation and I'm not really sure if it's what I am looking for, or really how to interpret it
Ffriction=μN
ΣFx=T-Ffriction-Fþverkraftur=0
ΣAFy=N-Fg⋅yA=0
Σ T-Fg⋅B= 0

FgA=mA*g*sin(θ)-mag*cos(θ)
FyB= mB*g

T-Ffriction-mA*g*sin(θ)=0
N-mAg*cos(θ)
T-mBg=0 → T=mBg

mb*g-Ffriction-mAg*sin(θ)=0
Ffriction=mBg-mAg*sin(θ)
N=mAg*cos(θ)

μN=μ*mAg*cos(θ)= Ffriction

__________________
1. Ffriction=g(mB-mA*sin(θ))= 9,8(0,83-2*sin(30))=-1,66 N
2. Ffmax = μ*mAg*cos(θ) = 0,18*2*9,8*cos(30)= 3,055

The first one is the frictional force needed to prevent the block from sliding. The second is the maximum frictional force. Since the first is less than the second that means the blocks do not move and that is why there is zero acceleration? And the tension in the string is equal to the gravitational pull of mass B? 8,134 N?

I hope this is understandable but I can't seem to even google what I want to know, well how to put it in english and icelandic sites don't give me any information, so am I on the right track at least? :)

5. Apr 24, 2015

### Staff: Mentor

You're on the right track, but your reasoning is a bit difficult to follow.

Answer this question. If there were no friction, which way would mass A accelerate? Up the incline or down the incline? That will help you determine the direction of the friction force that must be overcome.

6. Apr 25, 2015

### dean barry

Calculate the net force between the two blocks, is this greater than the friction force ? if not then no movement will take place.

7. Apr 26, 2015

### Brynhildur

Yes, that is exactly what I was looking for!! Thank you all very much!

8. Apr 26, 2015

### haruspex

Google translate gives "transverse force", which is reasonable, but it would be clearer to say force parallel to the plane, or force down the plane.