# Pulley problem 1

1. Jul 23, 2013

### Anthonyphy2013

1. The problem statement, all variables and given/known data
An object of mass m1 =45kg sits on a horizontal frictionless table. A rope is attached to it , which runs horizontally to a frictionless pulley , then down to a hanging mass m2 = 5kg
a) the acceleration of each mass
b) the tension in the rope

2. Relevant equations
a1=-a2
y component : F net = mg-t= ma2
x -component: F net : T=Ma1
3. The attempt at a solution
I just wonder there are differeent acceleration and that realtionship of them was the negative sign only ?

2. Jul 23, 2013

### Staff: Mentor

The sign just depends on the definition of your coordinate system, it has no physical meaning.
The length of the rope is constant, therefore the magnitude of the acceleration is the same for both objects.

3. Jul 23, 2013

### Anthonyphy2013

you mean the magnitude of the acceleration depends on the length . Could it be dependent on the tension as well ? Lets say , one rope is pulling the mass on the table on a horizontal frictionless pulley with the acceleration and that pulley is hanging two rope to the hanging mass going down. So the acceleration of the hanging mass is -1/2 a.
Is that true ?

4. Jul 24, 2013

### haruspex

No, that's not what mfb wrote.
You have to decide, separately for vertical and horizontal, what your sign convention is.
Since you have m2g-T= m2a2, it looks like a2 is positive down (as it has the same sign as g). If in the horizontal positive is towards the pulley then you should have a1=a2.

5. Jul 29, 2013

### physcrazy

pulley

acceleration of the system as a whole (sign convenctions can be assigned to our wish.usually up,right +tive down,left -tive)

a=total force acting/total mass = 5 g/(5+45)=1 newton.

tension in the string is due to the its elastic forces.T=m1*a=45*1=45 newton
equations: m2g-t=m2a
m1a=t

Last edited: Jul 29, 2013
6. Jul 30, 2013

### technician

a = 1 m/ s2 not 1 Newton .......minor typing error!