# Homework Help: Tension of a rope in a pulley system

1. Nov 12, 2009

### winthos

1. The problem statement, all variables and given/known data
A 1.0 kg mass and a 4.0 kg mass are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. What is the acceleration of the larger mass? What is the tension on the cord?

2. Relevant equations

F(net force)=F(tension)-weight

F=ma

3. The attempt at a solution
I found the acceleration using F=ma. The system has 3 kg of mass going down (4kg-1kg), which is equal to 29.4N of force (found by multiplying 3kg by 9.8m/s^2). The whole system has a mass of 5 kg, so I substituted in to F=ma.

29.4N=5.0 kg + a

The acceleration is 5.88 m/s^2

Now comes the part I am confused about. The tension of the cord should be
F(net force)=tension-weight, or rearranged to be:

F(tension)=F(net force)+weight or (Tension=ma+mg)

I can substitute in what i know:
Tension =m(5.88m/s^2)+m(9.8m/s^2)

I am at a loss of which mass to use. Do i use the mass that i found in working on acceleration (3 kg), or the mass of the whole system (5 kg)?

#### Attached Files:

• ###### forcediag.jpg
File size:
10.9 KB
Views:
2,184
Last edited: Nov 12, 2009
2. Nov 12, 2009

### indr0008

can u draw the force diagrams of each mass? that will help u a lot :)

3. Nov 12, 2009

### winthos

I attached a quick diagram, i don't know if that's detailed enough, but that is basically what's going on. (i'm bad at drawing with a mouse, so it probably doesn't look that spectacular)- the 4kg mass is pulling the whole system down at 5.88m/s^2, which is great since that is the first half of the question.

My problem is which mass do i use to find the tension? the mass of 3 kg (because of the pulley 4kg-1kg=3kg, the 3kg was then used to find the amount of downward force, which turned out to be 29.4N) or the mass of the whole system, 5kg?

Last edited: Nov 12, 2009
4. Nov 14, 2009

### indr0008

ok, u are done basically, u can use any of the masses actually, so let's say we concentrate on the heavier one

the heavy mass' equation of force will be:

-Mg+T=-Ma

-(4 kg)(9.8 m/s^2)+T=-(4kg)(5.88 m/S^2)

that's it, u've found it :)

can u find it now using the smaller mass? :)