# Pulley Problem

1. Mar 26, 2014

### Fresh Coffee

1. The problem statement, all variables and given/known data
Two masses are attached by a light string and looped over two frictionless pulleys. Object 1 (200g) accelerates upwards at a rate of 4.90 m/s squared. Determine mass of object 2.

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2. Relevant equations

3. The attempt at a solution
I tries doing this but don't think its right.

Fnet= F1 + F2
ma= m1 + m2
-m2g = m1g – ma
m2g=ma-m1g
m2g= m(a-g)
m2g/g=m(a-g)/g
m2=m(a)
m2=(.2kg)(4.90m/s 2)
m2=.98 kg or 980 grams

2. Mar 26, 2014

### dauto

Yes, that solution is not correct. The problem has only two masses but your equation "-m2g = m1g – ma" has three different masses in it. you must fix that. Besides, there is also a sign mistake on your net force calculation. Make sure you draw a diagram before attempting to solve any problem like that. Include all the relevant forces in your diagram.

Also (a-g)/g is not equal to a. That's just basic algebra

3. Mar 26, 2014

### paisiello2

You did not do this correctly.

Start by setting up two different equations by ∑F=ma for each mass.

4. Mar 26, 2014

### Fresh Coffee

Was anything right?

5. Mar 26, 2014

### paisiello2

The units look right.

6. Mar 26, 2014

### Fresh Coffee

Lol ok thanks. I shall endeavour to solve this enigmatic question.

7. Mar 27, 2014

### Fresh Coffee

Ok so would finding out the net force of object 1 help?

8. Mar 27, 2014

### paisiello2

It's a start.

9. Mar 27, 2014

### Fresh Coffee

Ok so I have set up two equations for each mass.

Fnet1=ma

Fnet2=ma

Fnet1=(.2kg)(4.90m/s squared)

Fnet1=.98 N

However I am stumped what to do with Fnet2. I go Fnet2=m(-4.90 m/s squared) then ma=m(-4.90m/s squared) but don't know what to do next.

10. Mar 27, 2014

### paisiello2

Did you do a free body diagram of each mass? Include all the forces acting on each mass. These will add up to Fnet for each case.

11. Mar 27, 2014

### Fresh Coffee

Yes but I can't seem to get much from that....I have free body diagram of object 1 and 2 which is a block with Fg acting down and Ft acting up.

12. Mar 27, 2014

### Fresh Coffee

I can't get it. I'v tried all sorts of ways but its that darn net force that I can't get! If I could just get that I'd have it.

13. Mar 27, 2014

### jbunniii

Are the gravity forces acting on object 1 and 2 the same? If not, then don't call them both Fg. What about Ft? Is that the same for both objects?

14. Mar 27, 2014

### paisiello2

Great, now sum them up to get Fnet.

15. Mar 27, 2014

### Fresh Coffee

mass of object 2 = 600 grams

16. Mar 27, 2014

### paisiello2

Yeah, that's good to know. But sum up all the forces you outlined above in an equation and set it equal to Fnet.

17. Mar 27, 2014

### Fresh Coffee

I don't need to. All I need is mass of object 2.

18. Mar 27, 2014

### paisiello2

that's not correct. You have to sum up all the forces and set it to ma.

19. Mar 27, 2014

### Fresh Coffee

20. Mar 27, 2014

### dauto

There is a minus sign missing in lines 1 through 3. Everything else seems correct. You should wait until the very last step before plugging in any values such as the m1 and g. Algebra is easier than arithmetic.