# Homework Help: Help on pulley problem!

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1. Jan 25, 2017

### soolights

<Mentor's note: moved from a technical section, therefore no template>

hey guys, i have a question about a tricky physics 1 problem our teacher gave us today, paraphrased as follows:

there is a mass M1 on a flat, frictionless horizontal table connected to a pulley, and on the other side of the pulley is another mass M2 the pulls downward (off the table). the system is frictionless. the tension of the string is T and the mass of the table mass is M1, so what is M2 in terms of M1 and T?

we have a test soon and i would really appreciate help :/

Last edited: Jan 25, 2017
2. Jan 25, 2017

### Arman777

You can use Newton's Laws to find an Solution.Draw a free body diagram for each object then equal them to the net force.I can try to help you If u give me something.You should try to do it yourself a bit at least

3. Jan 25, 2017

### soolights

i don't understand how to list these out with so few information. there's nothing given for acceleration, movement, or relationships between anything...?

Last edited: Jan 25, 2017
4. Jan 25, 2017

### Arman777

If you think theres no friction and both $M_1$ and $M_2$ are not moving..thats impossible I think.Does $M_1$ makes a circular motion ?

5. Jan 25, 2017

### soolights

yes i made a mistake, we don't know which way it moves

6. Jan 25, 2017

### Arman777

Whats this means ?

7. Jan 25, 2017

### Arman777

Ok I think $M_1$ makes a circular motion.If theres no friction this is the seems only possible way.Now What forces acting on $M_1$ and whats the total force on acting on $M_1$ ?

8. Jan 25, 2017

### soolights

no, it doesn't make a circular motion. we just don't know the acceleration of the system. everything is linear! (these problems should be solvable without a calculator)
the tension force is acting on M1 and pulling it toward the edge of the table, along with the gravitational weight force and normal force from the table. the net force for M1 is the tension force.

9. Jan 25, 2017

### Arman777

Ok,I understand,Now the idea is the Net force acting on the object should be equal the total force.and

$\vec F_{total}=m\vec a$ can you find an equation for M1 using this ?

10. Jan 25, 2017

### soolights

fnet = ma, so m = fnet/a? i don't know if this will help or not because it doesn't provide acceleration :/

11. Jan 25, 2017

### Arman777

As you said there's Tension acting on the Object, thats the net force , and this net force should be equal to Total force so...?

12. Jan 25, 2017

### Nidum

Deleted post .

13. Jan 25, 2017

### soolights

so the only force is the tension...but how does that link to the mass of m2 hanging off the table? m2's only force is the tension force holding it up..

14. Jan 25, 2017

### PetSounds

If M2 is pulling downward off the table, don't you already know its acceleration?

15. Jan 25, 2017

### soolights

oh gosh!!! since there's no friction and it pulls downward, it's just 9.81. fnet = t, so using fnet = ma, m2 = t/9.8???

16. Jan 25, 2017

### Nidum

Sorry but that is not correct .

Draw a diagram and show what forces are acting on m1 and m2 .

17. Jan 25, 2017

### soolights

this is the FBD i've been using--

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18. Jan 25, 2017

### Arman777

I am trying to that...M1 has T (thats the net force) and thats equal to M1a (total force) so,T=M1a.

You have to do this M2

Theres two forces acting on M2

19. Jan 25, 2017

### Nidum

Your diagram does not really show the forces acting .

20. Jan 25, 2017

### soolights

so, for tension and weight..?

21. Jan 25, 2017

### soolights

i'm not sure what else there is to put...normal, tension, gravitational forces are all there and i don't know of any other ones that are applicable?

22. Jan 25, 2017

### Arman777

yes,try to do what I did for M1

23. Jan 25, 2017

### Arman777

Write down the equation then you solved

24. Jan 25, 2017

### soolights

okay, so force in the y direction is both tension and gravity, so fnet = w-t (is necessarily w>t?). so w-t = m2a?

25. Jan 25, 2017

### Arman777

yep correct

Now we have two equations
T=M1a
M2g-T=m2a so...?