Sliding problems on a pulley

Homework Statement

A 10.0 g mass is tied to a string. The string is attached to a 500.0 g mass and stretched over a pulley , leaving the 10.0 g mass suspended above the floor. Determine the time it will take the 10.0 g mass to hit the floor if it starts out 1.50m above the floor.

Homework Equations

I've set up the problem, but it seems almost impossible. There's just simply not enough info to solve it. You are missing : Force tension , Force Friction , and the overall acceleration.

The Attempt at a Solution

My attempt isn't correct.

If someone could just tell me how to set it up in a simple way that would be great thanks

ehild
Homework Helper
Could you please show a picture of the arrangement of the masses and pulley? Is the bigger mass on a horizontal table?

ehild

gneill
Mentor
Isolate each mass and draw a free-body force diagram. Write the acceleration of each in terms of the tension in the string (an unknown force at this point).

Now, since they are connected by the (assumedly light, inextensible) string, the accelerations must be equal in magnitude. Solve for the tension.

What can you do once you've got the tension?

I honestly don't know, I was going to use the acceleration to find the time. In order to find the acceleration you need Force of friction; which I don't know how to find.

This is what I have once it is simplified :

Massb*g - Ff = a(Mass b + Mass A)

Which to find acceleration I need to know Ff. Once I know that I can plug the acceleration
into Vf^2 = Vi^2 + 2ad

Once I have found the Final Velocity I then can find time

Vf=Vi +at

gneill
Mentor
I honestly don't know, I was going to use the acceleration to find the time. In order to find the acceleration you need Force of friction; which I don't know how to find.

This is what I have once it is simplified :

Massb*g - Ff = a(Mass b + Mass A)

Which to find acceleration I need to know Ff. Once I know that I can plug the acceleration
into Vf^2 = Vi^2 + 2ad

Once I have found the Final Velocity I then can find time

Vf=Vi +at

So what happens if Ff = 0?

Then it would be unbalanced and accelerate based on the force of tension

gneill
Mentor
Isn't that what is expected?

Hmm, true, I guess I didn't think about that, so if Ff = o then I can solve it like the way I have above ?

I'm going to make another attempt at it, when ever I'm done could you check my answer?

gneill
Mentor
Hmm, true, I guess I didn't think about that, so if Ff = o then I can solve it like the way I have above ?

Well, to be fair you didn't show your work as to how you arrived at the acceleration expression, but the expression itself looked okay.

Alright, well I got 4 secs technically 3.9 . If you could check it I would really appreciate it.

gneill
Mentor
Alright, well I got 4 secs technically 3.9 . If you could check it I would really appreciate it.

Can you show your work? In particular, show the final expression and value you obtained for the acceleration, and your calculation of the time.

For Mass A aka 500G

the equation what setup like this :

Ft=MassA * a
(Orginally Ft-Ff=MassA * a ; til I knew to just forget friction)

Mass B was set up like so :

Mass B * G - Ft =Mass B * a

Substituting Mass A for Ft : Mass B * G - Mass A * a = Mass B * a
Arranged all acclamations on one side and all Gravity's on one side :
Mass B * G = Mass B * a + Mass A * a
Then factored
Mass B*G = a(Mass B + Mass A)
Found the acceleration and did the following :

Massb*g = a(Mass b + Mass A)

into Vf^2 = Vi^2 + 2ad

Once I have found the Final Velocity I then can find time

Vf=Vi +at

gneill
Mentor
Sorry, I'm not seeing your final value for the acceleration. Can you give me a number?

Taking the Mass B * G = A(Mass B + Mass A)

Mass B = 10g = 0.01 kg

Mass A = 500G = 0.5 kg

0.01*9.8 = A ( 0.01 + 0.5)
0.098 = A 0.51
A = 0.19 m/s^2

Vf^2 = Vi^2 + 2 ad
Vf^2 = o^2 + 2 0.19(1.50)

Vf = 0.76

Then use this formula
Vf = Vi + a T
0.76= 0+ 0.19(t)
T= 4 seconds

gneill
Mentor
Okay.

You can get to the time a bit quicker by going via d = (1/2)*a*t2, so that

$$t = \sqrt{\frac{2d}{a}}$$