How Does Mass Loss Affect Acceleration in Atwood's Machine?

[ScreamingIntoTheVoid]

Homework Statement

The figure shows Atwood's machine, in which two containers are connected by a cord (of negligible mass) passing over a frictionless pulley (also of negligible mass). At time t = 0 container 1 has mass 1.3 kg and container 2 has mass 2.7 kg, but container 1 is losing mass (through a leak) at the constant rate of 0.24 kg/s. At what rate is the acceleration magnitude of the containers changing at (a)t = 0 and (b)t = 3 s?

Homework Equations

f=ma[/B]

The Attempt at a Solution

**NOTE, a is not 0**
T= m1g - m2g = 26.46-12.74= 13.72
4a=13.72 (**4 is the mass of the entire system**) a= 3.43 m/s^2

Try 1:
d/dt 3.43 + 0.24t --> 0.24 m/s^3 for both a and b (which is obviously wrong)

Try 2: *Lets find a at t=3*
1.3 - 0.24(3)= 0.58

T= 26.46 - 5.684= 20.776
(3.28)a = 20.776 --> a= 6.331097561

(6.331097561m/s^2-3.43m/s^2)/3s = 0.9670325203 m/s^3 (which was also wrong)

Help please? I'm not looking for answers, I just want to know how to do this. Thank you in advance!

 

[barryj]

a = f/m You might consider first determining the force as a function of time and then the mass as a function of time and doing the division.

 

[ScreamingIntoTheVoid]

a = f/m You might consider first determining the force as a function of time and then the mass as a function of time and doing the division.

Thank you so much for responding.
So I got this:

a= [(26.46N) - (1.3kg - 0.24t)(9.8)]/ [(1.3kg - 0.24t) + 2.7]
a= [13.72N -2.352t]/[4 - 0.24t]

Assuming I've done this correctly, to find the rate of change of acceleration, would I just derive this and that would get me my answer or is there still something I'm forgetting?

 

[barryj]

I think you are good so far. You can do the derivative thing or use your calculator.

 

[ScreamingIntoTheVoid]

think you are good so far. You can do the derivative thing or use your calculator..

Cool! I found the acceleration at t=0 to be 3.43 m/s^2 and at t=3 to be a=2.031707317. Are you sure I didn't do something wrong though? If the mass of 1 is constantly increasing, shouldn't my value at t=3 be larger than my value at t=0?

 

[barryj]

You have the acceleration as a function of time, a= [13.72N -2.352t]/[4 - 0.24t] so just take the derivative and evaluate at t = 0 and t = 3.

Your numbers look reasonable for t = 0.

 

[ScreamingIntoTheVoid]

You have the acceleration as a function of time, a= [13.72N -2.352t]/[4 - 0.24t] so just take the derivative and evaluate at t = 0 and t = 3.

Your numbers look reasonable for t = 0.

Ok. I did that with my calculator and got -0.3822000014 m/s^3 at t=0 and -0.5684116628 at m/s^3

 

[barryj]

Try again. Graph the function and then use the derivative feature 2nd,calc,6 dy/dx and see what you get (if you are using a ti-84)

 

[TSny]

You have the acceleration as a function of time, a= [13.72N - 2.352t]/[4 - 0.24t]

Check the sign in the numerator.

 

[ScreamingIntoTheVoid]

Try again. Graph the function and then use the derivative feature 2nd,calc,6 dy/dx and see what you get (if you are using a ti-84)

Ok. When I did that it still gave me a negative slope.

 

[ScreamingIntoTheVoid]

Check the sign in the numerator.

When I put that into the calculator I got the rates to be 1.1805 m/s^3 (t=3) and 0.7938 m/s^3 (t=0). That got marked correct, but out of curiosity how did you get a positive in the numerator?

 

[ScreamingIntoTheVoid]

double post

 

[ScreamingIntoTheVoid]

TSny and barryj thank you both for taking time to help me out!

 

[TSny]

When I put that into the calculator I got the rates to be 1.1805 m/s^3 (t=3) and 0.7938 m/s^3 (t=0). That got marked correct, but out of curiosity how did you get a positive in the numerator?

a= [(26.46N) - (1.3kg - 0.24t)(9.8)]/ [(1.3kg - 0.24t) + 2.7]
a= [13.72N - 2.352t]/[4 - 0.24t]

"Two negatives make a positive"

- (1.3 - 0.24t) = -1.3 $\mathbf +$ 0.24t