What is the acceleration of a falling mass attached to a pulley system?

[conniechiwa]

A block of mass m1 = 3 kg rests on a table with which it has a coefficient of friction µ = 0.7. A string attached to the block passes over a pulley to a block of mass m3 = 5 kg. The pulley is a uniform disk of mass m2 = 0.4 kg and radius 15 cm. As the mass m3 falls, the string does not slip on the pulley.

With what acceleration does the mass m3 fall?
What is the tension in the vertical string, T3?

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys101/fall07/homework/09/07b/7.gif

I'm not sure how to do this...

 

[learningphysics]

It's like the regular pulley problems except that the pulley isn't massless... hence you have to take its moment of inertia into account.

net torque about the pulley is I*(angular acceleration).

 

[conniechiwa]

I found T1, which is 31.23 N, but I don't know how to find T3 from it since the tensions are not equal.

 

[learningphysics]

I found T1, which is 31.23 N, but I don't know how to find T3 from it since the tensions are not equal.

How did you get T1?

 

[conniechiwa]

umm i don't remember how i got T1, but that is the answer.

 

[learningphysics]

umm i don't remember how i got T1, but that is the answer.

ok. first get the acceleration of m1 to the right... that's the same acceleration that m3 has downward.

use that acceleration to get T3.

 

[conniechiwa]

I tried doing that, but I'm not getting the right answer:

T1-FF =ma
T1 - uk*N = ma
T1-uk*mg = ma
31.23N - (0.7*3kg*9.8m/s^2) = (10kg) a
a =1.065 m/s^2

 

[learningphysics]

I tried doing that, but I'm not getting the right answer:

T1-FF =ma
T1 - uk*N = ma
T1-uk*mg = ma
31.23N - (0.7*3kg*9.8m/s^2) = (10kg) a
a =1.065 m/s^2

why are you using 10kg?

 

[conniechiwa]

Oh oops, but even after using 3kg it doesn't work either..
T1-FF =ma
T1 - uk*N = ma
T1-uk*mg = ma
31.23N - (0.7*3kg*9.8m/s^2) = (3kg) a
a =3.55 m/s^2

 

[learningphysics]

Are you really sure about the 31.23 N? Did that answer get accepted by the computer?

 

[conniechiwa]

yeah the 31.23 N got accepted by the computer.

 

[conniechiwa]

i think what i did originally was substitute T3 in order to find T1 and I somehow got that answer and it worked. but i guess T1 and T3 are not equal.

 

[learningphysics]

i think what i did originally was substitute T3 in order to find T1 and I somehow got that answer and it worked. but i guess T1 and T3 are not equal.

"With what acceleration does the mass m3 fall?
What is the tension in the vertical string, T3?"

so the question didn't ask for T1... and you entered 31.23 for T3?

T1 and T3 are not equal... if you already have T3... then calculate the acceleration of m3 from that...

you just need the acceleration of m3 and T3 right?

 

[conniechiwa]

the question asked for T1 and I got it. But now I need T3 and m3.

 

[learningphysics]

the question asked for T1 and I got it. But now I need T3 and m3.

Your answer for T1 is close (letting it get accepted by the computer), but not exact which is why the other answers aren't going through... I recommend solving from scratch... and getting all the values from the beginning...