Calculating Kinetic Energy in a Pulley System

[Alt+F4]

A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.75. A string attached to the block passes over a pulley to a block of mass m3 = 4 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.


Pic Attatched

I have the free body diagram


Force of friction - T1 = Rotational energy of disk = mg-T3

I am trying to find T1, Acceleration and T3. do i need to find the total kinetic energy first of the system or can i somehow do it my way?? Thanks


Edit: Sorry about the title

 

[Da-Force]

A block of mass m1 = 2 kg rests on a table with which it has a coefficient of friction µ = 0.75. A string attached to the block passes over a pulley to a block of mass m3 = 4 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.


Pic Attatched

I have the free body diagram


Force of friction - T1 = Rotational energy of disk = mg-T3

I am trying to find T1, Acceleration and T3. do i need to find the total kinetic energy first of the system or can i somehow do it my way?? Thanks


Edit: Sorry about the title

Let's look at this. I assume its like the table situation, one mass attached to another and a pulley on the corner :-|

Draw 2 force-body diagrams for each.

Go ahead and label everything, force of friction, normal force, weight, and tension :-) That should simplify...

Note that the sum of all forces on a SINGLE force-body diagram must be zero...

Start with the one that falls to find tension, then figure out force friction, acceleration based on net force of the force-body diagram on block.

Hope that simplifies it a bit...?

 

[Alt+F4]

so i don't need to worry bout kinetic and all that other stuff

 

[Alt+F4]

Any help please, i tried ur idea and it is not the answer

 

[Hootenanny]

Okay, basically, you have a number of forces acting. You have the frictional force of m1 and the moment of inertia of the pully which will oppose the force generated by the weight of m3. Can you write down a formula to represent those variables? Don't forget, the tension is constant throughout the whole string.

-Hoot

 

[Alt+F4]

okay then, if tension is constant then why do i have to figure the tension for 1 and tension for 3? Are they basically the same answer.

So what u are trying to say is


Frictional Force - (moment of intertia of pully+Mass of M3*9.8) = 0

and just solve for Velocity, then that will help me get Acceleration?

 

[Hootenanny]

So what u are trying to say is

Frictional Force - (moment of intertia of pully+Mass of M3*9.8) = 0

No, what I'm saying is that the frictional force and the moment of inertia of the pully will oppose the weight of m3.

and just solve for Velocity, then that will help me get Acceleration?

No, using forces will give you the acceleration. I will start you off.

$$m_{3}g = F_{friction} + F_{pully}$$

Friction is given by $Fr = \mu R = \mu mg$. The moment of inertia of a disc is given by $\frac{1}{2}mr^2$.

Can you construct a formula from this information?

 

[Alt+F4]

i am soo sorry but i am lost, how is this formula going to help me find Acc or even tension. i am not getting where you want me to head. WIll torque get involved in this problem

 

[Hootenanny]

i am soo sorry but i am lost, how is this formula going to help me find Acc or even tension. i am not getting where you want me to head. WIll torque get involved in this problem

Its no problem, I'm not very good at explaining this type of question. For sum reason they always seemto confuse me It's probably best if we wait for someone else to asnwer you, because I'll end up confusing myself Sorry I've not been much help :shy:

 

[Hootenanny]

I may have figured it out. Here's my take on it;

As the particles are connected the acceleration must be equal, therefore

$$a = \frac{\Sigma F}{\Sigma m}$$

Can you sum the forces?

 

[Alt+F4]

nope that aint doing it, and i did some all forces + masses. :( I hate these kind of problems

 

[Alt+F4]

i still haven't figured it out, any help please. Does this involve torque cause i tried. Thanks a lot