Homework Help: Rigid object under net torque: 2 blocks, pulley, on incline

1. Jun 17, 2010

mickellowery

1. The problem statement, all variables and given/known data
A block with mass m1=2.00kg is on a flat surface and a block of mass m2=6.00kg is on an incline and they are connected by a massless string over a pulley in the shape of a solid disk with a radius R=0.250m and mass M=10.0kg. The angle of the incline is $$\Theta$$=30o and the blocks are allowed to move. The coefficient of kinetic friction is 0.360 for both blocks. Determine the acceleration of the two blocks and the tensions in the string on both sides of the pulley.

2. Relevant equations
$$\Sigma$$Fx=T-$$\mu$$k+Tcos30=ma
$$\Sigma$$Fy=m1g+m2g-Tsin30=ma
$$\Sigma$$$$\tau$$=RT1-RT2=I$$\alpha$$

I think I'm accounting for all of the forces I've come up with in my free body diagram, but I'm not sure if I should have a force of friction in the y direction.

3. The attempt at a solution

2. Jun 17, 2010

jwxie

The given said both blocks experience frictions (same friction value). One is on the incline, and one is on the flat surface.

Now think about a simple case with just a block on the incline. If Fk is present, where would the direction of the Fk go? There are many possibilities: (1) horizontal or vertical direction (up, down, right, left), or (2) along the incline (again, could be up, down, right, left). Only one is true. The rest are just apples and oranges.

Similarly, the one on the flat surface. Which direction would the Fk mostly likely be?

:) Then you will complete your free body diagram for all forces. Now recheck your equations.

3. Jun 17, 2010

kuruman

1. You don't have enough equations.
The mass on the flat surface generates one equation, the mass on the incline generates two and the pulley generates one.

2. Your notation is all messed up. When you write ma, which "m" is this? Same thing with the symbol "T" in the first two equations.

3. The force of kinetic friction is not just μk, it is μkN where N is the normal force. Each mass has its own normal force.

4. Define a coordinate system for each mass. Only then will "x component" and "y component" make sense.

Fix all of the above and then we will see what you come up with.

4. Jun 17, 2010

mickellowery

OK I think I'm onto something a little closer, but I'm sure I've still got a few mistakes.
$$\Sigma$$F1x=T1-$$\mu$$km1g=m1a
$$\Sigma$$F2y= -T2sin30-$$\mu$$k(m2g)cos30=m2a
$$\Sigma$$F2x= -T2cos30-$$\mu$$k(m2g)cos30=m2a
$$\Sigma$$$$\tau$$=RT1-RT2=1/2MR2$$\alpha$$

5. Jun 17, 2010

kuruman

Things would be easier for you if, for m2 you picked the x-axis down the incline and the y axis perpendicular to the incline. Then the acceleration will have only an x-component. The first and fourth equations look OK.

6. Jun 17, 2010

mickellowery

OK I should've realized that one. So will the equations be
$$\Sigma$$F2y= T2-$$\mu$$k(m2g)sin30=m2a
$$\Sigma$$F2y=T2-$$\mu$$k(m2g)cos30=m2a
I was thinking that there should still be a sin30 because of the normal force. Is this a correct thought?

7. Jun 17, 2010

kuruman

Did you follow my hint and take the x-direction down the incline and the y direction perpendicular to the incline? If so, what are the x and y components of T2? What about the friction and the acceleration? Draw a drawing and see in what direction all these vectors are.

8. Jun 17, 2010

mickellowery

OK I hate this problem!! I have both T and $$\mu$$k going in the negative X direction and the normal force will be in the positive y direction but then do I have to worry about mg having X and Y components? Or would the equations just be:
$$\Sigma$$F2y= -T2-$$\mu$$km2g=m2a
$$\Sigma$$F2x= -T2-$$\mu$$km2g=m2a

9. Jun 17, 2010

kuruman

You did not answer my question, but let's assume that the x-axis is down the incline and the y-axis is perpendicular to it. Please answer the following questions:

1. What is the component of the acceleration along x?
2. What is the component of the acceleration along y?
3. What is the component of the tension along x?
4. What is the component of the tension along y?
5. What is the component of the normal force along x?
6. What is the component of the normal force along y?
7. What is the component of friction along x?
8. What is the component of friction along y?

Eight questions - eight answers. Provide the answers first, then write the equations.

10. Jun 17, 2010

mickellowery

Alright I have:
ax=mgsin30
ay=mgcos30
Tx=-T
Ty=0
Nx=0
Ny=N
Fkx=$$\mu$$kN
Fky=0

11. Jun 18, 2010

kuruman

If the y-axis is perpendicular to the incline and there is a component of acceleration in that direction, this means that the velocity will change in that direction. In other words, the block will either sink into or fly off the incline. Does either one of these things happen?
These are correct.
Is friction positive or negative,i.e. in the same or opposite direction as tension?

In the same vein, you will need to figure out the components of the weight.

12. Jun 18, 2010

mickellowery

OK I see what you're saying but then how do I account for the fact that there is gravitational acceleration not in the x or y plane? would it be x=mgsin30 and y= -mgcos30?

13. Jun 18, 2010

kuruman

You have to find the components of m2g in a direction down the incline and perpendicular to the incline. Be sure to put minus signs where they belong.

14. Jun 18, 2010

mickellowery

Yea stupid mistake. Thanks so much for all your help. I'm completely physics illiterate, and I think that it's abundantly clear that I too know nothing.