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

[mickellowery]

Homework Statement

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 [tex]\Theta[/tex]=30^o 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.

Homework Equations

[tex]\Sigma[/tex]F_x=T-[tex]\mu[/tex]_k+Tcos30=ma
[tex]\Sigma[/tex]F_y=m₁g+m₂g-Tsin30=ma
[tex]\Sigma[/tex][tex]\tau[/tex]=RT₁-RT₂=I[tex]\alpha[/tex]

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.

The Attempt at a Solution

 

[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.

 

[kuruman]

Issues you need to address.

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 μ_k, it is μ_kN 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.

 

[mickellowery]

OK I think I'm onto something a little closer, but I'm sure I've still got a few mistakes.
[tex]\Sigma[/tex]F_1x=T₁-[tex]\mu[/tex]_km₁g=m₁a
[tex]\Sigma[/tex]F_2y= -T₂sin30-[tex]\mu[/tex]_k(m₂g)cos30=m₂a
[tex]\Sigma[/tex]F_2x= -T₂cos30-[tex]\mu[/tex]_k(m₂g)cos30=m₂a
[tex]\Sigma[/tex][tex]\tau[/tex]=RT₁-RT₂=1/2MR²[tex]\alpha[/tex]

 

[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.

 

[mickellowery]

OK I should've realized that one. So will the equations be
[tex]\Sigma[/tex]F_2y= T₂-[tex]\mu[/tex]_k(m₂g)sin30=m₂a
[tex]\Sigma[/tex]F_2y=T₂-[tex]\mu[/tex]_k(m₂g)cos30=m₂a
I was thinking that there should still be a sin30 because of the normal force. Is this a correct thought?

 

[kuruman]

OK I should've realized that one. So will the equations be
[tex]\Sigma[/tex]F_2y= T₂-[tex]\mu[/tex]_k(m₂g)sin30=m₂a
[tex]\Sigma[/tex]F_2y=T₂-[tex]\mu[/tex]_k(m₂g)cos30=m₂a
I was thinking that there should still be a sin30 because of the normal force. Is this a correct thought?

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.

 

[mickellowery]

OK I hate this problem! I have both T and [tex]\mu[/tex]_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:
[tex]\Sigma[/tex]F_2y= -T₂-[tex]\mu[/tex]_km₂g=m₂a
[tex]\Sigma[/tex]F_2x= -T₂-[tex]\mu[/tex]_km₂g=m₂a

 

[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.

 

[mickellowery]

Alright I have:
a_x=mgsin30
a_y=mgcos30
T_x=-T
T_y=0
N_x=0
N_y=N
F_kx=[tex]\mu[/tex]_kN
F_ky=0

 

[kuruman]

Alright I have:
a_x=mgsin30
a_y=mgcos30

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?

T_x=-T
T_y=0
N_x=0
N_y=N

These are correct.

F_kx=[tex]\mu[/tex]_kN
F_ky=0

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.

 

[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?

 

[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.

 

[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.