Free Body Diagram and Newton's 2nd

[odie5533]

Homework Statement

http://img527.imageshack.us/img527/2125/diagramkl0.png

The Attempt at a Solution

My attempt is in the picture. I am trying to get the diagram done before I continue on because if I mess up in the diagram, my work is guaranteed to be wrong. So basically what I am asking is is my diagram correct?

NOTE: the diagram in the bottom right was drawn by the professor.

 

[PhanthomJay]

When you draw a FBD of the system as a whole (both blocks together), you must identify only the weight of the system and the forces external to your FBD. The friction forces between the two blocks or normal force between the 2 blocks do not enter into the FBD of the system.

 

[odie5533]

http://img225.imageshack.us/img225/6476/diagram2yj3.png
Is that right?

 

[PhanthomJay]

http://img225.imageshack.us/img225/6476/diagram2yj3.png
Is that right?

Yes, more or less. You have the horizontal and vertical components of the applied force F, and the weight...which you should identify as the weight of BOTH blocks...and the Normal force of the floor on the bottom block. Your FBD should look more like your professor's diagram with those forces identified on it.

 

[odie5533]

Ok, I drew it on my paper. So for B:
\sum F = Fcos\theta = (m_{1} + m_{2})a
a = \frac{Fcos\theta}{m_{1} + m_{2}}

 

[odie5533]

And C:
\sum F_{m1} = Fcos\theta - \mu_{s}m_{1}g = m_{1}a
\sum F_{m2} = \mu_{s}m_{1}g = m_{2}a
a = \frac{\mu_{s}m_{1}g}{m_{2}}
Fcos\theta - \mu_{s}m_{1}g = \frac{m_{1}^2\mu_{s}g}{m_{2}}
Fcos\theta = \frac{m_{1}^2g}{m_{2}}\mu_{s} + m_{1}g\mu_{s}
\mu_{s} = \frac{Fcos\theta}{\frac{m_{1}^2g}{m_{2}} + m_{1}g}

 

[PhanthomJay]

Ok, I drew it on my paper. So for B:
\sum F = Fcos\theta = (m_{1} + m_{2})a
a = \frac{Fcos\theta}{m_{1} + m_{2}}

Looks good!

 

[PhanthomJay]

And C:
\sum F_{m1} = Fcos\theta - \mu_{s}m_{1}g = m_{1}a
\sum F_{m2} = \mu_{s}m_{1}g = m_{2}a
a = \frac{\mu_{s}m_{1}g}{m_{2}}
Fcos\theta - \mu_{s}m_{1}g = \frac{m_{1}^2\mu_{s}g}{m_{2}}
Fcos\theta = \frac{m_{1}^2g}{m_{2}}\mu_{s} + m_{1}g\mu_{s}
\mu_{s} = \frac{Fcos\theta}{\frac{m_{1}^2g}{m_{2}} + m_{1}g}

A very nice attempt with your FBD's, but you fell a bit short. The friction force is mu_s times the normal force. You have not correcty identified the normal force between the 2 blocks. And don't lose sight of what the problem is asking for. It gets a tad confusing with so many letters and oh so few numbers.

 

[odie5533]

Had class about 3 hours ago (turned this in with the wrong answer). At least it wasn't lonely, I had a whole paper with wrong answers to keep it company.

Normal_{m2/m1} = m_{1}g + Fsin\theta
F_{friction} = \mu_{s}n = \mu_{s}(m_{1}g + Fsin\theta)

\sum F_{m2} = F_{friction} = m_{2}a
a = \frac{F_{friction}}{m_{2}}\sum F_{m1} = Fcos\theta - F_{friction} = m_{1}{a}
Fcos\theta - F_{friction} = m_{1}\frac{F_{friction}}{m_{2}}
F = \frac{\frac{m_{1}F_{friction}}{m_{2}} + F_{friction}}{cos\theta}

It still looks wrong, and I'm still laughing at me solving for \mu_{s} and turning that in.

 

[PhanthomJay]

Had class about 3 hours ago (turned this in with the wrong answer). At least it wasn't lonely, I had a whole paper with wrong answers to keep it company.

Normal_{m2/m1} = m_{1}g + Fsin\theta
F_{friction} = \mu_{s}n = \mu_{s}(m_{1}g + Fsin\theta)

\sum F_{m2} = F_{friction} = m_{2}a
a = \frac{F_{friction}}{m_{2}}


\sum F_{m1} = Fcos\theta - F_{friction} = m_{1}{a}
Fcos\theta - F_{friction} = m_{1}\frac{F_{friction}}{m_{2}}
F = \frac{\frac{m_{1}F_{friction}}{m_{2}} + F_{friction}}{cos\theta}

It still looks wrong, and I'm still laughing at me solving for \mu_{s} and turning that in.

Your ability to draw FBD's is exceeded only by your abilty to use LaTex, good job at both. I don't see anything wrong with what you've done; I'm wondering if you used the result of the acceleration from the 'system' FBD if the result for F would be simpler. If F exceeds a certain value, then the block on top will slide. Too many letters for me to figure it out.

 

[odie5533]

Thanks for the help. I just realized F_{friction} has Fsin\theta in it. Even using the system acceleration, I get this:
F = \frac{\mu_{s}m_{1}g}{\frac{m_{2}cos\theta}{m_{1} + m_{2}} - \mu_{s}sin\theta}
That is after quite a bit of simplification... I'm surprised this question was so complicated. The other questions did not take nearly as long to complete.