Two Blocks with rope and Pulley on Inclined Plane

  • Thread starter Dahaka14
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  • #1
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Homework Statement


This is a problem that I am trying to help my friend with. I understand how to solve this problem the traditional way by rotating the axes of the block on the inclined plane, but then my friend asked me to show him how the answer is the same with non-rotated axes.
There are two blocks connected by a rope that goes over a pulley, one block of mass m[tex]_{2}[/tex] hanging by the rope over the edge, and the other block of mass m[tex]_{1}[/tex] on the inclined plane. Given the angle of the plane with the horizontal, and without any type of friction, find the tension in the rope and the acceleration of the two blocks.


Homework Equations


m[tex]_{1}[/tex]=5kg
m[tex]_{2}[/tex]=10kg
[tex]\theta[/tex]=40[tex]^{0}[/tex]
[tex]\sum[/tex]F[tex]_{x:m[tex]_{1}[/tex]}[/tex]=Tcos[tex]\theta[/tex]+Ncos[tex]\theta[/tex]=-m[tex]_{1}[/tex]acos[tex]\theta[/tex]
[tex]\sum[/tex]F[tex]_{y:m[tex]_{1}[/tex]}[/tex]=Tsin[tex]\theta[/tex]+Nsin[tex]\theta[/tex]-m[tex]_{1}[/tex]g=m[tex]_{1}[/tex]asin[tex]\theta[/tex]
[tex]\sum[/tex]F[tex]_{x:m[tex]_{2}[/tex]}[/tex]=0
[tex]\sum[/tex]F[tex]_{y:m[tex]_{2}[/tex]}[/tex]=T-m[tex]_{2}[/tex]g=-m[tex]_{2}[/tex]a


The Attempt at a Solution


I somehow get the Tsin[tex]\theta[/tex]'s to cancel when dividing the force equations to eliminate the m[tex]_{1}[/tex]a's. Maybe it's just the end of the day and I'm tired and suck at this. Can someone please help?[tex]_{}[/tex]
 

Answers and Replies

  • #2
Doc Al
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Careful with the signs of the components of N and the acceleration. (Make sure your signs for the acceleration are consistent.)
 
  • #3
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Well, i made the acceleration of m1 positive y (+sine) and negative x (-cosine) because it is moving "up and to the right." My normal force is acting "up and to the left," so it has a positive y and positive x orientation. The difference is that m2 has an acceleration in the negative y direction. Does this sound like it would be correct? If you need me to, I can explicitly write out what I did, but if you don't need me to I won't because the tex can get tedious.
 
  • #4
Doc Al
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45,188
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Well, i made the acceleration of m1 positive y (+sine) and negative x (-cosine) because it is moving "up and to the right."
OK. For some reason, you take "to the left" as positive. No problem, just be consistent. That means that the x-component of the tension must be negative.
 

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