- #1

Dahaka14

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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]