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September 16, 2014

September 16, 2014

Posted by **Eric (please help me)** on Thursday, February 15, 2007 at 9:12am.

ans: 2.20 m/s^2, 27.4 N

I have no clue how to approach problem. I start with a free body diagram. Since the blocks are resting on a surface, then there is no acceleration along the y direction. The acceleration and the tension are going to be the same because they are connected by the same pulley.

sigma Fx = T + m1cos theta

ahh I'm so lost.

the graph: imagine a triangle, at the top is the pulley and on left incline (of triangle) is the 3.50 kg and on right incline is 5.00 kg. Both angles pointing toward each corner are 35 degrees.

First, note the component of weight parallel to the incline is mg*sinTheta.This is the force pulling each side of the rope.

Net force= (8-3.5)g sinTheta

this has to equal total mass*acceleration

8-3.5)g sinTheta= (3.5+5) * acceleration

solve for acceleration

For tension, consider the left block.

t-weightdown= ma

tension= weight down + ma

= 3.5*g*sinTheta+3.5*acceleartion

- Physics (Newton's laws) -
**Temesgen Bekele**, Tuesday, June 12, 2012 at 8:10am2.20m/s 27.4N

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