PHYSiiCS ANY0NE? [MACHiiNE PR0BLEMS]?

Question

ahhh i hate physics!..i have never had to come up with equations myself before!!, if someone could show me how to work these problems so i can use it for future ones i would greatly appreciate it

1. Find equations for the acceleration and the tensions in the ropes. For your variables use "m1", "m2", "m3" and "g". Assume all surfaces are frictionless, and the pulley and rope are massless and frictionless. Let the left rope have tension T1 and the right rope have tension T2.

Diagram: http://www.webassign.net/userimages/48537?db=v4net

**I kinda started this one and got a=((m3-m2)*g)/(m1+m2+m3) but webassign said it was wrong!!

2. Calculate the acceleration and the tension in the rope. For your variables use "m1", "m2", "theta" and "g". Assume all surfaces are frictionless, and the pulley and rope are massless and frictionless.

Diagram: http://www.webassign.net/userimages/48538?db=v4net

Thanks for your help! You were right about webassign but i tried to get the tension in problem one from the acceleration and I got T1=(2*m2*m3*g)/(m1+m2+m3) which is wrong so I didn't even attempt T2!! And in problem two can you please rewrite the tension so that it only uses the variables "m1", "m2", "theta" and "g"...webassign won't accept the equation with "a" in it!

Answer 1

1. Please do not panic---I am sure you are right--Maybe m2> m3 and webassign assumed that right-ward movement is positive. Please check if that is the case. Also, with "a" calculated out, find T1 and T2. I would not interfere you---you can do it!

2. Partition the weight m2g into a component parallel and the other perpendicular to the slope. The one perpendicular to the slope inserts a normal force to the slope, but does not help the movement. The one parallel to the slope does cause the movement along the slope. Hence:
F = m2g*sin(theta) = a(m1+m2)
a = m2g*sin(theta)/ (m1+m2)
T = a*m1