Recent content by savelieffd

Aha, revelation. Looks like i figured it out! T1 = (m1*m2)/(2m1 + .5m2)*g then T2 = (m1*m2)/(m1 + .25m2)*g yes? and using the equations of tension, a1 = t1/m1 = (m2*g)/(2m1 +.5m2) and so a2 = half of a1 which is = (m2*g)/(4m1+m2) and in the end we get a correct answer...

Because of the frictionless surface, would the net force be just the tension? If so, would a1 = ((m1*m2*(Sin(theta)+1)*g)/(m1+m2))/m1) and Sin(theta) equals 1 becuase theta is equal to 90?

Would m1 accelerate at 2*g? BEcause of the pulley?

well i know that a1 = 2 *a2, and i know that T2 = 2 *T1. But i don't know how i can put these together to work with Newtons 2nd law.

hmmm, I am sorry bout that, do you know what they mean by multiplicative factor?

well i got (0.5*mg*sin(x2))/cos(x1))=T1.

Hope someone can help me with this one. I have two blocks with m1 and m2 connected to two pulleys. I need to figure out T1, T2, a1, and a2 using only the values of g, m1, and m2. But I've been unable to come up with an equation that will work with it. I have a picture attached. I tried using...