Recent content by The___Kid

Which 2nd equation :confused:

3(-m1*gsinθ)- 3(m1a1) = 2(-m2g)-2(m2a2) 3(-m1*gsinθ)- 3(m1*-1.5a2) = 2(-m2g)-2(m2a2) I'm not sure what what do after this point.

Alright, I'm a bit confused when it comes to the algebra. I got a2 = (2(m2)+3g(sinTheta)-2g)/6.5

Ok. So now m1a1 = 2T-m1g(sin(theta)) m2a2 = 3T-m2g a1= 1.5(a2)? T = (-m1g(sin(theta))-m1a1)/2 T = (-m2g-m2a2)/3 3(-m1g)(sin(theta))-3(m1a1) = 2(-m2g)-2(m2a2) and substitute a1 for a2 and vice versa to solve?

m1 will move up 1/2X and m2 will move down 2/3X?

If m1 goes up the plane by 1 meter then m2 will go down by 1/3 of a meter?

so if it's parallel then the summation of forces on m1 just becomes the other way around with mg(cos theta) and mg(sin theta). How is that simpler? If there's two forces of tension acting on m1 and three acting on m2 shouldn't that be the acceleration?

Homework Statement Frictionless, massless, pulley system. Cables attached to m1 are parallel to the tabletop. Cables attached to m2 are parallel to each other, and perpendicular to the ground. I'm trying to solve for the acceleration of the objects in terms of the variables, m1, m2, g, and...