Two rollers, of masses 2kg and 3kg, are connected over a pulley by a massless string. They rest in equilibrium on two inclined slopes, as shown in the diagram to the left. Friction is negligible. What is the distance X?
Fg = mg
Component of Fg parallel to the slope = mgsin(theta)
The Attempt at a Solution
Since they are in equilibrium, the forces on both sides must be equal.
So I created a separate coordinate system for each side and found the forces acting along the slope.
On the left side, the force of gravity parallel to the slope is (2kg)(10 m/s^2)(sin(theta)). To find theta, I did arctan(.3/.4) to get 36.87 degrees. So therefore (2kg)(10)(sin(36.87)) = 12 N.
So the right side force must also be equal to 12 N.
12 N = (3kg)(10 m/s^2)(sin(theta)).
So now I found this theta by solving the equation and got 23.58 degrees.
And finally, to find x, I did sin(23.58) = 0.3/x --> x = 0.75 m.
But the answer is 0.69 m. What did I do wrong?