Balancing two rollers on a hill

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
2 replies · 1K views · last reply
rasen58
Messages
71
Reaction score
2

Homework Statement


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?
Diagram of two rollers on inclined slopes joined over a pulley by a string: a 2 kg roller on the left slope and a 3 kg roller on the right, with a vertical height of 0.3 m and horizontal distance X marked along the base.


Homework Equations


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?
 
Physics news on Phys.org
rasen58 said:
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?
 
  • Like
Likes   Reactions: rasen58
Wow, I don't even know why I thought it was sin for so long.
Thanks, it should be tan.