Balancing two rollers on a hill

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SUMMARY

The discussion centers on solving a physics problem involving two rollers of masses 2kg and 3kg connected by a massless string over a pulley on inclined slopes. The user initially calculated the forces acting on each roller using the formula for gravitational force (Fg = mg) and the component of gravitational force parallel to the slope (mgsin(theta)). After determining the angles of inclination, the user incorrectly applied the sine function instead of the tangent function to find the distance X, leading to an incorrect answer of 0.75 m instead of the correct 0.69 m.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with trigonometric functions, specifically sine and tangent
  • Basic knowledge of inclined plane physics
  • Ability to solve equations involving angles and forces
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  • Review the principles of equilibrium in physics
  • Study the application of trigonometric functions in physics problems
  • Learn about the dynamics of pulleys and connected objects
  • Practice solving inclined plane problems with varying angles and masses
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Students studying physics, particularly those focusing on mechanics and inclined planes, as well as educators looking for examples of equilibrium problems involving pulleys.

rasen58
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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?
upload_2015-2-9_19-28-38.png


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?
 
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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?
 
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Wow, I don't even know why I thought it was sin for so long.
Thanks, it should be tan.
 

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