A metal ball lies on a flat horizontal surface. It is attracted by two magnets placed around it.
The first magnet's force on the ball is
\[1\text{ N}\] in a direction that is a
\[60\degree\] rotation from east.
The second magnet's force on the ball is
\[2\text{ N}\] in a direction that is a
\[160\degree\] rotation from east.
Given the combined force of the magnets, in what direction do they pull the ball?
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We have a metal ball on a horizontal surface attracted by two magnets. The first magnet exerts a force of 1 Newton at 60 degrees from east, and the second magnet exerts 2 Newtons at 160 degrees from east. We need to find the direction of the combined force.
Now we break down each force into x and y components. For Force 1 at 60 degrees: the x-component is 1 times cosine 60 degrees equals 0.5 Newtons, and the y-component is 1 times sine 60 degrees equals 0.866 Newtons. For Force 2 at 160 degrees: the x-component is 2 times cosine 160 degrees equals negative 1.879 Newtons, and the y-component is 2 times sine 160 degrees equals 0.684 Newtons.
Now we add the corresponding components to find the resultant force. For the x-direction: F-x equals F-1-x plus F-2-x, which is 0.5 plus negative 1.879, giving us negative 1.379 Newtons. For the y-direction: F-y equals F-1-y plus F-2-y, which is 0.866 plus 0.684, giving us 1.55 Newtons. The resultant vector is shown in green using the parallelogram method.
To find the magnitude of the resultant force, we use the Pythagorean theorem. The magnitude equals the square root of F-x squared plus F-y squared. Substituting our values: the square root of negative 1.379 squared plus 1.55 squared, which equals the square root of 1.901 plus 2.403, giving us the square root of 4.304, which equals 2.075 Newtons. This forms the hypotenuse of our right triangle.
Finally, we calculate the direction using the arctangent function. Theta equals arctangent of F-y over F-x, which is arctangent of 1.55 over negative 1.379, giving us arctangent of negative 1.124, which equals negative 48.3 degrees. Since F-x is negative and F-y is positive, we're in the second quadrant, so we add 180 degrees to get 131.7 degrees from east. The final answer is a force of 2.075 Newtons at 131.7 degrees from east, pulling the ball toward the northwest direction.