One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 60.2o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.
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To solve this problem, we need to use the principle of torque balance, which states that the sum of the clockwise torques is equal to the sum of the counterclockwise torques.
First, we need to determine the torques generated by the two forces. Torque is calculated by multiplying the force by the perpendicular distance from the axis of rotation. Since we are given the magnitude and direction of the forces, we can use the equation:
Torque = Force * Distance * sin(θ),
where θ is the angle between the force and the lever arm.
Let's calculate the torques generated by each force:
Torque1 = 2.00 N * (1.00 m) * sin(90°) = 2.00 N * 1.00 m * 1 = 2.00 N m.
Torque2 = 6.00 N * Distance * sin(60.2°).
Now, since the net torque is zero, the counterclockwise torques must equal the clockwise torques. Therefore, we have:
2.00 N m = 6.00 N * Distance * sin(60.2°).
Now, we can solve for Distance:
Distance = (2.00 N m) / (6.00 N * sin(60.2°)).
Using a calculator, we find that sin(60.2°) ≈ 0.885.
Substituting this value into the equation, we have:
Distance = (2.00 N m) / (6.00 N * 0.885).
Simplifying further, we get:
Distance ≈ 0.379 m.
Therefore, the 6.00-N force is applied at a distance of approximately 0.379 m from the end of the stick that is pinned.