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 27.8o 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.
To solve this problem, we can use the principle of torque equilibrium. Torque is a measure of the force's ability to cause rotational motion. In equilibrium, the net torque acting on an object must be zero.
The torque exerted by a force on an object is given by the formula: Torque = Force * Distance * sin(θ), where θ is the angle between the force vector and the lever arm distance.
Let's start by analyzing the first force, which has a magnitude of 2.00 N and is applied perpendicularly to the length of the stick at the free end. Since it is perpendicular to the length of the stick, the angle between the force and the lever arm distance is 90 degrees. Hence, the torque exerted by this force is zero (sin(90) = 1).
Now, let's focus on the second force, which has a magnitude of 6.00 N and acts at a 27.8 degrees angle with respect to the length of the stick. To find the torque exerted by this force, we need to determine the lever arm distance.
Let's assume that the distance along the stick, where the 6.00 N force is applied, is 'x'. The lever arm distance for this force would then be the length of the meter stick minus 'x'. Therefore, the torque exerted by the 6.00 N force is: Torque = (6.00 N) * ((1.00 m - x) * sin(27.8)).
Since the net torque is zero, the torque exerted by the 6.00 N force must be equal and opposite in direction to the torque exerted by the 2.00 N force. So we have:
(6.00 N) * ((1.00 m - x) * sin(27.8)) = (-2.00 N) * (0 m) [since the torque of the first force is zero]
Simplifying this equation:
(6.00 N) * (1.00 m - x) * sin(27.8) = 0
Since the sine of an angle can be zero, the only possibility for this equation to hold is if 1.00 m - x = 0. So we have:
1.00 m - x = 0
Solving for 'x', we find:
x = 1.00 m
Therefore, the 6.00 N force is applied at a distance of 1.00 meter along the meter stick, with respect to the end that is pinned.