The left end of a 1-kg, 1.0 m long uniform thin rod is attached to a pivot, about which it can rotate freely. A force F is applied at the right end of the rod at 30 degree angle to hold the rod stationary in a horizontal position. What is the magnitude of the force F?
I think the answer is 10N but I have no clue how to solve it.
kim
To solve this problem, you can use the principle of torque equilibrium, which states that the sum of the torques acting on an object must be zero for it to be in rotational equilibrium.
In this case, the rod is stationary, so the sum of the torques acting on it must be zero.
The torque τ produced by a force about a pivot point is given by the equation:
τ = rFsinθ
where r is the perpendicular distance from the pivot point to the line of action of the force F, and θ is the angle between the force and the line connecting the pivot point to the point where the force is applied.
In this problem, the pivot point is at the left end of the rod, and the force F is applied at the right end of the rod. The perpendicular distance from the pivot point to the line of action of the force is the length of the rod, which is 1.0 m. The angle θ is given as 30 degrees.
Plugging these values into the equation for torque, we get:
τ = (1.0 m)(F)(sin 30°)
τ = (1.0 m)(F)(0.5)
τ = 0.5F
Since the rod is stationary, the sum of the torques acting on it must be zero. In this case, there is only one torque acting on the rod, which is produced by the force F. Therefore, we can set τ equal to zero:
0.5F = 0
Solving for F, we find:
F = 0 / 0.5
F = 0 N
Therefore, the magnitude of the force F required to hold the rod stationary in a horizontal position is 0 N, not 10 N as you initially suggested.