Question

A long, massless rod of length L= 1.92 m has three forces perpendicular to it but is in static equilibrium. F_1 acts on one end and has a magnitude of 383 N. F_2 acts on the other end, and has a magnitude of 245 N Measured relative to where Facts, how far from the end of the rod does F_3 act (i.e., what is the distance x)?. Express your answer in metres.

Answer 1

The distance x where
`F_3 \)` acts is approximately 0.68 m. This is determined by balancing torques in static equilibrium using the equation
`\( F_1 \cdot (L - x) = F_2 \cdot x \)`.

Step-by-step explanation:

In static equilibrium, the sum of torques acting on the rod is zero. Taking torques about one end of the rod, the torques due to
`\( F_1 \)` and
`\( F_2 \)` must be balanced by the torque due to
`F_3 \)`.

The equation is
`\( F_1 \cdot (L - x) = F_2 \cdot x \)`, where
`\( F_1 \)` is the force at one end,
`\( F_2 \)` is the force at the other end, ( L ) is the length of the rod, and ( x ) is the distance from the end where
`\( F_3 \)` acts.

Solving for x , we find that x is approximately 0.68 m.

The principles of torque, static equilibrium, and force distribution in physics, especially when dealing with extended objects and multiple forces.