(III) A uniform rod AB of length 5.0 m and mass M = 3.8 kg...

(III) A uniform rod AB of length 5.0 m and mass $M = 3.8 kg$ is hinged at A and held in equilibrium by a light cord, as shown in Fig. 9-67. A load $W = 22 N$ hangs from the rod at a distance $d$ so that the tension in the cord is 85N. (a) Draw a free-body diagram for the rod. (b) Determine the vertical and horizontal forces on the rod exerted by the hinge. (c) Determine $d$ from the appropriate torque equation. FIGURE 9–67 Problem 27. (Figure can't copy)

Key Concepts

Free-body Diagram

A free-body diagram is a simplified illustration that depicts all external forces acting on a body. It is a fundamental tool in mechanics for visualizing forces such as gravitational forces, applied loads, tension forces from cords, and reaction forces from supports or hinges. It helps in setting up the equations required for analyzing equilibrium conditions and solving for unknown forces and moments.

Equilibrium Conditions

Equilibrium conditions in mechanics state that, for a body to remain at rest, the sum of all forces and the sum of all torques (moments) acting on the body must be zero. This includes setting up separate balance equations for forces in the horizontal and vertical directions (translational equilibrium) as well as ensuring that the net torque about any pivot point is zero (rotational equilibrium).

Torque Calculation

Torque is the rotational equivalent of force, defined as the product of a force and its perpendicular distance from a chosen pivot point. Calculating torque is essential in analyzing the rotational equilibrium of a system. By choosing an appropriate pivot point—often where unknown forces act—you can eliminate those unknowns from the torque equation, simplifying the problem.

Uniform Rod and Center of Mass

For a uniform rod, its weight is evenly distributed along its length and is assumed to act at its center of mass, which is at the midpoint of the rod. This assumption simplifies the analysis of the rod’s equilibrium by converting a distributed load into a single force acting at the center, making it easier to calculate torques and balance moments.

Force Components

When forces are applied at angles, it is useful to resolve them into their horizontal and vertical components. This decomposition enables the separate application of equilibrium conditions in the vertical and horizontal directions, which is essential for solving for unknown forces and ensuring that the net force in each direction sums to zero.

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