A 350-N, uniform, 1.50-m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tension of 500.0 N without breaking, and cable B can support up to 400.0 N. You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

Verified step by step guidance

1. First, understand that the bar is in static equilibrium, meaning the sum of forces and the sum of torques (moments) around any point must be zero.
2. Identify the forces acting on the bar: the weight of the bar (350 N) acting at its center, the tension in cable A (T_A), the tension in cable B (T_B), and the weight of the additional object (W) that we want to place on the bar.
3. Set up the equation for the sum of vertical forces: T_A + T_B = 350 N + W. This ensures that the bar remains in vertical equilibrium.
4. Choose a pivot point to set up the torque equation. A convenient choice is one of the cable attachment points, say at cable A. The torque due to the weight of the bar is (350 N) * (0.75 m) since it acts at the center of the bar. The torque due to the weight W is W * d, where d is the distance from cable A. The torque due to T_B is (T_B) * (1.50 m). Set the sum of torques around cable A to zero: (350 N * 0.75 m) + (W * d) - (T_B * 1.50 m) = 0.
5. Solve the system of equations: Use the maximum tension values for T_A and T_B to find the maximum W and the corresponding distance d. Ensure that neither cable exceeds its maximum tension by substituting back into the force and torque equations.

Key Concepts

Torque and Equilibrium

Torque is the rotational equivalent of force, calculated as the product of force and the distance from the pivot point. For an object in equilibrium, the sum of torques around any point must be zero. In this problem, understanding torque is crucial to determine how the weight distribution affects the tension in the cables.

Tension in Cables

Tension is the force conducted along a cable or rope when it is pulled tight by forces acting from opposite ends. Each cable in this scenario has a maximum tension it can withstand before breaking. Calculating the tension in each cable helps determine the maximum weight that can be added without exceeding these limits.

Center of Mass and Balance

The center of mass is the point where the mass of an object is concentrated and balanced. For the bar to remain in equilibrium, the torques due to the bar's weight and the additional weight must balance around the pivot point. Understanding how to position the weight relative to the center of mass is key to solving where to place the weight.