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. Step 1: Begin by understanding the forces acting on the bar. The bar is in equilibrium, meaning the sum of forces and the sum of torques (moments) must be zero. The weight of the bar acts at its center, and the tension in cables A and B act vertically upwards at each end.
2. Step 2: Calculate the torque due to the weight of the bar. The torque (τ) is given by τ = F × d, where F is the force and d is the distance from the pivot point. Choose one end of the bar as the pivot point to simplify calculations. The weight of the bar (350 N) acts at the midpoint, 0.75 m from either end.
3. Step 3: Set up the torque equilibrium equation. If you choose the left end as the pivot, the torque due to the bar's weight is τ_bar = 350 N × 0.75 m. The torque due to the weight you add (W) will depend on its position x from the left end, τ_weight = W × x. The tension in cable B will also create a torque, τ_B = T_B × 1.5 m.
4. Step 4: Use the maximum tension values for cables A and B to find the limits for the weight. Cable A can support up to 500 N, and cable B can support up to 400 N. The sum of vertical forces must be zero, so T_A + T_B = 350 N + W. Use this equation along with the torque equilibrium to solve for W and x.
5. Step 5: Solve the equations simultaneously to find the maximum weight (W) and its position (x) that will not exceed the tension limits of either cable. Ensure that the calculated tensions do not exceed 500 N for cable A and 400 N for cable B.

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 the bar to remain in equilibrium, the sum of torques around any point must be zero. This concept helps determine how the weight distribution affects the tension in the cables, ensuring neither cable exceeds its maximum tension.

Tension in Cables

Tension is the force exerted by a cable or rope when it is subjected to a load. In this scenario, each cable has a maximum tension it can withstand before breaking. Understanding how the weight affects the tension in each cable is crucial to ensure the weight is distributed such that neither cable exceeds its limit.

Center of Mass

The center of mass is the point where the mass of an object is concentrated and balanced. For a uniform bar, it is at its midpoint. When adding weight, the position relative to the center of mass affects the distribution of forces and torques, influencing which cable bears more load. Proper placement of the weight ensures balanced tension in both cables.