Question

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?

Short answer

The heaviest weight without breaking cables is approximately 700 N, placed near cable B.

Step-by-step solution

Identify forces and moments

The bar has a weight of 350 N acting downward at its center. Therefore, the bar's weight creates a torque (moment) about any pivot point. Additional weight will also create torque depending on its position on the bar. The forces in the cables and the positions need to be calculated such that neither cable exceeds its tension limit.

Set up equilibrium equations

For the bar to remain in equilibrium, two conditions must be met: the sum of all vertical forces must equal zero, and the sum of all moments (torques) must equal zero. We define the weight to be added as \( W \), cable tensions as \( T_A \) and \( T_B \), and the distance from the left end of the bar to the added weight as \( x \). The length of the bar is 1.50 m.

Apply force equilibrium

The sum of vertical forces: \( T_A + T_B = 350 + W \). This equation expresses the balance of vertical forces on the bar.

Calculate moment about one end

Calculate moments about cable A to find \( T_B \). The moment arm for the bar's weight is 0.75 m (half of the bar's length), and for the added weight is \( x \). \[ T_B \times 1.5 = 350 \times 0.75 + W \times x \]

Solve for tensions without exceeding limits

Substitute the tension limit constraints: 1. \( T_A \leq 500 \) N2. \( T_B \leq 400 \) NSolve these inequalities along with the equilibrium equations to find the maximum \( W \) and its position \( x \).

Solve system of equations

From Step 4, express \( W \) in terms of \( x \) and find constraints based on tension limits. Use the condition where one cable is at its maximum tension to determine \( W \): Set \( T_B = 400 \) N as it is more restrictive and substitute into \[ 400 \times 1.5 = 350 \times 0.75 + W \times x \] and solve for \( W \) and \( x \).

Calculate final values

Plug limitations and rearrange to find: \[ W = \frac{400 \times 1.5 - 350 \times 0.75}{x} \]Substitute the critical cable limits back to verify constraints and solve for largest permissible \( x \).

Conclusion

The heaviest weight, \( W \), is determined both by finding the maximum consistent \( T_B \leq 400 \) and solving for optimal \( x \). Position \( x \) can be calculated via trial by maintaining equilibrium equations and balance close to \( B \) until \( A \) is at 500 N tension.

Key concepts

Torque

In physics, torque is the ability of a force to cause an object to rotate around a pivot point or axis. Think of it as the rotational equivalent of force. For the problem you are dealing with, torque is generated when weights are placed on the bar. This torque needs to be calculated to ensure the bar stays balanced and doesn't rotate, which is vital for maintaining equilibrium.

• The moment arm is the perpendicular distance from the pivot point to the line of action of the force.
• Torque is calculated as the force multiplied by the moment arm: \( \text{Torque} = \text{Force} \times \text{Distance} \).
• In this exercise, placing a weight on the bar introduces additional torque, which needs to be counterbalanced by the cables to prevent rotation.

Balancing these torques is crucial for the bar to remain steady. By understanding how forces create torque, you can determine where to position a weight to maintain equilibrium without exceeding cable tension limits.

Tension in Cables

Tension is the force exerted by a cable or rope when it is used to support a weight. In the context of this problem, tension plays a key role in determining how different forces keep the bar suspended without breaking the cables.

• Each cable at the bar’s ends supports a portion of the total weight, including the weight of the bar itself and any additional weight added.
• The tension in the cables must not exceed their maximum capacity of 500 N for cable A and 400 N for cable B to prevent them from breaking.

When more weight is added, it's crucial to understand how this changes the distribution of tension between the cables. By carefully calculating these forces, you can ensure that the cables are used efficiently and safely without failing.

Force Equilibrium

Force equilibrium is achieved when all the forces acting on an object are balanced, resulting in no net force and no acceleration. In simpler terms, it's when the object is in a stable position and not moving.

• For a system in equilibrium, the sum of all vertical forces acting on the bar must equal zero.
• This principle is applied in this scenario by ensuring that the tension in cables A and B, combined with the bar's and added weight's forces, equilibrate perfectly.

By setting up equations for the forces involved, you can solve for unknowns such as the weight and position of added loads. This ensures the system's stability, making force equilibrium analysis a key step in solving statics problems.

Statics

Statics is a branch of physics that studies objects at rest or in a state of constant velocity, implying no net forces or torques act on them. In the current exercise, applying principles of statics ensures that the bar does not move or rotate under the influence of various forces.

• Statics involves analyzing forces and torques acting on a body to determine conditions of equilibrium.
• Here, you must account for the forces from the bar's weight, the added weight, and the tension in cables to ensure the system is in a stable equilibrium.

Statics helps in understanding and predicting how structures like beams, bridges, and bars behave under load. It's crucial for ensuring the bar remains in a stationary horizontal position safely.

Mechanical Equilibrium

Mechanical equilibrium refers to a state where the sum of forces and the sum of torques are zero, resulting in an object that is neither translating (moving linearly) nor rotating.

• For the bar in the problem, achieving mechanical equilibrium means balancing all forces and torques so that neither end of the bar tilts up or down.
• By ensuring that both conditions of equilibrium are met, you can place the maximum weight possible without risking cable failure.

Understanding mechanical equilibrium is fundamental to solving problems in statics and ensuring that structures can safely support the loads they encounter. This balance of forces and torques is critical for maintaining safety in structural engineering scenarios.