6.12: End-of-Chapter Problem Set

End-of-Chapter Problem Set

Apply the complete five-step workflow to each problem. Show the Define table, symbolic Model equation, Analyze steps with units, Interpret paragraph, and Communicate sentence.

1. A 750 kg load is supported symmetrically by two cables at 25° from horizontal. Compute tension in each cable.
2. A 15 V source is connected to a 470 Ω resistor. Compute current and power. Is a standard 0.25 W resistor safe?
3. A 12 V source is connected to a 1.2 kΩ resistor. Compute current in mA and power in mW.
4. For \(T = \dfrac{mg}{2\sin\theta}\), determine what happens as \(\theta \rightarrow 0°\). Explain the physical implication for cable installation.
5. A resistor rated at 0.5 W is connected to a 9 V battery. What is the minimum safe resistance? Show the full workflow.
6. If the cable angle in Problem 1 decreases from 25° to 15°, does tension increase by roughly 50% or more than 100%? Compute both values and explain the nonlinearity.
7. A 24 V source is connected to a 4.7 kΩ resistor. Compute current and power. Interpret safety if the resistor rating is 0.125 W.
8. [Challenge] A bracket supports a hanging sign weighing 200 N via a horizontal strut and a cable at 45°. Using equilibrium \(\sum F_x = 0\), \(\sum F_y = 0\), find the tension in the cable and the compressive force in the strut. Apply the full workflow.

Answers - click to expand

1. 750 kg load supported by two cables at 25°

   Model: For symmetric loading, the vertical components of the two cable tensions support the weight: \(2T\sin\theta = mg\). Solving for tension gives \(T = \dfrac{mg}{2\sin\theta}\).

   Analyze: \(T = \dfrac{(750 \text{ kg})(9.81 \text{ m/s}^2)}{2\sin(25°)} = \dfrac{7357.5 \text{ N}}{0.845} \approx 8705 \text{ N}\).

   Interpret: Each cable carries about \(8.71 \text{ kN}\), which is greater than half the load weight because each cable is angled. Only the vertical components support the load.

   Communicate: The required tension is approximately \(8.71 \text{ kN}\) per cable, before applying any safety factor.

2. 15 V source with 470 Ω resistor

   Model: Ohm's Law gives \(I = \dfrac{V}{R}\). Resistor power can be found from \(P = VI\) or \(P = \dfrac{V^2}{R}\).

   Analyze: \(I = \dfrac{15 \text{ V}}{470 \ \Omega} = 0.0319 \text{ A} = 31.9 \text{ mA}\). Power is \(P = \dfrac{(15 \text{ V})^2}{470 \ \Omega} = 0.479 \text{ W}\).

   Interpret: The resistor dissipates about \(0.479 \text{ W}\), which is greater than \(0.25 \text{ W}\). A standard quarter-watt resistor is not safe for this condition.

   Communicate: The current is approximately \(31.9 \text{ mA}\) and the power is approximately \(0.479 \text{ W}\); use a resistor rated above this value, such as \(0.5 \text{ W}\) or higher with appropriate margin.

3. 12 V source with 1.2 kΩ resistor

   Model: Use \(I = \dfrac{V}{R}\) and \(P = VI\).

   Analyze: \(I = \dfrac{12 \text{ V}}{1200 \ \Omega} = 0.010 \text{ A} = 10 \text{ mA}\). Power is \(P = (12 \text{ V})(0.010 \text{ A}) = 0.120 \text{ W} = 120 \text{ mW}\).

   Interpret: The current is modest and the resistor dissipates \(120 \text{ mW}\). This would be below a \(0.25 \text{ W}\) rating, though margin should still be considered.

   Communicate: The circuit current is \(10 \text{ mA}\) and the resistor power is \(120 \text{ mW}\).

4. Behavior of \(T = \dfrac{mg}{2\sin\theta}\) as \(\theta \rightarrow 0°\)

   Model: The cable tension is \(T = \dfrac{mg}{2\sin\theta}\).

   Analyze: As \(\theta \rightarrow 0°\), \(\sin\theta \rightarrow 0\). Therefore, the denominator \(2\sin\theta\) approaches zero, so \(T\) grows without bound.

   Interpret: A perfectly horizontal cable cannot support a vertical load through vertical force components because the vertical component of tension becomes zero. The equation predicts theoretically infinite tension, which signals an impossible or unsafe physical design.

   Communicate: Cable systems must avoid very shallow angles; designers should specify minimum cable angles, adequate anchor strength, and appropriate safety factors.

5. Minimum safe resistance for a 0.5 W resistor on a 9 V battery

   Model: Use \(P = \dfrac{V^2}{R}\). To keep power at or below the rating, solve for resistance: \(R_\text{min} = \dfrac{V^2}{P_\text{max}}\).

   Analyze: \(R_\text{min} = \dfrac{(9 \text{ V})^2}{0.5 \text{ W}} = \dfrac{81}{0.5} = 162 \ \Omega\).

   Interpret: A resistor smaller than \(162 \ \Omega\) would dissipate more than \(0.5 \text{ W}\) and exceed the rating. In practice, a designer would choose a standard value above this minimum and include margin.

   Communicate: The minimum safe resistance is \(162 \ \Omega\); a practical design should use a higher standard resistance or a higher power rating.

6. Cable angle decreases from 25° to 15°

   Model: Use \(T = \dfrac{mg}{2\sin\theta}\).

   Analyze: At \(25°\), \(T_{25} = \dfrac{7357.5 \text{ N}}{2\sin(25°)} \approx 8705 \text{ N}\). At \(15°\), \(T_{15} = \dfrac{7357.5 \text{ N}}{2\sin(15°)} \approx 14215 \text{ N}\).

   The percent increase is \(\dfrac{14215 - 8705}{8705} \times 100\% \approx 63\%\).

   Interpret: The increase is closer to \(50\%\) than to more than \(100\%\), but it is still a large increase. The relationship is nonlinear because \(\sin\theta\) is in the denominator. As the angle becomes shallow, small changes in angle cause increasingly large changes in tension.

   Communicate: Reducing the cable angle from \(25°\) to \(15°\) increases the tension from about \(8.71 \text{ kN}\) to about \(14.2 \text{ kN}\), an increase of about 63%.

7. 24 V source with 4.7 kΩ resistor

   Model: Use \(I = \dfrac{V}{R}\) and \(P = \dfrac{V^2}{R}\).

   Analyze: \(I = \dfrac{24 \text{ V}}{4700 \ \Omega} = 0.00511 \text{ A} = 5.11 \text{ mA}\). Power is \(P = \dfrac{(24 \text{ V})^2}{4700 \ \Omega} = 0.1226 \text{ W}\).

   Interpret: The calculated power is slightly below \(0.125 \text{ W}\), so it is technically under the rating. However, the margin is extremely small. Component tolerance, supply variation, temperature, and continuous operation could push the resistor beyond its safe rating.

   Communicate: The current is approximately \(5.11 \text{ mA}\) and the power is approximately \(0.123 \text{ W}\); a \(0.125 \text{ W}\) resistor is too close to the limit for a robust design.

8. Challenge: bracket with 200 N hanging sign, horizontal strut, and 45° cable

   Model: At the joint, use equilibrium. Vertical equilibrium gives \(\sum F_y = 0\), so \(T\sin(45°) - 200 \text{ N} = 0\). Horizontal equilibrium gives \(\sum F_x = 0\), so \(F_s - T\cos(45°) = 0\), assuming the strut provides the opposing horizontal force.

   Analyze: From vertical equilibrium, \(T = \dfrac{200 \text{ N}}{\sin(45°)} = \dfrac{200 \text{ N}}{0.707} \approx 283 \text{ N}\). From horizontal equilibrium, \(F_s = T\cos(45°) = (283 \text{ N})(0.707) \approx 200 \text{ N}\).

   Interpret: The cable tension is larger than the sign weight because only the vertical component of the cable tension supports the sign. The horizontal component of the cable tension must be balanced by the strut, so the strut carries a compressive force of about \(200 \text{ N}\).

   Communicate: The cable tension is approximately \(283 \text{ N}\), and the horizontal strut carries approximately \(200 \text{ N}\) in compression.