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3. A 70 kg package is suspended by two strings as shown below. The angle of the strings with respect to the ceiling are equal and both are 15 degrees (theta). This entire system is inside an elevator that is accelerating upwards at a rate of 3 m/s².

3. A 70 kg package is suspended by two strings as shown below. The angle of the strings with respect to the ceiling are equal and both are 15 degrees (theta). This entire system is inside an elevator that is accelerating upwards at a rate of 3 m/s².

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## Problem Description 3. A 70 kg package is suspended by two strings as shown below. The angle of the strings with respect to the ceiling are equal and both are 15 degrees (\(\theta\)). This entire system is inside an elevator that is accelerating upwards at a rate of 3 m/s\(^2\). **Question:** What is the tension in one of the strings? ## Solution Steps ### a. Draw a Properly Labeled Free Body Diagram - The package is suspended by two strings. - Each string makes an angle \(\theta = 15^\circ\) with the ceiling. - The tension in each string is represented as \(T\). - The weight of the package is \(mg\) (where \(m = 70 \, \text{kg}\)). - The upward acceleration \(a = 3 \, \text{m/s}^2\). ### b. Show All Equations Used and Calculations Made 1. **Newton's Second Law:** - The net force in the vertical direction (\(F_{net}\)) is \(F_{net} = ma\). 2. **Vertical Forces:** - The forces acting are the weight (\(mg\)) downward and the vertical components of tension (\(2T \sin \theta\)) upward. - Equation: \(2T \sin \theta - mg = ma\). 3. **Substitute Values:** - Mass (\(m\)) = 70 kg - Gravitational acceleration (\(g\)) = 9.8 m/s\(^2\) - Acceleration (\(a\)) = 3 m/s\(^2\) - Angle (\(\theta\)) = 15 degrees 4. **Calculate Tension:** - Convert the angle to radians if necessary: \(\theta = 15^\circ\). - Rearrange the equation: \(T = \frac{m(g + a)}{2 \sin \theta}\). ### c. Give the Answer in the Correct Units - Tension is measured in Newtons (N). ### d. Give the Correct Numerical Answer 1. **Calculation:** - Calculate \(g + a = 9.8 + 3 = 12.8\) m/s\(^2\). - \(\sin 15^\circ \approx 0
Transcribed Image Text:## Problem Description 3. A 70 kg package is suspended by two strings as shown below. The angle of the strings with respect to the ceiling are equal and both are 15 degrees (\(\theta\)). This entire system is inside an elevator that is accelerating upwards at a rate of 3 m/s\(^2\). **Question:** What is the tension in one of the strings? ## Solution Steps ### a. Draw a Properly Labeled Free Body Diagram - The package is suspended by two strings. - Each string makes an angle \(\theta = 15^\circ\) with the ceiling. - The tension in each string is represented as \(T\). - The weight of the package is \(mg\) (where \(m = 70 \, \text{kg}\)). - The upward acceleration \(a = 3 \, \text{m/s}^2\). ### b. Show All Equations Used and Calculations Made 1. **Newton's Second Law:** - The net force in the vertical direction (\(F_{net}\)) is \(F_{net} = ma\). 2. **Vertical Forces:** - The forces acting are the weight (\(mg\)) downward and the vertical components of tension (\(2T \sin \theta\)) upward. - Equation: \(2T \sin \theta - mg = ma\). 3. **Substitute Values:** - Mass (\(m\)) = 70 kg - Gravitational acceleration (\(g\)) = 9.8 m/s\(^2\) - Acceleration (\(a\)) = 3 m/s\(^2\) - Angle (\(\theta\)) = 15 degrees 4. **Calculate Tension:** - Convert the angle to radians if necessary: \(\theta = 15^\circ\). - Rearrange the equation: \(T = \frac{m(g + a)}{2 \sin \theta}\). ### c. Give the Answer in the Correct Units - Tension is measured in Newtons (N). ### d. Give the Correct Numerical Answer 1. **Calculation:** - Calculate \(g + a = 9.8 + 3 = 12.8\) m/s\(^2\). - \(\sin 15^\circ \approx 0
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