An elevator car has two equal masses attached to the ceiling as shown. (Assume m = 3.00 kg.) T T2 (a) The elevator ascends with an acceleration of magnitude 1.80 m/s?. What are the tensions in the two strings? (Enter your answers in N.) 1 = 58.8 Draw free-body diagrams for each mass. What forces act on each mass? What are the directions of the forces due to each rope on the upper mass? How are the forces that the lower rope exerts on the two masses related? Apply Newton's second law to each mass. You might find it easier to solve for T, first, then use your result when solving for T,. N T, = 58.8 This is the sum of the weight of the two masses. Draw a free-body diagram of the lower mass. What forces are acting on the lower mass? Apply Newton's second law and solve for T,. N (b) The maximum tension the strings can withstand is 73.1 N. What is the maximum acceleration of the elevator so that a string does not break? (Enter the magnitude in m/s2.) 24.367 It looks like you divided the maximum tension by the mass. Note that the tension is greater in the upper string, so this string will break first. You are looking for a when T, = Tmay: m/s²

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**Understanding Tension in an Elevator System**

An elevator car has two equal masses attached to the ceiling as shown in the illustration. Here, we assume each mass \( m = 3.00 \, \text{kg} \).

**(a) Calculating Tensions in the Strings During Elevator Ascent**

When the elevator ascends with an acceleration of \( 1.80 \, \text{m/s}^2 \), we need to determine the tensions in the two strings, \( T_1 \) and \( T_2 \).

- **\( T_1 \) Calculation Attempt**: The initial value entered is 58.8 N, but this is incorrect. It's important to draw free-body diagrams for each mass to understand the forces acting on them. Consider the directions of forces due to each rope and the relation between the forces exerted by the ropes. Apply Newton’s second law to solve for \( T_2 \) first, then use that result to find \( T_1 \).

- **\( T_2 \) Calculation Attempt**: The initial value entered is 58.8 N, which is incorrect. This value represents only the sum of the weights of the two masses. Build a free-body diagram for the lower mass and apply Newton's second law to find the correct \( T_2 \).

**(b) Evaluating Maximum Elevator Acceleration Before String Breaks**

The strings have a maximum tension threshold of 73.1 N. To find the maximum acceleration the elevator can sustain without breaking the strings, we use the tension limit.

- **Calculation Attempt**: An initial value of 24.367 was entered, which is incorrect. This mistake comes from dividing the maximum tension by the mass. Note that tension is greater in the upper string \( T_1 \), which will break first. Calculate the acceleration when \( T_1 = T_{\text{max}} \).

This exercise explores how forces and acceleration relate through Newton's laws, providing a better understanding of mechanical systems like elevators.
Transcribed Image Text:**Understanding Tension in an Elevator System** An elevator car has two equal masses attached to the ceiling as shown in the illustration. Here, we assume each mass \( m = 3.00 \, \text{kg} \). **(a) Calculating Tensions in the Strings During Elevator Ascent** When the elevator ascends with an acceleration of \( 1.80 \, \text{m/s}^2 \), we need to determine the tensions in the two strings, \( T_1 \) and \( T_2 \). - **\( T_1 \) Calculation Attempt**: The initial value entered is 58.8 N, but this is incorrect. It's important to draw free-body diagrams for each mass to understand the forces acting on them. Consider the directions of forces due to each rope and the relation between the forces exerted by the ropes. Apply Newton’s second law to solve for \( T_2 \) first, then use that result to find \( T_1 \). - **\( T_2 \) Calculation Attempt**: The initial value entered is 58.8 N, which is incorrect. This value represents only the sum of the weights of the two masses. Build a free-body diagram for the lower mass and apply Newton's second law to find the correct \( T_2 \). **(b) Evaluating Maximum Elevator Acceleration Before String Breaks** The strings have a maximum tension threshold of 73.1 N. To find the maximum acceleration the elevator can sustain without breaking the strings, we use the tension limit. - **Calculation Attempt**: An initial value of 24.367 was entered, which is incorrect. This mistake comes from dividing the maximum tension by the mass. Note that tension is greater in the upper string \( T_1 \), which will break first. Calculate the acceleration when \( T_1 = T_{\text{max}} \). This exercise explores how forces and acceleration relate through Newton's laws, providing a better understanding of mechanical systems like elevators.
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