Question 3-a, Numerical: A crane is used to carry a heavy container of 1000 kg as shown in the figure. ↑a If the strings can withstand a maximum tension of 14800 N, what maximum acceleration can the crane have (measured in m/s2) before the string breaks? Newl Devious
Question 3-a, Numerical: A crane is used to carry a heavy container of 1000 kg as shown in the figure. ↑a If the strings can withstand a maximum tension of 14800 N, what maximum acceleration can the crane have (measured in m/s2) before the string breaks? Newl Devious
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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If the strings can withstand a maximum tension of 14800 N, what maximum acceleration can the crane have (measured in m/s²) before the string breaks?
**Answer Box:**
(Place your numerical answer here)
### Explanation of Diagram:
The diagram shows a crane lifting a container. The container is connected to the crane via a string that can withstand a maximum tension of 14800 N. The objective is to determine the maximum acceleration the crane can sustain without breaking the string.
### Solving the Problem:
To solve this problem, we need to use Newton's second law of motion.
\[ T = m(a + g) \]
Where:
- \( T \) is the tension in the string,
- \( m \) is the mass of the container,
- \( a \) is the acceleration of the crane,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
Given:
- \( T = 14800 \, \text{N} \),
- \( m = 1000 \, \text{kg} \),
- \( g = 9.8 \, \text{m/s}^2 \).
Plug these values into the equation and solve for \( a \):
\[ 14800 = 1000(a + 9.8) \]
\[ 14800 = 1000a + 9800 \]
\[ 14800 - 9800 = 1000a \]
\[ 5000 = 1000a \]
\[ a = \frac{5000}{1000} \]
\[ a = 5 \, \text{m/s}^2 \]
Therefore, the maximum acceleration the crane can have before the string breaks is \( 5 \, \text{m/s}^2 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcde4937d-e883-446c-8941-8d2ade4d99b0%2Fd424ff0b-3745-4572-ad1c-a03711e2f1f8%2Fxenkbab_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 3-a. Numerical:
A crane is used to carry a heavy container of 1000 kg as shown in the figure.
If the strings can withstand a maximum tension of 14800 N, what maximum acceleration can the crane have (measured in m/s²) before the string breaks?
**Answer Box:**
(Place your numerical answer here)
### Explanation of Diagram:
The diagram shows a crane lifting a container. The container is connected to the crane via a string that can withstand a maximum tension of 14800 N. The objective is to determine the maximum acceleration the crane can sustain without breaking the string.
### Solving the Problem:
To solve this problem, we need to use Newton's second law of motion.
\[ T = m(a + g) \]
Where:
- \( T \) is the tension in the string,
- \( m \) is the mass of the container,
- \( a \) is the acceleration of the crane,
- \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).
Given:
- \( T = 14800 \, \text{N} \),
- \( m = 1000 \, \text{kg} \),
- \( g = 9.8 \, \text{m/s}^2 \).
Plug these values into the equation and solve for \( a \):
\[ 14800 = 1000(a + 9.8) \]
\[ 14800 = 1000a + 9800 \]
\[ 14800 - 9800 = 1000a \]
\[ 5000 = 1000a \]
\[ a = \frac{5000}{1000} \]
\[ a = 5 \, \text{m/s}^2 \]
Therefore, the maximum acceleration the crane can have before the string breaks is \( 5 \, \text{m/s}^2 \).
#### Task:
If the crane has an upward acceleration \( a = 1.60 \, \text{m/s}^2 \), find the tension \( T \) in its strings measured in Newtons (N).
#### Explanation:
In the image, we see a crane lifting a container using its lifting mechanism. The container is hanging by a string or cable which experiences tension. The following forces act on the container:
1. Gravitational Force (\( F_g \)):
\[ F_g = m \cdot g \]
Where \( m = 1000 \, \text{kg} \) is the mass of the container and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.
2. Tension (\( T \)) in the string is counteracting the gravitational force and providing the upward acceleration of the container:
\[ T - F_g = m \cdot a \]
#### Calculation:
1. Calculate the gravitational force (\( F_g \)):
\[ F_g = 1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 9810 \, \text{N} \]
2. Calculate the net force providing the upward acceleration (\( F_{\text{net}} \)):
\[ F_{\text{net}} = m \cdot a = 1000 \, \text{kg} \times 1.60 \, \text{m/s}^2 = 1600 \, \text{N} \]
3. Since the tension has to counteract both the gravitational force and provide the upward acceleration, the tension \( T \) in the string is:
\[ T = F_g + F_{\text{net}} \]
\[ T = 9810 \, \text{N} + 1600 \, \text{N} = 11410 \, \text{N} \]
#### Answer:
The tension \( T \) in the strings is \( 11410 \, \text{N} \).
---
### Graphical Explanation
The diagram shows a crane with an extended arm lifting a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcde4937d-e883-446c-8941-8d2ade4d99b0%2Fd424ff0b-3745-4572-ad1c-a03711e2f1f8%2Fxzlr8q9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 3-a: Numerical
#### Problem Statement:
A crane is used to carry a heavy container of 1000 kg as shown in the figure.
#### Task:
If the crane has an upward acceleration \( a = 1.60 \, \text{m/s}^2 \), find the tension \( T \) in its strings measured in Newtons (N).
#### Explanation:
In the image, we see a crane lifting a container using its lifting mechanism. The container is hanging by a string or cable which experiences tension. The following forces act on the container:
1. Gravitational Force (\( F_g \)):
\[ F_g = m \cdot g \]
Where \( m = 1000 \, \text{kg} \) is the mass of the container and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.
2. Tension (\( T \)) in the string is counteracting the gravitational force and providing the upward acceleration of the container:
\[ T - F_g = m \cdot a \]
#### Calculation:
1. Calculate the gravitational force (\( F_g \)):
\[ F_g = 1000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 9810 \, \text{N} \]
2. Calculate the net force providing the upward acceleration (\( F_{\text{net}} \)):
\[ F_{\text{net}} = m \cdot a = 1000 \, \text{kg} \times 1.60 \, \text{m/s}^2 = 1600 \, \text{N} \]
3. Since the tension has to counteract both the gravitational force and provide the upward acceleration, the tension \( T \) in the string is:
\[ T = F_g + F_{\text{net}} \]
\[ T = 9810 \, \text{N} + 1600 \, \text{N} = 11410 \, \text{N} \]
#### Answer:
The tension \( T \) in the strings is \( 11410 \, \text{N} \).
---
### Graphical Explanation
The diagram shows a crane with an extended arm lifting a
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