Twe mases mes9Ka and my13 Kg are connected with itring and pulled with a force F30 N that makes in angle of 53 with a imooth horizontal surface ihown in the ngure. Find the tention in the string in a unit of Newton. F m2 ml A1.306 B 1044 C. None D. 14.77 1.3.093
Twe mases mes9Ka and my13 Kg are connected with itring and pulled with a force F30 N that makes in angle of 53 with a imooth horizontal surface ihown in the ngure. Find the tention in the string in a unit of Newton. F m2 ml A1.306 B 1044 C. None D. 14.77 1.3.093
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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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![### Problem Statement:
Two masses \( m_1 = 9 \, \text{kg} \) and \( m_2 = 13 \, \text{kg} \) are connected with string and pulled with a force \( F = 30 \, \text{N} \) that makes an angle of \( 52^\circ \) with a smooth horizontal surface as shown in the figure. Find the tension in the string in units of Newtons.
### Diagram Explanation:
The diagram depicts two blocks on a smooth horizontal surface. The blocks are labeled \( m_1 \) and \( m_2 \). Block \( m_1 \) is being pulled to the right by a string attached to block \( m_2 \). An external force \( F \) of 30 Newtons is applied at an angle \( \theta \) of 52 degrees above the horizontal on block \( m_2 \).
### Multiple Choice Options:
- A. 7.386
- B. 10.44
- C. None
- D. 14.77
- E. 3.669
### Solution Approach:
To solve this problem, you will need to apply Newton's laws of motion, considering the forces acting on each mass and the geometry of the applied force. The following steps might be helpful:
1. **Resolve the Force \( F \)**: Decompose the force \( F \) into horizontal and vertical components.
\[
F_x = F \cos(\theta)
\]
\[
F_y = F \sin(\theta)
\]
2. **Calculate Acceleration**: Since the surface is smooth (frictionless), sum the horizontal forces and use Newton's second law to find the acceleration \( a \). Both masses will have the same acceleration since they are connected by a string.
\[
F_x - T = m_2 a
\]
\[
T = m_1 a
\]
3. **Solve for Acceleration**:
\[
F \cos(52^\circ ) = (m_1 + m_2) a
\]
\[
a = \frac{F \cos(52^\circ )}{m_1 + m_2}
\]
4. **Solve for Tension**: Use the acceleration to find the tension \( T \) in](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbeae604a-2e4b-47ef-907a-313aaf16b611%2F167356ba-4c7e-406f-94ee-167df36ac20a%2F3l0zad4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Two masses \( m_1 = 9 \, \text{kg} \) and \( m_2 = 13 \, \text{kg} \) are connected with string and pulled with a force \( F = 30 \, \text{N} \) that makes an angle of \( 52^\circ \) with a smooth horizontal surface as shown in the figure. Find the tension in the string in units of Newtons.
### Diagram Explanation:
The diagram depicts two blocks on a smooth horizontal surface. The blocks are labeled \( m_1 \) and \( m_2 \). Block \( m_1 \) is being pulled to the right by a string attached to block \( m_2 \). An external force \( F \) of 30 Newtons is applied at an angle \( \theta \) of 52 degrees above the horizontal on block \( m_2 \).
### Multiple Choice Options:
- A. 7.386
- B. 10.44
- C. None
- D. 14.77
- E. 3.669
### Solution Approach:
To solve this problem, you will need to apply Newton's laws of motion, considering the forces acting on each mass and the geometry of the applied force. The following steps might be helpful:
1. **Resolve the Force \( F \)**: Decompose the force \( F \) into horizontal and vertical components.
\[
F_x = F \cos(\theta)
\]
\[
F_y = F \sin(\theta)
\]
2. **Calculate Acceleration**: Since the surface is smooth (frictionless), sum the horizontal forces and use Newton's second law to find the acceleration \( a \). Both masses will have the same acceleration since they are connected by a string.
\[
F_x - T = m_2 a
\]
\[
T = m_1 a
\]
3. **Solve for Acceleration**:
\[
F \cos(52^\circ ) = (m_1 + m_2) a
\]
\[
a = \frac{F \cos(52^\circ )}{m_1 + m_2}
\]
4. **Solve for Tension**: Use the acceleration to find the tension \( T \) in
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