Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here F, = 444 N and F, = 343 N and friction is negligible. In the diagram below, the mass of the car = 3500 kg, 8, = -25° and e, = 12°. (Assume the car faces the positive x-axis before the forces are applied.) F, F, (b) What is the acceleration (in m/s?) of the car? magnitude m/s? direction (counterclockwise from the +x-axis)
Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here F, = 444 N and F, = 343 N and friction is negligible. In the diagram below, the mass of the car = 3500 kg, 8, = -25° and e, = 12°. (Assume the car faces the positive x-axis before the forces are applied.) F, F, (b) What is the acceleration (in m/s?) of the car? magnitude m/s? direction (counterclockwise from the +x-axis)
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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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I just need part B explained. Thank you!
In the diagram:
- \( \vec{F_1} \) is a force of 444 N applied at an angle \( \theta_1 = -25^\circ \) from the positive x-axis.
- \( \vec{F_2} \) is a force of 343 N applied at an angle \( \theta_2 = 12^\circ \) from the positive x-axis.
- The car is represented by a top-down view, indicating the direction of the applied forces.
**(b) What is the acceleration (in m/s\(^2\)) of the car?**
- Magnitude: \[ \boxed{ \dots } \, \text{m/s}^2 \ ]
- Direction (counterclockwise from the +x-axis): \[ \boxed{ \dots }^\circ \]
To calculate the acceleration:
1. Resolve each force into its x and y components.
2. Sum the components to find the net force.
3. Use Newton's second law, \( \vec{F} = m \vec{a} \), to find the acceleration.
Calculation steps:
1. For \( F_1 \):
- \( F_{1x} = F_1 \cos(\theta_1) \)
- \( F_{1y} = F_1 \sin(\theta_1) \)
2. For \( F_2 \):
- \( F_{2x} = F_2 \cos(\theta_2) \)
- \( F_{2y} = F_2 \sin(\theta_2) \)
3. Sum the components:
- Net force in the x-direction: \( F_{net,x} = F_{1x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F219f3226-c940-46da-afab-f83d4af1a26e%2Ff566ae3a-1ede-4394-ba2d-d8d26ad2d54f%2F7hpwuuj_processed.png&w=3840&q=75)
Transcribed Image Text:Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here \( F_1 = 444 \, \text{N} \) and \( F_2 = 343 \, \text{N} \) and friction is negligible. In the diagram below, the mass of the car is 3500 kg, \( \theta_1 = -25^\circ \) and \( \theta_2 = 12^\circ \). (Assume the car faces the positive x-axis before the forces are applied.)
In the diagram:
- \( \vec{F_1} \) is a force of 444 N applied at an angle \( \theta_1 = -25^\circ \) from the positive x-axis.
- \( \vec{F_2} \) is a force of 343 N applied at an angle \( \theta_2 = 12^\circ \) from the positive x-axis.
- The car is represented by a top-down view, indicating the direction of the applied forces.
**(b) What is the acceleration (in m/s\(^2\)) of the car?**
- Magnitude: \[ \boxed{ \dots } \, \text{m/s}^2 \ ]
- Direction (counterclockwise from the +x-axis): \[ \boxed{ \dots }^\circ \]
To calculate the acceleration:
1. Resolve each force into its x and y components.
2. Sum the components to find the net force.
3. Use Newton's second law, \( \vec{F} = m \vec{a} \), to find the acceleration.
Calculation steps:
1. For \( F_1 \):
- \( F_{1x} = F_1 \cos(\theta_1) \)
- \( F_{1y} = F_1 \sin(\theta_1) \)
2. For \( F_2 \):
- \( F_{2x} = F_2 \cos(\theta_2) \)
- \( F_{2y} = F_2 \sin(\theta_2) \)
3. Sum the components:
- Net force in the x-direction: \( F_{net,x} = F_{1x
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