1. A 90.0 kg crate is being pushed across a level floor at a constant speed by a 260 N force at an angle of 28.0° below the horizontal. The surface is not frictionless. a) Draw a free body diagram that includes all of the forces exerted on the crate, including a labeled system coordinate. FN FC0328 mg b) What are the algebraic expressions for the normal force (FN) and the coefficient of kinetic friction between the crate and the floor? EF $28

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### Physics Problem: Force and Motion on a Crate

#### Problem Statement
A 90.0 kg crate is being pushed across a level floor at a constant speed by a 260 N force at an angle of 28.0° below the horizontal. The surface is not frictionless.

#### Question a
**a)** Draw a free-body diagram that includes all of the forces exerted on the crate, including a labeled system coordinate.

**Free-Body Diagram:**

- **Diagram Description:** 
  - There is a rectangular block labeled "m" to represent the crate.
  - A force, \( F \), is applied at an angle of 28° below the horizontal.
  - The components of the applied force are broken down into \( F \cos 28° \) (horizontal component) and \( F \sin 28° \) (vertical component pointing downwards).
  - The normal force, \( F_N \), is acting upwards.
  - The gravitational force, \( mg \), is acting downwards.

#### Question b
**b)** What are the algebraic expressions for the normal force (\( F_N \)) and the coefficient of kinetic friction between the crate and the floor?

\[ \sum F = 0 \]

#### Question c
**c)** Calculate the magnitude of the normal force and the coefficient of kinetic friction between the crate and the floor.

#### Solution Explanation

1. **Free-Body Diagram Analysis:**
   - From the free-body diagram, list all forces acting on the crate:
     - Normal Force (\( F_N \))
     - Gravitational Force (\( mg \))
     - Applied Force (\( F \)) divided into \( F \cos 28° \) and \( F \sin 28° \)

2. **Equilibrium Conditions:**
   - Since the crate is moving at a constant speed, the acceleration is zero, meaning the sum of forces in both the x (horizontal) and y (vertical) directions must be zero.
   - Horizontal Direction:
     \[ \sum F_x = F \cos 28° - F_{\text{friction}} = 0 \]
     - So, the frictional force \( F_{\text{friction}} = F \cos 28° \).
     
   - Vertical Direction:
     \[ \sum F_y = F_N + F \sin 28° - mg = 0 \]
Transcribed Image Text:### Physics Problem: Force and Motion on a Crate #### Problem Statement A 90.0 kg crate is being pushed across a level floor at a constant speed by a 260 N force at an angle of 28.0° below the horizontal. The surface is not frictionless. #### Question a **a)** Draw a free-body diagram that includes all of the forces exerted on the crate, including a labeled system coordinate. **Free-Body Diagram:** - **Diagram Description:** - There is a rectangular block labeled "m" to represent the crate. - A force, \( F \), is applied at an angle of 28° below the horizontal. - The components of the applied force are broken down into \( F \cos 28° \) (horizontal component) and \( F \sin 28° \) (vertical component pointing downwards). - The normal force, \( F_N \), is acting upwards. - The gravitational force, \( mg \), is acting downwards. #### Question b **b)** What are the algebraic expressions for the normal force (\( F_N \)) and the coefficient of kinetic friction between the crate and the floor? \[ \sum F = 0 \] #### Question c **c)** Calculate the magnitude of the normal force and the coefficient of kinetic friction between the crate and the floor. #### Solution Explanation 1. **Free-Body Diagram Analysis:** - From the free-body diagram, list all forces acting on the crate: - Normal Force (\( F_N \)) - Gravitational Force (\( mg \)) - Applied Force (\( F \)) divided into \( F \cos 28° \) and \( F \sin 28° \) 2. **Equilibrium Conditions:** - Since the crate is moving at a constant speed, the acceleration is zero, meaning the sum of forces in both the x (horizontal) and y (vertical) directions must be zero. - Horizontal Direction: \[ \sum F_x = F \cos 28° - F_{\text{friction}} = 0 \] - So, the frictional force \( F_{\text{friction}} = F \cos 28° \). - Vertical Direction: \[ \sum F_y = F_N + F \sin 28° - mg = 0 \]
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