### Problem 5: Determining the Normal Force on a BoxA large box is pulled by a person with a force $ \vec{F} $ as illustrated in the figure below. The angle $ \theta $ between the force vector and the horizontal surface is $ 25^\circ $. Given that the mass of the box is $ m = 23.5 $ kg and the coefficient of static friction between the box and the floor is $ 0.64 $, calculate the magnitude of the normal force exerted on the box at the moment when it is just about to move.#### Details and Diagram- **Mass of the box (m)**: 23.5 kg- **Coefficient of static friction ($ \mu_s $)**: 0.64- **Angle ($ \theta $)**: $ 25^\circ $![Diagram Showing Force $ \vec{F} $ Applied at an Angle $ \theta $](image_link)In the diagram, the force $ \vec{F} $ is applied at an angle $ \theta $ of $ 25^\circ $ above the horizontal. The forces acting on the box need to be analyzed to determine the normal force acting on it just as it is about to move.To find the normal force,\[ N = mg - F \sin(\theta) \]Where:- $ N $ is the normal force.- $ m $ is the mass of the box.- $ g $ is the acceleration due to gravity (approximately $ 9.81 \, \text{m/s}^2 $).- $ F $ is the applied force.- $ \theta $ is the angle of the applied force from the horizontal.Given this setup, the applied force $ \vec{F} $ can be further analyzed including its horizontal and vertical components. The normal force can be precisely determined by resolving all the vertical components acting on the box including the effect of $ \vec{F} $. This equation explains the reduction in normal force due to the vertical component of the applied force $ F $.This problem is an excellent example of static equilibrium and the application of Newton's laws in two dimensions. Students should consider breaking down the forces into their components and applying the equilibrium conditions to solve for the normal force.

Given: The pulling force is F. The direction of the pulling force with horizontal is 25o. The mass…

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ssf60 ssf60 ssfo ´0 ssf60 ssf60 ssf60 ssf60 ssf6° 5. A large box is pulled by a person with a force F as shown in the figure. The angle 8 = 25°. What is the magnitude of the normal force at the moment when the box is about to move if the mass of the box is m = 23.5 kg and the coeff of static friction between the box and the floor is 0.64? 8 TG ssf60 ssf60 ssf60 ssfó ƒ60 ssf60 ssf60 ssf60 s ssf60f50 s$£60sf60 ssf60 ssf6 50 s 6sfossf60 ssf50 ssf60 st 09Jss

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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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Question

### Problem 5: Determining the Normal Force on a Box

A large box is pulled by a person with a force $ \vec{F} $ as illustrated in the figure below. The angle $ \theta $ between the force vector and the horizontal surface is $ 25^\circ $. Given that the mass of the box is $ m = 23.5 $ kg and the coefficient of static friction between the box and the floor is $ 0.64 $, calculate the magnitude of the normal force exerted on the box at the moment when it is just about to move.

#### Details and Diagram

- **Mass of the box (m)**: 23.5 kg
- **Coefficient of static friction ($ \mu_s $)**: 0.64
- **Angle ($ \theta $)**: $ 25^\circ $

![Diagram Showing Force $ \vec{F} $ Applied at an Angle $ \theta $](image_link)

In the diagram, the force $ \vec{F} $ is applied at an angle $ \theta $ of $ 25^\circ $ above the horizontal. The forces acting on the box need to be analyzed to determine the normal force acting on it just as it is about to move.

To find the normal force,
\[ N = mg - F \sin(\theta) \]

Where:
- $ N $ is the normal force.
- $ m $ is the mass of the box.
- $ g $ is the acceleration due to gravity (approximately $ 9.81 \, \text{m/s}^2 $).
- $ F $ is the applied force.
- $ \theta $ is the angle of the applied force from the horizontal.

Given this setup, the applied force $ \vec{F} $ can be further analyzed including its horizontal and vertical components. The normal force can be precisely determined by resolving all the vertical components acting on the box including the effect of $ \vec{F} $.

This equation explains the reduction in normal force due to the vertical component of the applied force $ F $.

This problem is an excellent example of static equilibrium and the application of Newton's laws in two dimensions. Students should consider breaking down the forces into their components and applying the equilibrium conditions to solve for the normal force.

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