The force applied by the student

Oct 13, 2025 by sorusgentrification.com 135 views

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Introduction

When a student pushes on a box, several forces come into play, including the force applied by the student, the normal force exerted by the floor on the box, and the force of friction. In this scenario, we are given the mass of the box, the force applied by the student, the angle of the force, and the acceleration of the box. Our goal is to determine the value of the normal force on the box.

The Forces Acting on the Box

The box is subject to several forces, including:

The force applied by the student: This is the force that the student exerts on the box, given as 50 N at an angle of $30^{\circ}$ below the horizontal.
The normal force: This is the force exerted by the floor on the box, perpendicular to the surface of the floor.
The force of friction: This is the force that opposes the motion of the box, acting in the opposite direction to the force applied by the student.
The weight of the box: This is the force due to gravity, acting downward on the box.

The Components of the Force Applied by the Student

The force applied by the student can be resolved into its horizontal and vertical components. The horizontal component of the force is given by:

F_{\text{horizontal}} = F \cos \theta

where $F$ is the magnitude of the force and $\theta$ is the angle below the horizontal.

The vertical component of the force is given by:

F_{\text{vertical}} = F \sin \theta

The Acceleration of the Box

The acceleration of the box is given as $0.5 \, \text{m/s}^2$ . This acceleration is due to the net force acting on the box, which is the sum of the horizontal component of the force applied by the student and the force of friction.

The Normal Force

The normal force is the force exerted by the floor on the box, perpendicular to the surface of the floor. This force is equal in magnitude and opposite in direction to the vertical component of the force applied by the student.

The Weight of the Box

The weight of the box is the force due to gravity, acting downward on the box. This force is given by:

W = mg

where $m$ is the mass of the box and $g$ is the acceleration due to gravity.

The Net Force Acting on the Box

The net force acting on the box is the sum of the horizontal component of the force applied by the student, the force of friction, and the normal force. This net force is responsible for the acceleration of the box.

The Equations of Motion

The equations of motion for the box can be written as:

F_{\text{horizontal}} - f = ma

F_{\text{vertical}} + N - W = 0

where $f$ is the force of friction, $N$ is the normal force, and $a$ is the acceleration of the box.

The Solution

To solve for the normal force, we can use the second equation:

F_{\text{vertical}} + N - W = 0

Substituting the values given in the problem, we get:

50 \sin 30^{\circ} + N - 20 \times 9.8 = 0

Simplifying the equation, we get:

25 + N - 196 = 0

Solving for $N$ , we get:

N = 171 \, \text{N}

Conclusion

In this scenario, the student pushes on a 20.0 kg box with a force of 50 N at an angle of $30^{\circ}$ below the horizontal. The box accelerates at a rate of $0.5 \, \text{m/s}^2$ across a horizontal floor. The value of the normal force on the box is 171 N.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
[2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Discussion

The normal force is an important concept in physics, as it plays a crucial role in determining the motion of objects. In this scenario, the normal force is equal in magnitude and opposite in direction to the vertical component of the force applied by the student. The value of the normal force can be determined using the equations of motion and the given values of the force applied by the student and the acceleration of the box.

Q: What is the normal force on the box?

A: The normal force on the box is 171 N. This is the force exerted by the floor on the box, perpendicular to the surface of the floor.

Q: What is the force applied by the student?

A: The force applied by the student is 50 N at an angle of $30^{\circ}$ below the horizontal.

Q: What is the acceleration of the box?

A: The acceleration of the box is $0.5 \, \text{m/s}^2$ . This acceleration is due to the net force acting on the box, which is the sum of the horizontal component of the force applied by the student and the force of friction.

Q: What is the weight of the box?

A: The weight of the box is $20 \times 9.8 = 196 \, \text{N}$ . This is the force due to gravity, acting downward on the box.

Q: What is the horizontal component of the force applied by the student?

A: The horizontal component of the force applied by the student is $50 \cos 30^{\circ} = 43.3 \, \text{N}$ .

Q: What is the vertical component of the force applied by the student?

A: The vertical component of the force applied by the student is $50 \sin 30^{\circ} = 25 \, \text{N}$ .

Q: What is the force of friction?

A: The force of friction is not given in the problem, but it can be determined using the equation:

f = \mu N

where $\mu$ is the coefficient of friction and $N$ is the normal force.

Q: What is the coefficient of friction?

A: The coefficient of friction is not given in the problem, but it can be determined using the equation:

\mu = \frac{f}{N}

Q: How can the normal force be determined?

A: The normal force can be determined using the equation:

N = F_{\text{vertical}} + W

where $F_{\text{vertical}}$ is the vertical component of the force applied by the student and $W$ is the weight of the box.

Q: What is the relationship between the normal force and the force of friction?

A: The normal force and the force of friction are related by the equation:

f = \mu N

This equation shows that the force of friction is proportional to the normal force.

Q: What is the significance of the normal force?

A: The normal force is an important concept in physics, as it plays a crucial role in determining the motion of objects. The normal force is equal in magnitude and opposite in direction to the vertical component of the force applied by the student.

Q: What are some real-world applications of the normal force?

A: The normal force has many real-world applications, including:

Designing buildings and bridges: The normal force is an important consideration in the design of buildings and bridges, as it helps to determine the stability of the structure.
Understanding the motion of objects: The normal force is essential in understanding the motion of objects, including the forces that act on them and the resulting motion.
Developing new technologies: The normal force is an important concept in the development of new technologies, including robotics and artificial intelligence.

Q: What are some common misconceptions about the normal force?

A: Some common misconceptions about the normal force include:

The normal force is always equal to the weight of the object: This is not true, as the normal force can be greater or less than the weight of the object, depending on the forces acting on it.
The normal force is always perpendicular to the surface: This is not true, as the normal force can be at an angle to the surface, depending on the forces acting on it.

Q: What are some tips for understanding the normal force?

A: Some tips for understanding the normal force include:

Visualize the forces acting on the object: Visualizing the forces acting on the object can help to understand the normal force and its relationship to the other forces.
Use diagrams and graphs: Diagrams and graphs can help to visualize the forces acting on the object and the resulting motion.
Practice problems: Practicing problems can help to reinforce understanding of the normal force and its relationship to the other forces.