**Physics Problem on Normal Force Calculation****Problem Statement:**A horizontal force \( F \) is being used to push an object up a smooth inclined plane. The angle between the plane and the horizontal is \( \theta \). The normal force is:**Options:**(a) \( mg \cos \theta + F \cos \theta \) (b) \( mg \cos \theta \) (c) \( mg \cos \theta - F \sin \theta \) (d) \( mg \cos \theta + F \sin \theta \) (e) \( mg \cos \theta - F \cos \theta \)**Explanation:**In this problem, we are asked to determine the normal force exerted by the inclined plane on an object being pushed up by a horizontal force \( F \). The angle between the inclined plane and the horizontal is given as \( \theta \).To solve this, we need to analyze the forces in the direction perpendicular to the inclined plane. The normal force is the component of the contact force perpendicular to the surface of the incline. Typically, the forces to consider include the gravitational force and any applied forces.The gravitational force \( mg \) acts vertically downwards, and its component perpendicular to the inclined plane is \( mg \cos \theta \). The horizontal force \( F \) has a component acting parallel to the incline and one acting perpendicular to the plane. The perpendicular component of this horizontal force \( F \) is \( F \sin \theta \).Summarizing the forces perpendicular to the inclined plane:- The downward gravitational component perpendicular to the plane: \( mg \cos \theta \).- The component of the applied force \( F \) perpendicular to the plane: \( F \sin \theta \).Therefore, the normal force \( N \) is the sum of these components:\[ N = mg \cos \theta + F \sin \theta \]So, the correct answer is:(d) \( mg \cos \theta + F \sin \theta \)**Important Notes:**- This is a key example of resolving forces into components along the directions defined by inclines, which is a common technique in physics to simplify and solve problems involving inclined planes. - Understanding how to break down forces and correctly add their contributions along specific directions is crucial for correctly determining reactions such as normal force.

Answered: Image /qna-images/answer/a7d3637c-db94-4a09-818c-f905eb9904d1.jpg

A horizontal force F is being used to push an object up a smooth inclined plane. The angle between the plane and the horizontal is 8. The normal force is (a) mg cos+ F cos (b) mg cos (c) mg cos - F sin 0 (d) mg cos 0+F sin 0 (e) mg cos - F cos 0

A horizontal force F is being used to push an object up a smooth inclined plane. The angle between the plane and the horizontal is 8. The normal force is (a) mg cos+ F cos (b) mg cos (c) mg cos - F sin 0 (d) mg cos 0+F sin 0 (e) mg cos - F cos 0

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Question

**Physics Problem on Normal Force Calculation**

**Problem Statement:**

A horizontal force \( F \) is being used to push an object up a smooth inclined plane. The angle between the plane and the horizontal is \( \theta \). The normal force is:

**Options:**

(a) \( mg \cos \theta + F \cos \theta \)
(b) \( mg \cos \theta \)
(c) \( mg \cos \theta - F \sin \theta \)
(d) \( mg \cos \theta + F \sin \theta \)
(e) \( mg \cos \theta - F \cos \theta \)

**Explanation:**

In this problem, we are asked to determine the normal force exerted by the inclined plane on an object being pushed up by a horizontal force \( F \). The angle between the inclined plane and the horizontal is given as \( \theta \).

To solve this, we need to analyze the forces in the direction perpendicular to the inclined plane. The normal force is the component of the contact force perpendicular to the surface of the incline. Typically, the forces to consider include the gravitational force and any applied forces.

The gravitational force \( mg \) acts vertically downwards, and its component perpendicular to the inclined plane is \( mg \cos \theta \). The horizontal force \( F \) has a component acting parallel to the incline and one acting perpendicular to the plane. The perpendicular component of this horizontal force \( F \) is \( F \sin \theta \).

Summarizing the forces perpendicular to the inclined plane:

- The downward gravitational component perpendicular to the plane: \( mg \cos \theta \).
- The component of the applied force \( F \) perpendicular to the plane: \( F \sin \theta \).

Therefore, the normal force \( N \) is the sum of these components:

\[ N = mg \cos \theta + F \sin \theta \]

So, the correct answer is:

(d) \( mg \cos \theta + F \sin \theta \)

**Important Notes:**

- This is a key example of resolving forces into components along the directions defined by inclines, which is a common technique in physics to simplify and solve problems involving inclined planes.
- Understanding how to break down forces and correctly add their contributions along specific directions is crucial for correctly determining reactions such as normal force.

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