6. In a Figure below, the block of mass m is at rest on an inclined plane that makes an angle with the horizontal. The force of static friction f must be such that: 0 a. f>mg b.f > mg cose. c. f> mg sine. d. f = mg cose = mg sin0. De.f

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**Understanding Static Friction on an Inclined Plane**

Consider the scenario depicted in the diagram below. A block of mass \( m \) is at rest on an inclined plane. The plane forms an angle \( \theta \) with the horizontal. To maintain the block at rest, the force of static friction \( f \) must counteract certain components of the gravitational force acting on the block.

### Diagram Explanation
The diagram illustrates:
- An inclined plane forming an angle \( \theta \) with the horizontal.
- A block of mass \( m \) placed on the plane. 

The diagram helps in visualizing the forces acting on the block:

- The gravitational force \( mg \) acting downward.
- The normal force exerted by the plane perpendicular to its surface.
- The force of static friction \( f \), which acts parallel to the surface of the inclined plane, preventing the block from sliding down.

### Question and Answer Choices
In this scenario, we are asked to determine the condition that the force of static friction \( f \) must satisfy. Here are the options:

- **a. \( f > mg \)** 
- **b. \( f > mg \cos \theta \)** 
- **c. \( f > mg \sin \theta \)** 
- **d. \( f = mg \cos \theta \)** 
- **e. \( f = mg \sin \theta \)** 

**Correct Answer:** 

The static friction must counteract the component of the gravitational force parallel to the inclined plane for the block to remain stationary. This component is given by \( mg \sin \theta \). Therefore, the force of static friction \( f \) must equal this component for the block to stay at rest. Hence, the correct answer is:

**e. \( f = mg \sin \theta \)**
Transcribed Image Text:**Understanding Static Friction on an Inclined Plane** Consider the scenario depicted in the diagram below. A block of mass \( m \) is at rest on an inclined plane. The plane forms an angle \( \theta \) with the horizontal. To maintain the block at rest, the force of static friction \( f \) must counteract certain components of the gravitational force acting on the block. ### Diagram Explanation The diagram illustrates: - An inclined plane forming an angle \( \theta \) with the horizontal. - A block of mass \( m \) placed on the plane. The diagram helps in visualizing the forces acting on the block: - The gravitational force \( mg \) acting downward. - The normal force exerted by the plane perpendicular to its surface. - The force of static friction \( f \), which acts parallel to the surface of the inclined plane, preventing the block from sliding down. ### Question and Answer Choices In this scenario, we are asked to determine the condition that the force of static friction \( f \) must satisfy. Here are the options: - **a. \( f > mg \)** - **b. \( f > mg \cos \theta \)** - **c. \( f > mg \sin \theta \)** - **d. \( f = mg \cos \theta \)** - **e. \( f = mg \sin \theta \)** **Correct Answer:** The static friction must counteract the component of the gravitational force parallel to the inclined plane for the block to remain stationary. This component is given by \( mg \sin \theta \). Therefore, the force of static friction \( f \) must equal this component for the block to stay at rest. Hence, the correct answer is: **e. \( f = mg \sin \theta \)**
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