Case I: if [μs > tanφ ]

Static friction will act. Block will not move.

frictional force=mgsinφ

μs: coefficient of Static friction.

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Given:
- Angle of the inclined plane = 30 degrees
- Coefficient of static friction = 0.8
- Frictional force on the block = 10N
- Acceleration due to gravity (g) = 10 m/s^2

To find:
- Mass of the block

Explanation:

1. Resolving forces:
When a block is placed on an inclined plane, there are two forces acting on it:
- The weight of the block acting vertically downwards (mg)
- The normal force acting perpendicular to the inclined plane (N)

These forces can be resolved into components parallel and perpendicular to the inclined plane.

2. Components of weight:
The weight of the block can be resolved into two components:
- The component perpendicular to the inclined plane is equal to the normal force (N).
- The component parallel to the inclined plane is mg sin(θ), where θ is the angle of the inclined plane.

3. Frictional force:
The frictional force (f) acting on the block opposes its motion and acts parallel to the inclined plane. It can be calculated using the equation:

f = μN

where μ is the coefficient of static friction and N is the normal force.

In this case, f = 10N and μ = 0.8.

4. Normal force:
To find the normal force, we need to consider the vertical equilibrium of forces. The weight component perpendicular to the inclined plane (mg cos(θ)) is balanced by the normal force (N).

N = mg cos(θ)

5. Substitute into the frictional force equation:
Substituting the value of N into the frictional force equation, we get:

f = μN
10 = 0.8 * mg cos(θ)

6. Solve for the mass of the block:
Now, we can solve the equation for the mass (m) of the block:

10 = 0.8 * mg cos(30)
10 = 0.8 * 10 * cos(30)
10 = 8 * 0.866
10 = 6.928

Since the equation does not balance, it means there is an error in the given values or calculations.

Please double-check the values and calculations to ensure accuracy.