A 2kg block is in equilibrium on an incline of 60 degrees. find fn of the incline on the block
To find the normal force (Fn) acting on the block on an incline, you'll need to consider the gravitational force (weight) acting on the block and the forces acting perpendicular and parallel to the incline.
First, let's analyze the forces acting on the block on the incline:
1. Gravitational force:
The weight of the block is given by the equation:
W = m * g
Where:
m = mass of the block = 2 kg
g = acceleration due to gravity ≈ 9.8 m/s^2
So, the weight of the block can be calculated as:
W = 2 kg * 9.8 m/s^2
W = 19.6 N
2. Forces parallel and perpendicular to the incline:
The force of gravity can be divided into two components:
- The component parallel to the incline (mg*sinθ)
- The component perpendicular to the incline (mg*cosθ)
θ = angle of the incline = 60 degrees
So, the force parallel to the incline (F_parallel) is given by:
F_parallel = mg * sinθ
F_parallel = 2 kg * 9.8 m/s^2 * sin(60°)
F_parallel ≈ 16.94 N
Here, the force perpendicular to the incline (F_perpendicular) is equal in magnitude but opposite in direction to the normal force (Fn). Therefore:
Fn = F_perpendicular = mg * cosθ
Fn = 2 kg * 9.8 m/s^2 * cos(60°)
Fn ≈ 9.8 N
Hence, the normal force (Fn) exerted by the incline on the block is approximately 9.8 N.