A 3.20 kg block is in equilibrium on an incline

of 29.6
The acceleration of gravity is 9.81 m/s2 .
What is Fn of the incline on the block?
Answer in units of N

To find the normal force (Fn) exerted by the incline on the block, we need to understand that in equilibrium, the forces acting on the object are balanced. In this case, the normal force exerted by the incline is equal in magnitude but opposite in direction to the component of the gravitational force acting perpendicular to the incline.

To calculate the normal force, we can first find the gravitational force acting on the block, and then determine the component of this force that is perpendicular to the incline.

The formula to calculate the gravitational force is given by:

Fg = m * g

where
Fg is the gravitational force,
m is the mass of the block, and
g is the acceleration due to gravity.

Given:
m = 3.20 kg (mass of the block)
g = 9.81 m/s² (acceleration due to gravity)

Substituting the values, we have:

Fg = 3.20 kg * 9.81 m/s²
= 31.392 N

Next, we need to determine the perpendicular component of the gravitational force. This component is given by:

Fn = Fg * cos(theta)

where
Fn is the normal force,
Fg is the gravitational force, and
theta is the angle between the incline and the vertical direction.

Given:
theta = 29.6° (angle of the incline)

Converting the angle to radians:
theta (in radians) = theta (in degrees) * (pi/180)
= 29.6° * (pi/180)
≈ 0.5164 radians

Substituting the values, we have:

Fn = 31.392 N * cos(0.5164 radians)
≈ 31.392 N * 0.8708
≈ 27.344 N

Therefore, the normal force (Fn) exerted by the incline on the block is approximately 27.344 N.