A 2.20 kg block is in equilibrium on an incline of 37.3 degrees. The acceleration of gravity is 9.81 m/s2 . What is Fn of the incline on the block?

Answer in units of N.

Fn=mg*cosTheta is force normal

jgh

27 To find the normal force (Fn) of the incline on the block, we need to consider the forces acting on the block in the vertical direction.

The forces acting on the block are:
1. The gravitational force (weight) Mg acting vertically downward.
2. The normal force (Fn) acting perpendicular to the incline.

Since the block is in equilibrium, the net force acting in the vertical direction must be zero. This means the magnitude of the normal force (Fn) must be equal in magnitude and opposite in direction to the gravitational force (weight) Mg.

We can start by finding the magnitude of the gravitational force (weight). The weight (W) is given by the formula:

W = Mg

M is the mass of the block in kg and g is the acceleration due to gravity in m/s^2.

Given:
Mass of the block, M = 2.20 kg
Acceleration due to gravity, g = 9.81 m/s^2

Using the formula, we can calculate the weight (W) of the block as:

W = 2.20 kg * 9.81 m/s^2

Now, to find the normal force (Fn), we need to consider the vertical component of the weight. Since the incline is at an angle of 37.3 degrees, the vertical component of the weight is:

Vertical Component of Weight = W * cos(37.3 degrees)

Finally, the normal force (Fn) is equal in magnitude but opposite in direction to the vertical component of the weight. Therefore, the magnitude of the normal force (Fn) is:

Fn = - Vertical Component of Weight

Now, we can substitute the values into the equations and calculate the normal force (Fn).