What is the normal force exerted on a block of mass M by a surface which is inclined at an angle theta above the horizontal?

Answer

To find the normal force exerted on the block, we analyze the forces acting perpendicular to the inclined surface. 1. Define the Coordinate System: Let the x-axis be parallel to the inclined plane and the y-axis be perpendicular to the plane. 2. Identify the Forces: - The gravitational force (weight), Fg = M*g, acts vertically downward. - The normal force, FN, acts perpendicular to the surface in the positive y-direction. 3. Decompose the Gravitational Force: The weight component perpendicular to the surface (along the y-axis) is M*g*cos(theta), and the component parallel to the surface (along the x-axis) is M*g*sin(theta). 4. Apply Newton's Second Law: In the direction perpendicular to the surface (y-axis), there is no acceleration (ay = 0). Therefore, the sum of the forces in the y-direction must be zero: ΣFy = FN - M*g*cos(theta) = 0 5. Solve for Normal Force: By Rearranging the equation, we find that: FN = M*g*cos(theta). Thus, the normal force exerted on the block is M*g*cos(theta).