Question

A block of mass M is placed on an inclined surface. The incline makes an angle theta with respect to the horizontal. What are the magnitude and direction of the normal force exerted on the block by the incline? Group of answer choices magnitude =LaTeX: Mgcos\theta M g c o s θ , direction: perpendicular to the incline surface magnitude = LaTeX: Mgsin\theta M g s i n θ , direction: perpendicular to the incline surface magnitude =LaTeX: Mgcos\theta M g c o s θ , direction: perpendicular to the surface of the earth magnitude = Mg, direction: perpendicular to the incline surface magnitude = Mg, direction: perpendicular to the surface of the earth

Answer 1

Answer:

$N = Mg cos\theta$

Explanation:

As we know that the block is placed on the inclined plane

So here we know that the components of weight of the block are

$F_x = Mgsin\theta$ along the inclined plane

above component of the weight will be counter balanced by friction force as it will be at rest

$F_y = Mgcos\theta$ perpendicular to the inclined plane

above component is perpendicular to inclined plane surface and it will be counter balanced by the normal force

So here correct answer for normal force will be

$N = Mg cos\theta$