Question

Block on inclined plane experience a force due to gravity of 300N straight down. If the slope is inclined at 67.8°to the horizontal. What is the component of the force due to gravity perpendicular and parallel to the slope?

Answer 1

Answer:

The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.

Explanation:

Force is any cause capable of modifying the state of motion or rest of a body or of producing a deformation in it. Any force can be decomposed into two vectors, so that the sum of both vectors matches the vector before decomposing. The decomposition of a force into its components can be done in any direction.

Taking into account the simple trigonometric relations, such as sine, cosine and tangent, the value of their components and the value of the angle of application, then the parallel and perpendicular components will be:

• Fparallel = F*sinα =300 N*sin 67.8° =300 N*0.926⇒ Fparallel =277.8 N
• Fperpendicular = F*cosα =  300 N*cos 67.8° = 300 N*0.378 ⇒ Fperpendicular= 113.4 N

The component of the force due to gravity perpendicular and parallel to the slope is  113.4 N and 277.8 N respectively.