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a 100 kg box is sitting on a 10 degree incline. find a. the parallel and perpendicular components of the force due to gravity. b. the normal force. c. the force of friction that keeps the block from sliding

User FazeL
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(a) The parallel and perpendicular components of the force due to gravity is 170.2 N and 965.11 N respectively.

(b) The normal force is 980 N

(c) The force of friction that keeps the block from sliding is 170.2 N.

How to calculate the component's of the block's weight?

(a) The parallel and perpendicular components of the force due to gravity is calculated as;

F(parallel) = mg sinθ

F (parallel) = 100 kg x 9.8 m/s² x sin (10)

F(parallel) = 170.2 N

F (perpendicular) = mg cosθ

F (perpendicular) = 100 kg x 9.8 m/s² x cos (10)

F (perpendicular) = 965.11 N

(b) The normal force is calculated as;

Fn = mg

Fn = 100 kg x 9.8 m/s²

Fn = 980 N

(c) The force of friction that keeps the block from sliding is calculated as;

F = μ mg cosθ

F = (tan θ) mg cosθ

F = (tan θ) F(perpendicular)

F = (tan 10) x 965.11 N

F = 170.2 N

User Edcs
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