Question

Resolve a weight of 10 N in a direction parallel to a slope inclined at 45° to the horizontal.

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Spam = 10 answers report

Answer 1

Let the force 10N makes an angle θ with the inclined line

parallel component = 10×sinθ N ; perpendicular component = 10×cosθ N

Answer 2

Answer:

Let the force 10N makes an angle θ with the inclined line

parallel component = 10×sinθ N

perpendicular component = 10×cosθ N

Explanation:

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