Resolve a weight of 10 N in a direction parallel to a slope inclined at 45° to the horizontal.
✖️ No spam ✖️
Spam = 10 answers report
Answers
Answered by
1
Let the force 10N makes an angle θ with the inclined line
parallel component = 10×sinθ N ; perpendicular component = 10×cosθ N
Similar questions