14)
F vector at 40 degrees to motion direction Call Unit vector up ramp parallel to ramp Au.
The component of F in the direction of Au is
F dot Au = |F| * 1 cos (40) = 125(.766) = 95.8 Newtons
15) The 125 N is at 20+40 = 60 degrees from horizonal, so at 30 degrees from vertical
125 cos 30 = 108.3 N
15) In Question 14, if the ramp makes an angle of 20 degrees with the level ground. Find the magnitude of the force tending to lift the crate vertically.
++++Textbook Answer for Question 15: 108.3 N
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14) A crate is being dragged up a ramp by a 125 N force applies at an angle of 40 degrees to the ramp. Find the magnitude of the force in the direction of motion (Answer is 95.8 N).
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+++++Please provide a step by step solution to this problem (question 15)! I really don't know what I'm doing! I know the answer, but I don't know how to get that answer! Thank you.
1 answer
1 answer