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Q: The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 650 N.…
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A: Fx=-400kNFy=150 kNFA=(-400)2+1502=427.20kNθ=tan-1150-400=-20.6°Since vector is in second quadrant…
Q: The upward normal force exerted by the floor is 620 N on an elevator passenger who weighs 65O N.…
A: The upward normal force exerted by the floor is, N = 620 N weight of the person, W = 650 N
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Solve for the resultant force (magnitude) and (direction) of the concurrent forces shown. The two downward forces are inclined at 1:1.
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