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靜力學(解答)
發問:
最佳解答:
1. Resolve the 30N in the direction parallel and perpendicular to the 40N force. For direction parallel to the 40N force, force componet = 30cos(60) N For direction perpendicualr to the 40N force, force component = 30.sin(60) N Hence, sum of force in direction of 40N force = (40 + 30.cos(60)) N = 55 N force in direction perpendicualr to 40N force = 30.sin(60) N = 25.98 N Therefore, resultant force = square-root[55^2 + 25.98^2] N = 60.8 N 2. Force difference in direction parallel to 40N force = [40 - 30.cos(60)] N = 25 N Therefore, resultant force = square-root[25^2 + 25.98^2] N = 36.1 N
其他解答:
靜力學(解答)
發問:
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1. F1=30N,F2=40N,兩力夾角60度,則F1+F2之大小為多少?(求試算過程) 2 承上題,F1-F2之大小為何?(求試算過程)最佳解答:
1. Resolve the 30N in the direction parallel and perpendicular to the 40N force. For direction parallel to the 40N force, force componet = 30cos(60) N For direction perpendicualr to the 40N force, force component = 30.sin(60) N Hence, sum of force in direction of 40N force = (40 + 30.cos(60)) N = 55 N force in direction perpendicualr to 40N force = 30.sin(60) N = 25.98 N Therefore, resultant force = square-root[55^2 + 25.98^2] N = 60.8 N 2. Force difference in direction parallel to 40N force = [40 - 30.cos(60)] N = 25 N Therefore, resultant force = square-root[25^2 + 25.98^2] N = 36.1 N
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