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Q.

Two forces F→1=500N due east and F→2=250N due north have their common initial point F→2−F→1 is

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a

2505N,tan−1⁡(2)W of N

b

250N,tan−1⁡(2)W of N

c

Zero

d

750N,tan−1⁡(3/4)N of W

answer is A.

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Detailed Solution

F→2−F→1=F→2+−F→1=250N due north +500N due west tan⁡θ=500200=2F→2−F→1=(500)2+(250)2=2505N
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Two forces F→1=500N due east and F→2=250N due north have their common initial point F→2−F→1 is