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Question
Two forces −→F1= 500 N due east and −→ F2=250 N due north have their common initial point. −→F2−−→F1 is
Two forces −→F1= 500 N due east and −→ F2=250 N due north have their common initial point. −→F2−−→F1 is
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Solution
The correct option is A 250√5 N, tan−1(2) west of north
−→F2−−→F1 = −→F2+(−−→F1)= 250 N due north + 500 N due west and
tanθ = 500250=2 ⇒θ = tan−1(2)
As the above two direction are West and North , So θ will be in direction of west of North and
(−→F2−−→F1∣∣∣ = √(500)2+(250)2=250√5 N
−→F2−−→F1 = −→F2+(−−→F1)= 250 N due north + 500 N due west and
tanθ = 500250=2 ⇒θ = tan−1(2)
As the above two direction are West and North , So θ will be in direction of west of North and
(−→F2−−→F1∣∣∣ = √(500)2+(250)2=250√5 N
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