Two forces of 3N and 4N are acting at a point such that the angle between them is 60degrees. Find the resultant force.
A
Mangnitude of R=8N and angle q=tan−10.472
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B
Mangnitude of R=6N and angle q=tan−10.572
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C
Mangnitude of R=6.08N and angle q=tan−10.472
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D
Mangnitude of R=6.08N and angle q=tan−10.872
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Solution
The correct option is C Mangnitude of R=6.08N and angle q=tan−10.472 magnitude of resultant R is given as
= √(3)2+(4)2+2×3×4cos60∘ =√37=6.08N
Direction of R is given by finding the angle q
Thus, tanq=3sin(60∘)4+3cos(60∘)=0.472 ⇒q=tan−10.472