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Three forces acting on a body are shown in the
fiqure. To have the resultant force only alono the
y-direction, the magnitude of the minimum
additonal force needed is (AIPMT (Prelims)-2008
UN
(1) root3N
(2) O.5N
(3) 1.5 N
(4) root 3/4
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20
The answer is (2)0.5N (towards x- axis )
It can be explained as:-
Consider the figure in attachment,
For motion only along y-axis,
x+2Nsin30°+1Ncos60° = 4Nsin30°
where x is additional force,
=>x +2×½+½ = 4×½
=>x+1+½ = 2
=>x+3/2 = 2
=>x = 1/2 = 0.5N
(ps:- since it is not mentioned that the question is talking about which y-axis that's negative or positive, so the answer is also anonymous of direction)
Cheers!
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