(16) Two equal forces acting at a point, at rightangles to each other

1.	(16) Two equal forces acting at a point, at rightangles to each other have a resultant of 14.14 NThe magnitude of each force is(a) 28.28 N(c) 10 N(b) 24.14 N(d) 7.07 N
Answer» Let the magnitudes of the two equal forces be x.Suppose they make a right triangle..with hypotenuse as 14.14 N..(as resultant is the closing side according to triangle law of vector addition). By using Pythagoras theorem 14.14^2=x^2+x^2 On solving ,we get x=9.906 N hence x~10N Letboththeequalforces beFF The resultant will be =√(F²+F²+2F.F.cos(θ))=√(F²+F²+2F.F.cos(θ))(θis the angle between the forces) =√(2F²+0)=√(2F²+0)(cos(90) is 0) =√2.F=√2.F =>√2.F=1414=>√2.F=1414 =>F=1.414∗1000/√2=>F=1.414∗1000/√2 =>F=1000N I think the correct option must be (d) sry in the above answer I have done some mistake.The correct answer is 10N

(16) Two equal forces acting at a point, at rightangles to each other have a resultant of 14.14 NThe magnitude of each force is(a) 28.28 N(c) 10 N(b) 24.14 N(d) 7.07 N

Answer»

Let the magnitudes of the two equal forces be x.Suppose they make a right triangle..with hypotenuse as 14.14 N..(as resultant is the closing side according to triangle law of vector addition).

By using Pythagoras theorem

14.14^2=x^2+x^2

On solving ,we get

x=9.906 N

hence x~10N

Letboththeequalforces beFF

The resultant will be

=√(F²+F²+2F.F.cos(θ))=√(F²+F²+2F.F.cos(θ))(θis the angle between the forces)

=√(2F²+0)=√(2F²+0)(cos(90) is 0)

=√2.F=√2.F

=>√2.F=1414=>√2.F=1414

=>F=1.414∗1000/√2=>F=1.414∗1000/√2

=>F=1000N

I think the correct option must be (d)

sry in the above answer I have done some mistake.The correct answer is 10N

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