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Grade 12th passGeneral Physics

The sum of magnitudes of two forces acting at a point is 16N. If the resultant force is 8N and its direction is perpendicular to smaller force,then the forces are:- (A)6N and 10N, (B)8N and 8N

Profile image of Zoya khan
10 Years agoGrade 12th pass
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4 Answers

Profile image of Vikas TU
9 Years ago
Let the force be A and B, then the given condittions are:
|A| + |B| = 16.......................(1)
and
|A|^2 + |B|^2 + 2ABcosthetha = 64..........................(2)
and let B be the smaller force then,’
given 
(A + B).B = 0 
or
A.B + |B|^2 = 0......................(3)
3 eqns, 3 variables solve them and get the Foces.
Profile image of Atin Vikram Singh
9 Years ago
A(vector)+B(vector)=R(vector)          |A|+|B|=16         
Since resultant is perpendicular to smaller force
R*R+A*A=B*B    (Traingle Law of Addition and PYTHAGORES theorm)
64+A*A=B*B
Option a satisfies this condition (we can also  solve it also using A=16-B)
Answer = a
Profile image of Prathmesh A Bharuka
9 Years ago
Let two forces be A & B, and the resultant be P.
A + B = 16
P2 = B2 – A2 = (B – A)(B + A)  = 16(B – A)
(B - A) = P2/16 = 4
B – A = 16
B + A = 4
2B = 12
B = 6N
A = 10N 
 
Profile image of Kushagra Madhukar
5 Years ago
Dear student,
Please find the solution to your question.
 
Let the force be A and B, then the given condittions are:
|A| + |B| = 16.......................(1)
and
|A|2 + |B|2 + 2A.B = 64..........................(2)
and let B be the smaller force then,’
given 
(A + B).B = 0 
or
A.B + |B|2 = 0......................(3)
Hence, A.B = – |B|2
Replacing in equation (2)
|A|2 + |B|2 + 2(– |B|2) = 64
or, |A|2 – |B|2 = 64
Hence, (|A| + |B|)(|A| – |B|) = 64
Hence, |A| – |B| = 64/16 = 4 …....(4)
Adding, eqn(1) and (2)
Hence, 2|A| = 20
or, |A| = 10 N
|B| = 6 N
 
Thanks and regards,
Kushagra

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