Question : The sum of the magnitudes of two forces acting at a point is 16N. The resultant of these forces is perpendicular to the smaller force and has a magnitude of 8N. If the smaller force is of magnitude x, then the value of x is (in Newton).
Doubt by Suhani
Solution :
R=8 N
A+B=16 (Given)
B=16-A — (1)
Let A be the smaller force
A = x
tanβ=Bsinθ/(A+Bcosθ)
tan90°=Bsinθ/(x+Bcosθ)
1/0=Bsinθ/(x+Bcosθ)
x+Bcoθ = 0 × Bsinθ
x+Bcosθ=0
Bcosθ=-x — (2)
R²=A²+B²+2ABcosθ
R²=A²+(16-x)²+2x(-x)
[Using Equation (1) and (2)]
R²=A²+(16)²+x²-2(16)(x)-2x²
(8)²=2x²+256-32x-2x²
64=256-32x
32x=256-64
32x=192
x=192/32
x=6
Hence, the required value of x is 6.
Similar Question :
The simple sum of two co-initial vectors is 16 units. Their vector sum is 8 units. The resultant of the vectors is perpendicular to the smaller vector. The magnitudes of the two vectors are a) 2 units and 14 units
b) 4 units and 12 units
c) 6 units and 10 units
d) 8 units and 8 units.
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c) 6 units and 10 units