- The resultentof two force F and 15 N is 20 inclined at 60 degree to the 15 N force.find the magnitude and direction and Force
Answers
Answer:
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Answer:
The magnitude of resultant force is 30.41 N and is at an angle of 27.45 deg to the 20 N force.
Explanation:
Given that,
First force f_{1}= 15 Nf 1 =15N
Second force f_{2}= 20 NNf2 =20N
We know,
The magnitude of the resultant force is
R = \sqrt{f_{1}^2+f_{2}^2+2f_{1}f_{2}\ cos\theta}R=
f 12 +f 22 +2f f2cosθR =\sqrt{15^{2}+20^{2}\times2\times15\times20\times cos\ 60^{0}}R= 152+202×2×15×20×cos 60
0
R = \sqrt{15^{2}+20^{2}\times2\times15\times10}R=152+20 2×2×15×10
R = \sqrt{925}R= 925
R = 30.41 NR=30.41N
The direction of the resultant force is
tan\theta = \dfrac{f_{1}sin\theta}{f_{1}+f_{2}cos\theta}tanθ= f 1 +f 2cosθf 1
sinθtan\theta = \dfrac{15\times sin\ 60^{0}}{15+20\ cos 60^{0}}tanθ=
15+20 cos60
0
15×sin 60
0
tan\theta = \dfrac{15\times\dfrac{\sqrt{3}}{2}}{15+20\times\dfrac{1}{2}}tanθ=
15+20× 21
15× 23tan\theta = 0.5196tanθ=0.5196\theta = tan^{-1}0.5196θ=tan −10.5196
\theta = 27.45^{0}θ=27.45 0
Hence, The magnitude of resultant force is 30.41 N and is at an angle of 27.45 deg to the 20 N for