Two concurrent forces of 12 newton and 18 newton and acting at an angle of 60 degree find the resultant force taking p = 12 newton and q = 18 newton
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duiofxcalico ignoring supposedly distinct
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Considering the angle given as the angle between P and Q
Now
![r = \sqrt[]{p {}^{2} + q {}^{2} + 2pq\cos( \alpha ) } r = \sqrt[]{p {}^{2} + q {}^{2} + 2pq\cos( \alpha ) }](https://tex.z-dn.net/?f=r+%3D++%5Csqrt%5B%5D%7Bp+%7B%7D%5E%7B2%7D++%2B+q+%7B%7D%5E%7B2%7D++%2B++2pq%5Ccos%28+%5Calpha+%29+%7D+)
![= \sqrt{12 {}^{2} + 18 {}^{2} + 2pq \cos(60) } = \sqrt{12 {}^{2} + 18 {}^{2} + 2pq \cos(60) }](https://tex.z-dn.net/?f=+%3D++%5Csqrt%7B12+%7B%7D%5E%7B2%7D+%2B+18+%7B%7D%5E%7B2%7D++%2B+2pq+%5Ccos%2860%29++%7D+)
![= \sqrt[]{144 + 324 + 2 \times 12 \times 18 \cos(60) } = \sqrt[]{144 + 324 + 2 \times 12 \times 18 \cos(60) }](https://tex.z-dn.net/?f=+%3D++%5Csqrt%5B%5D%7B144+%2B+324+%2B+2+%5Ctimes+12+%5Ctimes+18+%5Ccos%2860%29+%7D+)
=√684
=26.15 N
Hope it helps
:)
Now
=√684
=26.15 N
Hope it helps
:)
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