Question

Two forces of magnitude 10 n and 20 n acting on 120 degree. find the pull that will replace the given forces system?

Answer 1

Answer:

$4 \sqrt{30 \: }$

Explanation:

$\sqrt{ {10}^{2} + {20}^{2} + 2 \times 10 \times } \\ \\ = \sqrt{100 + 400 + 2 \times 10 \times 20 \times \cos(120) } \\ \\ = \sqrt{500 - 40 \div 2} \\ \\ = \sqrt{500 - 20} = \sqrt{480} \: newton \\ = 4 \sqrt{30 } \: newton$