Forces of magnitude 20 N, $10 \sqrt{3}$ N, $20 \sqrt{3}$ N and 10 N respectively are acting on a particle of mass 2 kg due east, due north, due south and due west. Find acceleration of the particle.

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Solution

Effective force along East $= 20 - 10 = 10 N$
Effective force along South $= 20 \sqrt{3} - 10 \sqrt{3} = 10 \sqrt{3} N$
Resultant force $= \sqrt{10^{2} + (10 \sqrt{3})^{2}}$
$= \sqrt{100 + 300}$
$= \sqrt{400}$
$= 20 N$

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Forces of magnitude 20 N, 10√3 N, 20√3 N and 10 N respectively are acting on a particle of mass 2 kg due east, due north, due south and due west. Find acceleration of the particle.

Effective force along East=20−10=10 NEffective force along South=20√3−10√3=10√3 NResultant force=√102+(10√3)2 =√100+300 =√400 =20 N

Forces of magnitude 20 N, $10 \sqrt{3}$ N, $20 \sqrt{3}$ N and 10 N respectively are acting on a particle of mass 2 kg due east, due north, due south and due west. Find acceleration of the particle.

Effective force along East $= 20 - 10 = 10 N$
Effective force along South $= 20 \sqrt{3} - 10 \sqrt{3} = 10 \sqrt{3} N$
Resultant force $= \sqrt{10^{2} + (10 \sqrt{3})^{2}}$
$= \sqrt{100 + 300}$
$= \sqrt{400}$
$= 20 N$