Question

two forces of magnitude 16N and 4N acts at ab angle of 120° with each other, then magnitude of their resultant is
​

Answer 1

Given :

Two forces of magnitude of 16N and 4N acts at an angle of 120° with each other.

To Find :

The magnitude of their resultant.

Solution :

By triangle law or parallelogram law of vector addition, the magnitude of resultant R at two vectors A and B inclined to each other at angle θ is given by

$\bigstar\:\underline{\boxed{\bf{\orange{R=\sqrt{A^2+B^2+2ABcos\Theta}}}}}$

By substituting the given values,

$:\implies\tt \: R = \sqrt{(16)^2+(4)^2+2(16)(4)cos120^{\circ}}$

• cos 120° = -1/2

$:\implies\tt\:R=\sqrt{256+16+\dfrac{(-128)}{2}}$

$:\implies\tt\:R=\sqrt{272-64}$

$:\implies\tt\:R=\sqrt{208}$

$:\implies\tt\:R=\sqrt{16\times 13}$

$:\implies\:\underline{\boxed{\bf{\purple{R=4\sqrt{13}\:N}}}}$

★ Additional Information :

• The latin word vector means carrier.
• A vector whose initial point is fixed is called a localised vector and whose initial point is not fixed is called non-localised vector.
• Null vector is a vector which has zero magnitude and an arbitrary direction.
• A negative vector of a given vector is a vector of same magnitude but acting in a direction opposite to that of a given vector.

Answer 2

$\huge\boxed{\texttt{\fcolorbox{black}{pink}{Answer}}}$

$\red{Heya ! \: Mate \: ur \: ans \: is \: as \: follows:-}$

$\bigstar\:\underline{\boxed{\bf{\green{R=\sqrt{A^2+B^2+2ABcos\Theta}}}}}$

By substituting the given values,

$:\implies\tt \: R = \sqrt{(16)^2+(4)^2+2(16)(4)cos120^{\circ}}$

$:\implies\tt\:R=\sqrt{256+16+\dfrac{(-128)}{2}}$

$:\implies\tt\:R=\sqrt{272-64}$

$:\implies\tt\:R=\sqrt{208}$

$:\implies\tt\:R=\sqrt{16\times 13}$

$:\implies\:\underline{\boxed{\bf{\blue{R=4\sqrt{13}\:N}}}}$

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