Find :
(i) Resultant force on block
(ii) Direction and magnitude of force is to be applied on block to make it in equilibrium.

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Solution

(i) $F = \sqrt{F_{x}^{2} + F_{y}^{2}}$
$= \sqrt{4^{2} + 3^{2}}$
$= \sqrt{16 + 9}$
$= \sqrt{25}$
$= 5 N$

(ii) Direction of the force = ${tan}^{- 1} \frac{F_{y}}{F_{x}} = {tan}^{- 1} \frac{3}{4} = {36.8}^{\circ}$ with the x-axis.
A force of 5 N making an angle of ${36.8}^{\circ}$ directed towards the origin will make the block in equilibrium.

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Find :(i) Resultant force on block (ii) Direction and magnitude of force is to be applied on block to make it in equilibrium.

(i)F=√F2x+F2y =√42+32 =√16+9 =√25 =5N (ii) Direction of the force = tan−1FyFx=tan−134=36.8∘ with the x-axis.A force of 5 N making an angle of 36.8∘ directed towards the origin will make the block in equilibrium.

Find :
(i) Resultant force on block
(ii) Direction and magnitude of force is to be applied on block to make it in equilibrium.