Question

a force of 40N has one rectangular component as 20root3N. The another rectangular component will be???​

Answer 1

Answer:

Given:

Resultant force has magnitude = 40 N

One of the perpendicular components has magnitude = 20√3 N

To find:

Magnitude of the other perpendicular component

Concept:

Any vector can be divided into 2 perpendicular components . One of the component is along x axis and the other is along y axis.

Calculation:

Let the other component be y

As per Parallelogram Law of Addition of Vector :

$\sqrt{ \bigg \{({20 \sqrt{3}) }^{2} + {y}^{2} \bigg\} } = 40$

Squaring on both sides :

$\implies \: 1200 + {y}^{2} = 1600$

$\implies \: {y}^{2} = 1600 - 1200$

$\implies \: {y}^{2} = 400$

$\implies \: y = \sqrt{400}$

$\implies \: y = 20 \: N$

So final answer is :

$\boxed{ \huge{ \green{ \sf{ \bold{\: y = 20 \: N}}}}}$

Answer 2

$\underline{ \boxed{\purple{ \huge{ \mathfrak{Answer}}}}}$

Given :

One perpendicular component of Force 40N is $\rm\:20\sqrt{3}N$.

To Find :

Another perpendicular component of given Force.

Concept :

Any vector (2D) has two perpendicular component.

1. One component on x-axis
2. Second component on y-axis

So, we can say that given value of force is the resultant value of both perpendicular components.

Formula :

As per parellalogram method

$\: \: \: \: \: \: \: \: \: \dag \: \: \underline{ \boxed{ \bold{ \pink{R = \sqrt{ {x}^{2} + {y}^{2} } }}}} \: \: \dag$

Calculation :

$\implies \rm \: 40 = \sqrt{ {x}^{2} + {(20 \sqrt{3}) }^{2} } \\ \\ \implies \rm \: 1600 = {x}^{2} + 1200 \\ \\ \implies \rm \: {x}^{2} = 400 \\ \\ \therefore \: \underline{ \boxed{ \bold{ \orange{y = 20 \: N}}}} \: \red{ \star}$