Question

Find the magnitude of two forces such that if they act as right angle theire resultant is root 10

Answer 1

Answer:

Let the two forces be denoted as 'A' and 'B' vectors.

We know that, by parallelogram law of addition, we can find the resultant of two vectors, given the angle between these 2 vectors.

According to this law, the resultant vector is given by the formula:

$\text{Resultant} = \sqrt{ A^2 + B^2 + 2AB\:Cos\:\theta}$

Substituting the angle to be 90 degrees in the formula, we get:

⇒ Resultant = √ ( A² + B² + 2AB.Cos(90) )

⇒ Resultant = √ ( A² + B² + 2AB (0) )

⇒ Resultant = √ ( A² + B² )   ...(1)

It is already given that, Resultant Vector is equal to √10. Equating (1) with √10 we get:

⇒ √ ( A² + B² ) = √ 10

Squaring on both sides we get:

⇒ A² + B² = 10

Since we have two unknowns and only 1 relation between the unknowns, it can't be solved further.

Hence the final relation between the two force is: A² + B² = 10

Answer 2

Question:-

• Find the magnitude of two forces such that if they act as right angle theire resultant is √10.

To Find:-

• Find unit magnitude.

Solution:-

Given ,

• Resultant is √10

$\sf\longrightarrow \: Resultant = \sqrt{ {A}^{2} + {B}^{2} + 2AB \cos(θ) }$

$\sf\longrightarrow \: \sqrt { 10 } = \sqrt { { A }^{ 2 } + { B }^{ 2 } + 2AB \cos( 90 ) }$

$\sf\longrightarrow \: \sqrt { 10 } = \sqrt { { A }^{ 2 } + { B }^{ 2 } + 2AB( 0 ) }$

$\sf\longrightarrow \: \sqrt { 10 } = \sqrt { { A }^{ 2 } + { B }^{ 2 } }$

$\sf\longrightarrow \: { A }^{ 2 } + { B }^{ 2 } = 10$

Here ,

We can't solve this further .

Hence ,

• $\sf \: { A }^{ 2 } + { B }^{ 2 } = 10$