Question

the sum of two forces acting at a point is 16 Newton if the resultant force is 8 Newton and its direction is perpendicular to minimum force then the force ​

Answer 1

Answer:

$\large{\underline{\boxed{\sf 6 \: N \: \: and \: \: 10 \: N}}}$

Explanation:

Let the magnitude of smaller force be x Newton. So, the magnitude of larger force be (16 - x) Newton, as their sum is 16 N.

And it is given that, the resultant force is 8N which is perpendicular to smaller force, i.e., x Newton,

so we can write it as;

(16 - x)² = x² + 8²

⇒ (16)² + (x)² - 2 * 16 * x = x² + 64

⇒ 256 - 32x = 64

⇒ 32x = 256 - 64

⇒ 32x = 192

⇒ x = $\sf \dfrac{192}{32}$

⇒ x = 6

_______________________

Therefore,

The magnitude of small force = x = 6 N

And, the magnitude if larger force is given by, (16 - x) = (16 - 6) = 10N.

Hence, the smaller force is 6N and the larger force is 10N.

Answer 2

Explanation:

Given:

Sum of two forces acting at a point = 16N

Resultant force ($\hat{R}$) = 8 N

Let the minimum force be of magnitude = k N

=> Refer attachment

Magnitude of one of the force = 16 N

then, Smaller force's magnitude = (16 - k )

Now proceeds as below:

Apply vector sum addition rule:

$\bold{\mathsf{\hat{R} = \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos(\theta)}}}$

Using $\bold{\mathsf{\cos(\theta)= \cos(90) = 0 }}$

Now formula reduces as

$\bold{\mathsf{\hat{R} = \sqrt{ {a}^{2} + {b}^{2} } }}$

Put in the values

$\bold{\mathsf{{16 - k}^{2} = {k}^{2} + {8}^{2} }}$

$\bold{\mathsf{{16 - k}^{2} = {k}^{2} + 64 }}$

$\bold{\mathsf{256 + k^{2} - 2(16){(k)}^{2} = {k}^{2} + 64}}$

=> 256 - 64 = 32k

=> 192 = 32k

=> k = 192/32

=> k = 6

Now Value of smaller will be 6 N

Now Value of Larger force will be = 16 - 6 = 10 N

$\rule{200}{2}$

★$\boxed{\underline{\underline{\bold{\mathfrak{Formula\:used:-}}}}}$:—

$\bold{\mathsf{\hat{R} = \sqrt{ {a}^{2} + {b}^{2} + 2ab \cos(\theta) } }}$

here , a & b are respective magnitude of verctors .