Question

The sum of magnitude of two forces acting at a point is 16N.if the resultant force is 8N and its direction is perpendicular to small force,then the forces are?

Answer 1

let the force of smaller magnitude be A and the force of larger magnitude be B.

here,it is clearly given that the resultant force is R perpendicular to small force. hence, we can say that b is the hyopetenuse .

therefore, b² = R² + A²
=> R² = b² - A²
= 8²
= 64 .......................... ( 1 ).

here, it is also given that
A + B = 16
therefore , B = 16 - A ..........................( 2 ).

Now, by substituting ( 2 ) in ( 1 ).we get,

( 16 - A ) ²- A² = 64
=> 256 - 32A + A² - A² = 64
32A = 256 - 64 = 192
A = 192 / 32
= 6

hence, we get the result
B = 16 - A = 16 - 6 = 10

therefore, the magnitude of the two vectors are 6 and 10.

Answer 2

Answer:

8N,8N

Explanation:

There is given that the resultant vector is perpendicular with minimum vector.

We know that if we want to find the angle between resultant and a vector then we use

tan€=Qsin€\P+Qcos€

As already given that

€ is 90

So tan90`=Qsin€\P+Qcos€

So P+Qcos€ is equals to 0

P+Qcos€= 0

P= -Qcos€

Now use formula

R^2=P^2 + Q^2 + 2pqcos€

8^2 = P^2 + Q^2 -2Q^2

64= P^2 + Q^2

P+Q)(P-Q) = 64

WE KNOW P+Q = 16

P-Q= 4

SO 2P= 20

P= 10

ANDQ= 6