Question

the sum of magnitudes of two forces acting at a point is 16N if the resultant force is 8N and its direction is perpendicular to smaller 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.