Question

The sum of magnitudes of two forces acting at a point is 16 N. If their resultant is normal to the smaller force and has a magnitude of 8 N, the forces are (A) 6N, 10N (B) 8N, 8N (C) 4N, 12N (D) 2N, 14N

Answer 1

Answer:

Option A is correct.

Explanation:

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