2The sum of magnitudes of two forces actinpoint is 16N. If the resulta

1.	2The sum of magnitudes of two forces actinpoint is 16N. If the resultant force is 8Ndirection is perpendicular to smaller force, tforces are(1) 6N and 10N(3) 4N and 12N(2) 8N and 8N(4) 2N and 14
Answer» 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 thatA + B = 16 therefore , B = 16 - A ..........................( 2 ). Now, by substituting ( 2 ) in ( 1 ).we get, ( 16 - A ) ²- A² = 64=> 256 - 32A + A² - A² = 6432A = 256 - 64 = 192A = 192 / 32 = 6 hence, we get the resultB = 16 - A = 16 - 6 = 10 therefore, the magnitude of the two vectors are 6 and 10.

2The sum of magnitudes of two forces actinpoint is 16N. If the resultant force is 8Ndirection is perpendicular to smaller force, tforces are(1) 6N and 10N(3) 4N and 12N(2) 8N and 8N(4) 2N and 14

Answer»

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 thatA + B = 16 therefore , B = 16 - A ..........................( 2 ).

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

( 16 - A ) ²- A² = 64=> 256 - 32A + A² - A² = 6432A = 256 - 64 = 192A = 192 / 32 = 6

hence, we get the resultB = 16 - A = 16 - 6 = 10

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

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