Question

4. The simple sum of two forces acting at a point is 16 N and their vector sum is 8 N and its
direction is perpendicular to the smaller force, then the forces are
a. 6 N and 10 N
C. 8 N and 8 N
b. 4 N and 12 N
d. 2 N and 14 N
5. A particle moves in a circularpath with decreasing speed. Choose the correct statement​

Answer 1

Answer:

8 Ñ and 8 N

mark me please

I think this is the correct answer any way

Answer 2

Given : The simple sum of two forces acting at a point is 16 N and their vector sum is 8 N and its direction is perpendicular to the smaller force,

To Find :  the forces are

a. 6 N and 10 N

C. 8 N and 8 N

b. 4 N and 12 N

d. 2 N and 14 N

Solution:

Let say two forces are

(16 - A)  N  along  x axis

and   A  N    at angle  α

A > 16 - A

As Resultant direction is perpendicular to the smaller force,

Hence

A sin α  = 8  

16 -  A   + A Cosα = 0  

=>  A Cosα  = A - 16

A sin α  = 8

 A Cosα  = A - 16

Squaring and adding

A² = 64 + A²  + 256 - 32A

=>   32A  = 320

=>  A = 10

Hence 16- A = 6

Forces are 10 N   and 6  N

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