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) 2 N, 4N b) 3N, 6 N c) 6N, 10 N d) 7 N, 14 N​

Answer 1

Answer:

Let the smaller of the two forces is of magnitude be

F

N.

Hence the larger one will be

(

16

−

F

)

N. As the resultant

8

N is perpendicular to

F

we can write.

Answer 2

Answer:

c) 6N, 10N

Explanation:

Let the smaller of the two forces is of magnitude be  F  N.

Hence the larger one will be  ( 16 − F )  N.

As the resultant  8  N is perpendicular to  F.

So, we form a right angle triangle.

By pythagoras theorem,

$(16-F)^2-F^2=8^2\\\\(16-F)^2-F^2=64\\\\16^2+F^2-32F-F^2=64\\\\16(16-2F)=64\\\\256-32F=64\\\\F=\frac{256-64}{32} \\\\F=6N$

The smaller force is 6 N.

The larger force will be 16 - F

= 16 - 6

= 10 N

Hope this helps....