Question

Two forces of magnitudes 10N and 5N are
applied on a body simultaneously at 53 N
of E and at 37 W of N respectively. Find the
resultant force
(A) (153)^1/2 (B) (129)^1/2
(C) (125)^1/2 (D) (150)^1/2

Answer 1

Concept:

Forces can be divide into two component, one is x component along x axis and one is y component along y axis.

Given:

Force of 10 Newton at 53° N of E

Force of 5 Newton at 37° W of N

Find:

Magnitude of the resultant force

Solution:

Taking east as x axis and north as y axis

For Force of 10 Newton at 53° N of E :-

If we divide it in components it will become

10 cos 53° Î + 10 sin 53° Ĵ

10 × ( 3 / 5 ) Î + 10 × ( 4 / 5 ) Ĵ

6 Î + 8 Ĵ  

For Force of 5 Newton at 37° W of N :-

means direction of forces making (90 + 37)° angle from east.

means direction of forces making (90 + 37)° angle from east.

If we divide it in components it will become

5 cos (90 + 37)° Î + 5 sin (90 + 37)° Ĵ

- 5 sin 37° Î + 5 cos 37° Ĵ

- 5 × ( 3 / 5 ) Î + 5 × ( 4 / 5 ) Ĵ

- 3  Î + 4 Ĵ    

Now adding both Forces

( 6 + (- 3)  ) Î + ( 8 + 4 ) Ĵ  

3 Î + 12 Ĵ    ( approximately )

Resultant Force will be

R = √(3² + 12²)

R = √(9 + 144)

R = √153

Hence, the magnitude of resultant force is √153 or (153)^1/2.

#SPJ3