Question

A particle is acted upon by 3 forces in one plane equal to 2,2(2)^1/2,1 N respectively.First is horizontal,second acts at 45 degree to the horizontal and 3rd is vertical.What is the magnitude of resultant force?

Answer 1

A particle is acted upon by 3 forces in one plane equal to 2, $2(2)^\wedge 1/2$ , 1 N respectively. First is horizontal, second acts at 45 degree to the horizontal and 3rd is vertical. What is the magnitude of resultant force?

Solution:

To find resultant force we should add all forces, which are acting on the body.

Firstly, we should add $F_{2}$ and $F_{3}$ . To do that we need to project $F_{3}$ on the axis $X$ :

$$F _ {x} = F _ {3} \cdot \cos 4 5 {}^ {\circ} = 2 \sqrt {2} \cdot \frac {1}{\sqrt {2}} = 2 N$$

So the sum is:

$$F _ {2, 3} = F _ {2} + F _ {3} = 2 + 2 = 4 N$$

Secondly, we should add $F_{1}$ and $F_{3}$ . Now we project $F_{3}$ on the axis $Y$ :

$$F _ {y} = F _ {3} \cdot \sin 4 5 {}^ {\circ} = 2 \sqrt {2} \cdot \frac {1}{\sqrt {2}} = 2 N$$

So the sun is:

$$F _ {1, 3} = F _ {1} + F _ {3} = 1 + 2 = 3 N$$

To find the magnitude of resultant force we should add $F_{2,3}$ and $F_{1,3}$ . To do that just use Pythagoras' theorem:

$$F _ {r e s u l t a n t} = \sqrt {F _ {1 , 3} ^ {2} + F _ {2 , 3} ^ {2}} = \sqrt {4 ^ {2} + 3 ^ {2}} = \sqrt {1 6 + 9} = \sqrt {2 5} = 5 N$$

Answer: $F_{res} = 5$