The x -component of a force of 50N is 30N, then what will be the y-component of the same applied force?

The correct answer is 40N

CONCEPT:​

• Resolution of vectors into components: We have a vector (F) where the magnitude of the vector is F and the angle with horizontal is θ.

The vector has two components: 1. Vertical component and 2. Horizontal component

Vertical component (F_y) = F Sinθ

Horizontal component (F_x) = F Cosθ 

Here \(F = \sqrt {F_x^2 + F_y^2}\)

CALCULATION:

Here F₁ and F₂ are along X- and Y- direction.

Let the applied force F = 50

And the x-component of the applied force F_x = 30

The y-component of the applied force F_y = ?

We know that the vector sum of the force

\(F = \sqrt {F_x^2 + F_y^2}\)

\(50N = \sqrt {{{30}^2} + {F^2}}\)

Now squaring both sides

2500 = 900 + F²

\({F_y} = \sqrt {2500 - 900} = \sqrt {1600}\)

\({F_y} = 40N\)

So option 3 is correct.