A force is inclined at an angle of `60^(@)` from the horizontal. If the horizontal component of the force is 40N,calculate the vertical component.

Explanation

`A_(x)=40N, A_(y)=?, " "theta = 60^(@)`
As `A_(x)= A cos theta`
`therefore 40 =A cos 60^(@) or 40 = (A)/(2) or A=80N`

Now `A_(y) = A sin 60^(@)= (A sqrt(3))/(2)`
`=(80sqrt(3))/(2) =40 sqrt(3)N`