Question

the vector diagram above represents the horizontal component, $f_h$, and the vertical component, $f_v$, of a 24 - newton force acting at 35° above the horizontal. what are the magnitudes of the horizontal and vertical components?
a) $f_h = 14 n$ and $f_v = 20 n$
b) $f_h = 20 n$ and $f_v = 14 n$
c) $f_h = 4.9 n$ and $f_v = 3.5 n$
d) $f_h = 3.5 n$ and $f_v = 4.9 n$

Explanation:

Step1: Find horizontal - component formula

The formula for the horizontal component of a force $F$ with an angle $\theta$ with the horizontal is $F_H = F\cos\theta$. Here, $F = 24\ N$ and $\theta=35^{\circ}$.
$F_H=24\times\cos(35^{\circ})$
Since $\cos(35^{\circ})\approx0.819$, then $F_H = 24\times0.819 = 19.656\ N\approx20\ N$.

Step2: Find vertical - component formula

The formula for the vertical component of a force $F$ with an angle $\theta$ with the horizontal is $F_V = F\sin\theta$. Here, $F = 24\ N$ and $\theta = 35^{\circ}$.
$F_V=24\times\sin(35^{\circ})$
Since $\sin(35^{\circ})\approx0.574$, then $F_V=24\times0.574 = 13.776\ N\approx14\ N$.

Answer:

B. $F_H = 20\ N$ and $F_V = 14\ N$