Question

how much is the horizontal component of force? (your answer should just be a number, and round to the nearest hundredth position) 60° 23 n how much is the vertical component of force? (your answer should just be a number, and round to the nearest hundredth position)

Explanation:

Step1: Recall force - component formula

The horizontal component of a force $F$ with an angle $\theta$ with the horizontal is $F_x = F\cos\theta$, and the vertical component is $F_y=F\sin\theta$. Here $F = 23\ N$ and $\theta = 60^{\circ}$.

Step2: Calculate the horizontal component

$F_x=23\times\cos(60^{\circ})$. Since $\cos(60^{\circ})=\frac{1}{2}$, then $F_x = 23\times\frac{1}{2}=11.50\ N$.

Step3: Calculate the vertical component

$F_y = 23\times\sin(60^{\circ})$. Since $\sin(60^{\circ})=\frac{\sqrt{3}}{2}\approx0.866$, then $F_y=23\times0.866 = 19.92\ N$.

Answer:

Horizontal component: 11.50
Vertical component: 19.92