Question

The horizontal component of a force of $10N$ inclined at ${30}^o$ to vertical is

• A. $3N$
• B. $5\sqrt{3}N$
• C. $5N$
• D. $\frac{10}{\sqrt{3}}N$

Answer 1

The correct option is C $5N$

Given, $F=10N,\theta ={30}^o$


Horizontal component ($F_H$)=$F\sin \theta$
$F_H=10\sin 30=10\times \frac{1}{2}=5N$