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Vector Component

The horizonta...

Question

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

A

3 N

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B

5 \sqrt{3} N

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C

5 N

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D

\frac{10}{\sqrt{3}} N

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Solution

The correct option is C $5 N$
Given, $F = 10 N, θ = 30^{o}$

Horizontal component ( $F_{H}$ )= $F sin θ$
$F_{H} = 10 sin 30 = 10 \times \frac{1}{2} = 5 N$

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