Question

The horizontal component of a force 10N inclined at 30 degree to the vertical is

Answer 1

30 degree angle to the vertical= 60 degree angle to the horizontal.

Horizontal component of force= 10x cos60= 10x1/2= 5 N

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Answer 2

The horizontal component of the given force of 10 N, is 5 N.

Given,

A force of 10 N is inclined at 30° to the vertical.

To find,

The horizontal component of the force.

Solution,

It can be seen that a force of 10 N is given to be inclined at an angle of 30° to the vertical.

To determine the horizontal component of the 10 N force, we need to know its inclination with the horizontal direction.

Refer figure.

We may observe that when the force (10 N) is at 30° to the vertical (V), its inclination with the horizontal (H) will be 60°.

Now, its horizontal component can be determined using the relation

$F_H=F \cos \theta \hfill ...(1)$

where,

$F_H$ = the horizontal component of force F,

$\theta$ = inclination or the angle made by force F, with the horizontal.

Here,

F = 10 N, and

$\theta$ = 60°

Thus, from equation (1),

$F_H=10\cos 60$

$\implies F_H=10 \times \frac{1}{2}$

$\implies F_H=5$ N.

⇒ the horizontal component = 5 N.

Therefore, the horizontal component of the given force of 10 N, is 5 N.

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