A particle is subjected to two mutually perpendicular horizontal forces of magnitudes 15\t… — JAMB UTME

A particle is subjected to two mutually perpendicular horizontal forces of magnitudes $15\text{ N}$ acting due East and $20\text{ N}$ acting due South. What is the magnitude of the resultant force acting on the particle?

$25\text{ N}$ Answer
B
$35\text{ N}$
C
$5\text{ N}$
D
$13.2\text{ N}$

Answer

The magnitude of the resultant force acting on the particle is

25\text{ N}

Because the two forces act at right angles (

90^\circ

) relative to each other, vector addition requires applying the Pythagorean theorem. Squaring both component magnitudes (

15^2 = 225

and

20^2 = 400

), summing them to obtain

625

, and taking the square root gives the resultant magnitude of

25\text{ N}

Step-by-Step Solution

Identify the force components and their geometric orientation

The East force component

F_x = 15\text{ N}

and the South force component

F_y = 20\text{ N}

meet at an angle of

90^\circ

Perpendicular vectors form the adjacent and opposite sides of a right-angled vector triangle.

Apply the Pythagorean theorem to find the hypotenuse representing the resultant force

R = \sqrt{F_x^2 + F_y^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25\text{ N}

The resultant of two orthogonal vector quantities equals the square root of the sum of their individual squares.

Key Concept

Vector Addition of Perpendicular Forces

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