Question

A force vector points due east and has a magnitude of 140 newtons. A second force is added to . The resultant of the two vectors has a magnitude of 420 newtons and points along the (a) east/ (b) west line. Find the magnitude and direction of . Note that there are two answers. (a) Number newtons (b) Number newtons

Answer 1

Answer:

(a) When the resultant force is pointing along east line, the magnitude and direction of the second force is 280 N East

(b)  When the resultant force is pointing along west line, the magnitude and direction of the second force is 560 N West

Explanation:

Given;

a force vector points due east, $F_1$ = 140 N

let the second force = $F_2$

let the resultant of the two vectors = F

(a) When the resultant force is pointing along east line

the second force must be pointing due east

$F = F_1 + F_2\\\\F_2 = F - F_1\\\\F_2 = 420 \ N - 140 \ N\\\\F_2 = 280 \ N$

$F_2 = 280 \ East$

(b) When the resultant force is pointing along west line

the second force must be pointing due west and it must have a greater magnitude compared to the first force in order to have a resultant in west line.

$F = F_2 - F_1\\\\F_2 = F + F_1\\\\F_2 = 420 \ N + 140 \ N\\\\F_2 = 560 \ N$

$F_2 = 560 \ West$