Question

Four horizontal forces of magnitudes 1 N, 2 N, 3N and 4N act at a point in the direction whose bearings are 000, 060, 120 and 270 respectively. a Calculate the magnitude of their resultant. b. A 5th horizontal force of magnitude 3 N now acts at the same points so that the resultant of all five forces has a bearing of 090. Find the bearing of the 5th force

Answer 1

resultant = 0.356N 202.1°

Explanation:

Resultant force = √((x component)² + (y component)²)

X component= 1 cos 90 + 2 cos 30 + 3 cos 30 -4 cos 0

X component = 0 + 1.732 + 2.598 - 4

X component = 0.33

Y component = 1 sin 90 + 2 sin 60 -3sin 60 + 3 sin 0

Y component= 1+1.732-2.598

Y component= 0.134

Resultant = √( (0.33)² +(0.134)²)

Resultant= √(0.1089+0.017956)

Resultant= √ 0.126856

Resultant= 0.3562 N

Tan tita = 0.134/0.33

Tan tita = 0.406

Tita = 22.1°

Tab is positive In the third quadrant and first quadrant but the magnitude of the force lies mainly on the third so resultant = 0.356N 202.1°

For the fifth force.

X component =- 0.356 cos 67.9 +x

X component= -0.134 +x

Y component = 0.356sin22.1 +0

Y component= 0.1334

Tan tita = 0.1334/(-0.134+x)

Tita = tan^-1 0.1334/(-0.134+x)

90 = 0.1334/(-0.134+x)

Tan 90 is undefined so no more solution