Question

A pack of four Arctic wolves are exerting four different forces upon the carcass of a 500 kg dead polar bear. Wolf one is pulling on the bear with a force of 20 Newtons (N) due north. Wolf two is pulling on the bear with a force of 30 N, 30 degrees north of west. Wolf three is pulling on the bear with a force of 25 N due west, and wolf four is pulling on the bear with a force of 35 N, 10 degrees west of south. Determine the net force acting upon the polar bear and the direction of that force measured north of west.

Answer 1

Answer:

Net force is 86.135 N , N 82.255° W

Explanation:

Resolving all forces vertically and horizontally.

First vertically.

20N = 20sin90 = 20

30N = 30sin(90-30)= 30sin60

=25.981

25N = 25sin0 = 0

35N =- 35sin80= -34.368

Total of vertical =

20+25.981+0-34.368

= 11.613N

For horizontal

20N =- 20cos 90 = 0

30N =- 30cos 30 = -25.981

25N = -25cos0 = -25

35N = -35cos 10 = -34.468

Total horizontal = 0-25.981-25-34.368= -85.349

Resultant force = √(11.613)²+(-85.349)²

= √(134.862) +(7284.452)

= √ 7419.314

= 86.135 N

Tan(tita°) = 11.613/-85.349

Tan(tita°) =- 0.136

tita° = Tan^-1(-0.136)

tita° = -7.745

North of West = 90-7.745

= 82.255°

Answer 2

Answer:

F = 58.1N

θ = 10.4° South of East.

Explanation:

Let the situation be as represented in the diagram attached. To determine the net force acting on the bear, we need to resolve the forces into component forms.

Taking west-east as the +x axis and south-north as +y axis.

F1x = 0N (force is directed due north and so has not x component)

F2x = – F2cos30° = –30cos30° = –26N

F3x = – 25N (force is directed due west and so has only x component)

F4x = –F4sin10° = -35sin10° = –6.08N

Fx = F1x + F2x + F3x + F4x =0 – 26 –25 – 6.08 = –57.08N

F1y = 20N (force is directed due north and so has not x component)

F2y = F2sin30° = 30sin30° = 15N

F3y = 0N (force is directed due west and so has only x component)

Fy = –F4cos10° = -35cos10° = –34.5N

Fy = F1y + F2y + F3y + F4y = 20 + 15 + 0 –34.5 = –10.5N

F = √(Fx² + Fy²)

F = √((-57.1² + (-10.5)²) = 58.1N

θ = tan-¹(Fy/Fx) = tan-¹(-10.5/-57.1) = 10.4° South of East.

F = 58.1N

θ = 10.4° South of East.