Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 63.9 N63.9 N , Jill pulls with 79.1 N79.1 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 183 N183 N . (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey.

In this exercise we ask to find the net force, for which we will define a coordinate system fix the donkey and use trigonometry to decompose the forces

Jack F₁ₓ = 63.9 N

Jill F₂ = 79.1 N with direction 45º to the left

cos (180 -45) = F₂ₓ / F₂

sin 135 = [tex]F_{2y}[/tex] / F₂

F₂ₓ = F₂ cos 135

F_{2y} = F₂ sin 135

F₂ₓ = 79.1 cos 135 = -55.9 N

F_{2y} = 79.1 sin 135 = 55.9 N

Jane F₃ = 183 N direction 45th to the right

cos 45 = F₃ₓ / F3

sin 45 = F_{3y} / F3

F₃ₓ = F₃ cos 45 = 183 cos 45

F_{₃y} = F₃ sin 45 = 183 sin 45

F₃ₓ = 129.4 N

F_{3y} = 129.4 N

we add each component of the force

Fₓ = F₁ₓ + F₂ₓ + F₃ₓ

Fₓ = 63.9 + (-55.9) + 129.4

Fₓ = 137.4 N

F_{y} = F_{2y} + F_{3y}

F_{2y} = 55.9 + 129.4

F_{2y} = 185.3 N

we can give the result of the forms

a) F = (137.4 i ^ + 185 j ^) N

b) in the form of module and angle

F = RA (Fₓ² + F_{y}²)

F = Ra (137² + 185²)

F = 230.2 N

tan θ = F_{y} / Fₓ

θ = tan⁻¹ F_{y} / Fₓ

θ = tan⁻¹ (185/137)

θ = 53.5º