Question

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.

Answer 1

Answer:

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

b)    F = 230.2 N  ,  θ = 53.5º

Explanation:

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 = $F_{2y}$ / 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