A 20-kg object is subjected to three forces which produce an acceleration a = - 8 m. s^-2 i + 6.0 m. s^-2 j on the object. Two of the forces are:F1 = 3.0 N i + 16.0 N jF2 = - 12.0 N i + 8.0 N jFind the third force.

F₃ = - 151 N i + 96 N j

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

Forces acting on the object

F₁ = 3.0 N i + 16.0 N j

F₂ = - 12.0 N i + 8.0 N j

F₃ = F₃x N i + F₃ y N j

x component of the net force on the object

Fx=F₁x+F₂x+F₃ x

Fx = 3.0 N-12.0 N + F₃x

Fx = F₃x - 9 N

y component of the net force on the object

Fy=F₁y+F₂y+F₃ y

Fy = 16.0 N + 8.0 N + F₃y

Fy = F₃y + 24 N

Newton's second law to the object:

a = - 8 m/s² i + 6.0 m/s² j

∑Fx = m*ax m=20 kg, ax = - 8 m/s²

F₃x - 9 = 20 * (-8)

F₃x = - 160+9

F₃x = - 151 N

∑Fy = m*ay m=20 kg, ay = 6 m/s²

F₃y + 24 = 20 * (6)

F₃y = 120 - 24

F₃y = 96 N

F₃ = - 151 N i + 96 N j