Question

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.

Answer 1

Answer:

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