Answer:
1. The acceleration of the chair is 0.64 m/s²
2. The normal force acting on the chair is 523.6 N
Explanation:
Let's list out the parameters given us:
mass = 45 kg, F = 144 N, θ = 35°, F (f) = 89 N
Solving for the horizontal component of the Force, we use the formula
F (hor) = F * cos θ = 144 * cos 35°
F (hor) = 117.96 N
F (hor) = 118 N
The friction force is acting upon the chair in the opposite direction of the f (hor) & must be accounted for
F (net) = F (hor) - F (f)
F (net) = 118 - 89 = 29
F (net) = 29N
F (net) = mg
g = F (net) ÷ m = 29 ÷ 45
g = 0.64 m/s²
F (norm) = F (chair) + F (ver)
F (chair) = mg
where g = acceleration due to gravity = 9.8 m/s²
F (chair) = 45 * 9.8 = 441 N
F (ver) = F * sin θ = 144 * sin 35°
F (ver) = 82.6 N
F (norm) = 441 + 82.6 = 523.6
F (norm) = 523.6 N
