The drawing shows a top view of a door that is free to rotate about an axis of rotation that is perpendicular to the screen. Find the net torque (magnitude and direction) produced by the forces F1 and F2 about the axis.

τ = 23.34 Nm

Direction = Anti-Clockwise (Positive)

Explanation:

The image of the question is attached in the answer.

First, we will calculate the torque due to force F₁.

τ₁ = F₁ d₁ Sin θ₁

where,

τ₁ = Torque due to F₁ = ?

F₁ = 20 N

d₁ = moment arm of F₁ = 0.5 m

θ₁ = Angle between F₁ and d₁ = 90°

Therefore,

τ₁ = (20 N)(0.5 m)Sin 90°

τ₁ = - 10 Nm

Negative sign due to clockwise direction.

Now, we will calculate the torque due to force F₂.

τ₂ = F₂ d₂ Sin θ₂

where,

τ₂ = Torque due to F₂ = ?

F₂ = 35 N

d₂ = moment arm of F₂ = 1.10 m

θ₂ = Angle between F₂ and d₂ = 120°

Therefore,

τ₁ = (35 N)(1.1 m)Sin 120°

τ₁ = 33.34 Nm

Positive sign due to anti-clockwise direction.

Now, we find net torque on door:

τ = τ₁ + τ₂

τ = - 10 Nm + 33.34 Nm

τ = 23.34 Nm

Direction = Anti-Clockwise (Positive)

if they have a finite resultant