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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.

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hamzaahmeds

Answer:

Ο„ = 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 = 90Β°

Therefore,

τ₁ = (20 N)(0.5 m)Sin 90Β°

τ₁ = - 10 Nm

Negative sign due to clockwise direction.

Now, we will calculate the torque due to force Fβ‚‚.

Ο„β‚‚ = Fβ‚‚ dβ‚‚ Sin ΞΈβ‚‚

where,

Ο„β‚‚ = Torque due to Fβ‚‚ = ?

Fβ‚‚ = 35 N

dβ‚‚ = moment arm of Fβ‚‚ = 1.10 m

ΞΈβ‚‚ = Angle between Fβ‚‚ and dβ‚‚ = 120Β°

Therefore,

τ₁ = (35 N)(1.1 m)Sin 120Β°

τ₁ = 33.34 Nm

Positive sign due to anti-clockwise direction.

Now, we find net torque on door:

Ο„ = τ₁ + Ο„β‚‚

Ο„ = - 10 Nm + 33.34 Nm

Ο„ = 23.34 Nm

Direction = Anti-Clockwise (Positive)

Ver imagen hamzaahmeds