The magnitude of net torque – problems and solutions

1. A force P is applied to one end of a beam with a length of 2 m. What is the magnitude of the torque? The axis of rotation at point A.

Known :

Force (F) = 10 N

Length of AB (r_AB) = 2 m

Force F is perpendicular to the beam.

The lever arm (l) = r_AB sin 90^o = (2 m)(1) = 2 m

Wanted: The torque about the axis of rotation

Solution :

The torque :

τ = F l = (10 N)(2 m) = 20 N m

The plus sign because the beam rotates counterclockwise rotation.

2. The length of a beam AB is 2 m and the magnitude of force F is 10 N. What is the magnitude of the torque? The axis of rotation at point A.

Known :

Force (F) = 10 N

Length of AB (r_AB) = 2 m

The lever arm (l) = r_AB sin 60^o = (2 m)(0.5√3) = √3 m

Wanted: The torque about the axis of rotation

Solution :

The torque :

τ = F l = (10 N)(√3 m) = 10√3 N m

The plus sign because the force F causes the beam rotates counterclockwise rotation.

3. The length of a beam is 2 m. The magnitude of F₁ is 10 N and the magnitude of F₂ is 15 N. Determine the net torque about the center of the beam.

The axis of rotation at the center of the beam.

Known :

Force 1 (F₁) = 10 N

The distance between F₁ and the center of beam (r₁) = 1 m

The lever arm 1 (l₁) = r₁ sin 90^o = (1 m)(1) = 1 m

Force F₁ is perpendicular to the beam.

Force 2 (F₂) = 15 N

The distance between F₂ and the center of the beam (r₂) = 1 m

Force F₂ is perpendicular to the beam.

The lever arm 2 (l₂) = r₁ sin 90^o = (1 m)(1) = 1 m

Wanted : The net torque about the axis of rotation

Solution :

The torque 1 :

τ₁ = F₁ l₁ = (10 N)(1 m) = 10 N m

The plus sign because the force of F₁ causes the beam rotates counterclockwise rotation.

The torque 2 :

τ₂ = F₂ l₂ = (15 N)(1 m) = -15 N

The minus sign because the force F₂ causes the beam to rotates clockwise.

The net torque :

Στ = τ₁ – τ₂ = 10 – 15 = – 5 N m

The minus sign because the beam to rotates clockwise.

4. The length beam AB is 2 m, The magnitude of F₁ is 10 N and the magnitude of F₂ is 10 N. Determine the net torque about the center of the beam.

The axis of rotation at the center of the beam.

Known :

Force 1 (F₁) = 10 Nn

The distance between F₁ and the center of the beam (r₁) = 1 m

The lever arm 1 (l₁) = r₁ sin 60^o = (1 m)(0.5√3) = 0.5√3 m

Force 2 (F₂) = 10 N

The distance between F₂ and the center of the beam (r₂) = 1 m

Force F₂ is perpendicular to the beam.

The lever arm 2 (l₂) = r₂ sin 90^o = (1 m)(1) = 1 m

Wanted : The net torque about the axis of rotation

Solution :

The torque 1 :

τ₁ = F₁ l₁ = (10 N)(0.5√3 m) = 5√3 = 8.7 N.m

The plus sign because the force F₁ causes the beam to rotates clockwise.

The torque 2 :

τ₂ = F₂ l₂ = (10 N)(1 m) = -10 N m

The minus sign because the force F₂ causes the beam to rotates clockwise.

The net torque :

Στ = τ₁ – τ₂ = 8.7 – 10 = – 1.3 N m

The minus sign because the net force causes the beam to rotates clockwise.

5. The length of a beam is 10 m, the magnitude of F₁ is 10 N, the magnitude of F₂ is 10 N and the magnitude of F₃ is 15 N. The distance between point A and point C is 7.5 m. The Force F₂ located at the center of the beam. Determine the net torque about the point C located at 2.5 m from the point B.

The axis of rotation located at point C

Known :

Force 1 (F₁) = 10 N

The distance between F₁ and point C (r₁) = 2.5 m

Force F₁ is perpendicular to the beam.

The lever arm 1 (l₁) = r₁ sin 90^o = (2.5 m)(1) = 2.5 m

Force 2 (F₂) = 10 N

The distance between F₂ and point C (r₂) = 2.5 m

The force F₂ is perpendicular to the beam.

The lever arm 2 (l₂) = r₂ sin 90^o = (2.5 m)(1) = 2.5 m

Force 3 (F₃) = 15 N

The distance between F₃ and point C (r₃) = 7.5 m

The force F₃ is perpendicular to the beam.

The lever arm 3 (l₃) = r₃ sin 90^o = (7.5 m)(1) = 7.5 m

Wanted : The net torque about the axis of rotation

Solution :

The torque 1 :

τ₁ = F₁ l₁ = (10 N)(2.5 m) = 25 N m

The plus sign because the force F₁ causes the beam to rotates clockwise.

The torque 2 :

τ₂ = F₂ l₂ = (10 N)(2.5 m) = 25 N m

The plus sign because the force F₂ causes the beam to rotates clockwise.

The torque 3 :

τ₃ = F₃ l₃ = (15 N)(7.5 m) = -112..5 N m

The minus sign because the force F₃ causes the beam to rotates clockwise.

The net torque :

Στ = τ₁ + τ₂ – τ₃ = 25 + 25 – 112.5 = – 62.5 N m

The minus sign because the net force causes the beam to rotates clockwise.

6. The length of a beam is 10 m, the magnitude of F₁ is 10 N, the magnitude of F₂ is 10 N and the magnitude of F₃ is 10 N. Determine the net torque about point A, located 5 m from the point of application of force F₁.

The axis of rotation at point A.

Known :

Force 1 (F₁) = 10 N

The distance between F₁ and point A (r₁) = 5 m

The lever arm 1 (l₁) = r₁ sin 60^o = (5 m)(0.5√3) = 2.5√3 m

Force 2 (F₂) = 10 N

The distance between F₂ and point A (r₂) = 0

The force F₂ is perpendicular to the beam.

The lever arm 2 (l₂) = r₂ sin 90^o = (0)(1) = 0

The force 3 (F₃) = 10 N

The distance between F₃ and point A (r₃) = 10 m

The lever arm 3 (l₃) = r₃ sin 30^o = (10 m)(0.5) = 5 m

Wanted : the net torque about the axis of rotation

Solution :

The torque 1 :

τ₁ = F₁ l₁ = (10 N)(2.5√3 m) = 25√3 = 43.3 N m

The plus sign because the force F₁ causes the beam to rotates clockwise.

The torque 2 :

τ₂ = F₂ l₂ = (10 N)(0) = 0

The torque 3 :

τ₃ = F₃ l₃ = (10 N)(5 m) = -50 N m

The minus sign because the force F₃ causes the beam to rotates clockwise.

The net torque :

Στ = τ₁ + τ₂ – τ₃ = 43.3 + 0 – 50 = – 6.7 N m

The minus sign because the net force causes the beam to rotates clockwise.