# 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 (rAB) = 2 m

Force F is perpendicular to the beam.

The lever arm (l) = rAB sin 90o = (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.

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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 (rAB) = 2 m

The lever arm (l) = rAB sin 60o = (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.

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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 90o = (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 90o = (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.

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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 60o = (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 90o = (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.

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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 90o = (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 90o = (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 90o = (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.

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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 poi nt 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 60o = (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 90o = (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 30o = (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.