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.

The magnitude of net torque – problems and solutions 1Known :

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.

The magnitude of net torque – problems and solutions 2Known :

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.

The magnitude of net torque – problems and solutions 3Known :

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 magnitude of net torque – problems and solutions 4The 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

The magnitude of net torque – problems and solutions 5Known :

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 poiThe magnitude of net torque – problems and solutions 6nt 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.