Rotation of rigid bodies – problems and solutions

The moment of force

1. Three forces act on a beam with a length of 6 meters, as shown in the figure below. What is the net torque rotates the beam about the point O as the axis of rotation?

Known :Rotation of rigid bodies – problems and solutions 1

The axis of rotation at point O.

Force 1 (F1) = F

The distance between the line of action of F1 with the axis of rotation (r1) = 3 meters

Force 2 (F2) = 2F

The distance between the line of action of F2 with the axis of rotation (r2) = 2 meters

Force 3 (F3) = 2F

The distance between the line of action of F3 with the axis of rotation (r3) = 3 meters

Wanted: The magnitude of the moment of force

Solution :

The moment of force 1 :

τ1 = F1 r1 = (F)(3) = -3F

The moment of force 1 rotates beam clockwise so we assign a negative sign.

The moment of force 2 :

τ2 = F2 r2 = (2F)(2) = 4F

The moment of force 2 rotates beam counterclockwise so we assign positive sign.

The moment of force 3 :

τ3 = F3 r3 sin 30o = (2F)(3)(0.5) = 3F

The moment of force 2 rotates beam counterclockwise so we assign positive sign.

The resultant of the moment of force :

Στ = τ1 + τ2 + τ3

Στ = -3F + 4F + 3F

Στ = 4F

The magnitude of the moment of force is 4F Newton-meter. The resultant of the moment of force rotates beam counterclockwise so we assign a positive sign.

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2. α = 30o, length of AB = BC = 1 meter. What is the moment of force about the axis of rotation at point A?

Known :

The axis of rotation at point A. Rotation of rigid bodies – problems and solutions 2

Force 1 (F1) = 10 N

The distance between the line of action of F1 with the axis of rotation (r1) = 1 meter

Force 2 (F2) = 10 N

The distance between the line of action of F2 with the axis of rotation (r2) = 1 meter

Force 3 (F3) = 20 N

The distance between the line of action of F3 with the axis of rotation (r3) = 2 meters

Wanted: The resultant of the moment of force

Solution :

The moment of force 1 :

τ1 = F1 r1 sin 30o = (10)(1)(0.5) = 5 N m

The moment of force 1 rotates beam counterclockwise so we assign a positive sign.

The moment of force 2 :

τ2 = F2 r2 sin 30o = (10)(1)(0.5) = -5 Newton meter

The moment of force 2 rotates beam clockwise so we assign negative sign.

The moment of force 3 :

τ3 = F3 r3 sin 60o = (20)(2)(0.5√3) = -20√3 Newton meter

The moment of force 3 rotates beam clockwise so we assign a negative sign.

The resultant of the moment of force :

Στ = τ1 + τ2 + τ3

Στ = 5 – 5 – 20√3

Στ = – 20√3 N m

The magnitude of the moment of force is 20√3 N m. The resultant of the moment of force rotates beam clockwise so we assign negative sign.

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