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 :
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.
[irp]
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.
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.
[irp]