# Rotation of rigid bodies – problems and solutions

**The m****oment 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 (F_{1}) = F

The distance between the line of action of F_{1 }with the axis of rotation (r_{1}) = 3 meters

Force 2 (F_{2}) = 2F

The distance between the line of action of F_{2} with the axis of rotation (r_{2}) = 2 meters

Force 3 (F_{3}) = 2F

The distance between the line of action of F_{3} with the axis of rotation (r_{3}) = 3 meters

__Wanted:__ The magnitude of the moment of force

__Solution :__

The moment of force 1 :

τ_{1 }= F_{1} r_{1} = (F)(3) = -3F

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

The moment of force 2 :

τ_{2 }= F_{2} r_{2} = (2F)(2) = 4F

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

The moment of force 3 :

τ_{3 }= F_{3} r_{3 }sin 30^{o }= (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 = τ_{1 }+ τ_{2 }+ τ_{3}

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.

2. α = 30^{o}, 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 (F_{1}) = 10 N

*The distance between the line of action of F _{1 }with the axis of rotation (r_{1}) = 1 meter *

Force 2 (F_{2}) = 10 N

*The distance between the line of action of F _{2 }with the axis of rotation (r_{2}) = 1 meter *

Force 3 (F_{3}) = 20 N

*The distance between the line of action of F _{3 }with the axis of rotation (r_{3}) = 2 meters*

__Wanted:__ The resultant of the moment of force

__Solution :__

The moment of force 1 :

τ_{1 }= F_{1} r_{1} sin 30^{o }= (10)(1)(0.5) = 5 N m

*The **moment of force **1** rotates beam **counter**clockwise so we assig**n a positive **sign.*

The moment of force 2 :

τ_{2 }= F_{2} r_{2 }sin 30^{o }= (10)(1)(0.5) = -5 Newton meter

*The **moment of force 2 **rotates bea**m **clockwise so we assig**n negative **sign.*

The moment of force 3 :

τ_{3} = F_{3} r_{3 }sin 60^{o }= (20)(2)(0.5√3) = -20√3 Newton meter

*The **moment of force 3 **rotates bea**m **clockwise so we assig**n a negative **sign.*

The resultant of the moment of force :

Στ = τ_{1} + τ_{2 }+ τ_{3}

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.