# Moment of force – problems and solutions

1. If F_{R} is the net force of F_{1}, F_{2}, and F_{3}, what is the magnitude of force F_{2} and x?

__Known :__

Net force (F_{R}) = 40 N

Force 1 (F_{1}) = 10 N

Force (F_{3}) = 20 N

__Wanted:__ The magnitude of force F_{2} and distance of x

__Solution :__

Find the magnitude of force F_{2 }:

*Force points to upward, signed negative and force points to downward, signed negative.*

ΣF = 0

– F_{R} + F_{1} + F_{2} – F_{3} = 0

– 40 + 10 + F_{2 }– 20 = 0

– 30 + F_{2 }

– 50 + F_{2 }= 0

F_{2 }= 50 Newton.

Plus sign indicates that the direction of the force is upward.

Find x.

Choose A as the axis of rotation.

τ_{1 }= F_{1} l_{1} = (10 N)(1 m) = 10 Nm

*The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 3.*

τ_{2 }= F_{2} x = (50)(x) = 50x Nm

*The torque 1 rotates beam counterclockwise so we assign positive sign to the torque 3.*

τ_{3 }= F_{3} x = (20 N)(1.75 m) = -35 Nm

*The torque 2 rotates beam clockwise so we assign negative sign to the torque 2.*

The net of moment of force :

Στ = 0

10 + 50x – 35 = 0

50x – 25 = 0

50x = 25

x = 25/50

x = 0.5 m

2. Forces of F_{1}, F_{2}, F_{3}, and F_{4} acts on the rod of ABCD as shown in figure. If rod’s mass ignored, what is the magnitude of the moment of force, about point A.

The axis of rotation = points A.

__Known :__

Force F_{1 }= 10 N, the lever arm l_{1} = 0

Force F_{2 }= 4 N, the lever arm l_{2} = 2 meters

Force F_{3 }= 5 N, the lever arm l_{3} = 3 meters

Force F_{4 }= 10 N, the lever arm l_{4} = 6 meters

__Wanted :__ the moment of force about point A

__Solution :__

Moment of force 1 (τ_{1}) = F_{1} l_{1} = (10)(0) = 0

Moment of force 2 (τ_{2}) = F_{2} l_{2} = (4)(2) = -8 Nm

Moment of force 3 (τ_{3}) = F_{3} l_{3} = (5)(3) = 15 Nm

Moment of force 4 (τ_{4}) = F_{4} l_{4} = (10)(6) = -60 Nm

*If torque rotates rod counterclockwise then we assign positive sign. *

*If torque rotates rod clockwise then we assign negative sign.*

The resultant of the moment of force :

τ = 0 – 8 Nm + 15 Nm – 60 Nm

τ = -68 Nm + 15 Nm

τ = -53 Nm

Minus sign indicates that the moment of force rotates rod clockwise.

3. Three forces act on a rod, F_{A} = F_{C} = 10 N and F_{B} = 20 N, as shown in figure below. If distance of AB = BC = 20 cm, what is the moment of force about point C.

__Known :__

The axis rotation at point C.

Distance between F_{A} and the axis of rotation (r_{AC}) = 40 cm = 0,4 meters

Distance between F_{B} and the axis of rotation (r_{BC}) = 20 cm = 0.2 meters

Distance between F_{C} and the axis of rotation (r_{CC}) = 0 cm

F_{A} = 10 Newton

F_{B} = 20 Newton

F_{C} = 10 Newton

__Wanted :__ The resultant of the moment of force about point C.

__Solution :__

__Moment of force A :__

Στ_{A }= (F_{A})(r_{AC} sin 90^{o}

*Minus sign indicates that the moment of force rotates rod clockwise.*

__Moment of force B :__

Στ_{B } = (F_{B})(r_{BC} sin 90^{o}) = (20 N)(0,2 m)(1) = 4 N.m

*Plus sign indicates that the moment of force rotates rod counterclockwise.*

__Moment of force C :__

Στ_{C }= (F_{C})(r_{CC} sin 90^{o}) = (10 N)(0)(1) = 0

__The resultant of the moment of force :__

Στ = Στ_{1} + Στ_{2} + Στ_{3}

Στ = -4 + 4 + 0

Στ = 0 N.m

4. Length of a rod is 50 cm. Three forces act on the rod, as shown in figure below. If the axis of rotation is point C, what is the net of the moment of force.

__Known :__

The axis rotation at point C.

Distance between F_{1} and the axis of rotation is (r_{1}

Distance between F_{2} and the axis of rotation (r_{2}

Distance between F_{3} and the axis of rotation (r_{3}) = 20 cm = 0,2 meters

F_{1} = 10 Newton

F_{2} = 10 Newton

F_{3} = 10 Newton

__Wanted :__ Resultant of moment of force about point C.

__Solution :__

__Moment of force 1 :__

Στ_{1} = (F_{1})(r_{1} sin 90^{o}) = (10 N)(0,3 m)(1) = -3 N.m

*Minus sign indicates that the moment of force rotates rod clockwise.*

__Moment of force 2 :__

Στ_{2 }= (F_{2})(r_{2} sin 90^{o}) = (10 N)(0,1 m)(1) = 1 N.m

*Plus sign indicates that the moment of force rotates rod counterclockwise.*

__Moment of force 3 :__

Στ_{3 }= (F_{3})(r_{3} sin 30^{o}) = (10 N)(0,2 m)(0,5) = -1 N.m

*Minus sign indicates that the moment of force rotates rod clockwise.*

The resultant of the moment of force :

Στ = Στ_{1 } + Στ_{2} + Στ_{3}

Στ = -3 + 1 – 1

Στ = -3 N.m

*Minus sign indicates that the resultant of the moment of force rotates rod clockwise.*

5. Three forces F_{1}, F_{2}, and F_{3} act on a rod as shown in figure below. Length of rod is 4 meters. What is the moment of force about point C.

(sin 53^{o} = 0.8, cos 53^{o} = 0.6, AB = BC = CD = DE = 1 meter)

__Known :__

*The axis of rotation at point C. *

Force 1 (F_{1}) = 5 Newton

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

Force 2 (F_{2}) = 0.4 Newton

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

Force 3 (F_{3}) = 4.8 Newton

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

__Wanted:__ The moment of force about point C.

__Solution :__

Moment of force 1 :

τ_{1 }= F_{1} r sin 53^{o }= (5 N)(2 m)(0,8) = (10)(0,8) N = 8 N

*Plus sign indicates that the moment of force rotates rod counterclockwise.*

Moment of force 2 :

τ_{2 }= F_{2} r sin 90^{o }= (0,4 N)(1 m)(1) = -0,4 N

*Minus sign indicates that the moment of force rotates rod clockwise.*

Moment of force 3 :

τ_{3 }= F_{3} r sin 90^{o }= (4,8 N)(2 m)(1) = -9,6 N

*Minus sign indicates that the moment of force rotates rod clockwise.*

The resultant of the moment of force :

Στ = τ_{1 } – τ_{2} – τ_{3} = 8 – 0,4 – 9,6 = 8 – 10 = 2 N.m

*Plus sign indicates that the moment of force rotates rod counterclockwise.*

6. What is the resultant of the moment of force about the axis of rotation at point O by forces acts on the rod, as shown in the figure below?

__Known :__

*The axis of rotation at point O. *

Force 1 (F_{1}) = 6 Newton

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

Force 2 (F_{2}) = 6 Newton

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

Force 3 (F_{3}) = 4 Newton

*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 about point C

__Solution :__

Moment of force 1 :

τ_{1 }= F_{1} l_{1} = (6 N)(1 m) = 6 Nm

*Plus sign indicates that the moment of force rotates rod counterclockwise.*

Moment of force 2 :

τ_{2 }= F_{2} r_{2} sin 30^{o }= (6 N)(2 m)(0,5)= 6 Nm

*Plus sign indicates that the moment of force rotates rod counterclockwise.*

Moment of force 3 :

τ_{3 }= F_{3} l_{3} = (4 N)(2 m) = -8 Nm

*Minus sign indicates that the moment of force rotates rod clockwise.*

The resultant of the moment of force :

Στ = τ_{1 } + τ_{2} – τ_{3} = 6 + 6 – 8 = 4 N.m

*Plus sign indicates that the moment of force rotates rod counterclockwise.*