# Moment of force – problems and solutions

1. If FR is the net force of F1, F2, and F3, what is the magnitude of force F2 and x?

Known :

Net force (FR) = 40 N

Force 1 (F1) = 10 N

Force (F3) = 20 N

Wanted: The magnitude of force F2 and distance of x

Solution :

Find the magnitude of force F2 :

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

ΣF = 0

– FR + F1 + F2 – F3 = 0

– 40 + 10 + F2 – 20 = 0

– 30 + F2 – 20 = 0

– 50 + F2 = 0

F2 = 50 Newton.

Find x.

Choose A as the axis of rotation.

τ1 = F1 l1 = (10 N)(1 m) = 10 Nm

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

τ2 = F2 x = (50)(x) = 50x Nm

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

τ3 = F3 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

Read :  Transverse waves - problems and solutions

2. Forces of F1, F2, F3, and F4 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 F1 = 10 N, the lever arm l1 = 0

Force F2 = 4 N, the lever arm l2 = 2 meters

Force F3 = 5 N, the lever arm l3 = 3 meters

Force F4 = 10 N, the lever arm l4 = 6 meters

Wanted : the moment of force about point A

Solution :

Moment of force 1 (τ1) = F1 l1 = (10)(0) = 0

Moment of force 2 (τ2) = F2 l2 = (4)(2) = -8 Nm

Moment of force 3 (τ3) = F3 l3 = (5)(3) = 15 Nm

Moment of force 4 (τ4) = F4 l4 = (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

Read :  Optical instrument human eye – problems and solutions

3. Three forces act on a rod, FA = FC = 10 N and FB = 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 FA and the axis of rotation (rAC) = 40 cm = 0,4 meters

Distance between FB and the axis of rotation (rBC) = 20 cm = 0.2 meters

Distance between FC and the axis of rotation (rCC) = 0 cm

FA = 10 Newton

FB = 20 Newton

FC = 10 Newton

Wanted : The resultant of the moment of force about point C.

Solution :

Moment of force A :

ΣτA = (FA)(rAC sin 90o) = (10 N)(0,4 m)(1) = -4 N.m

Moment of force B :

ΣτB = (FB)(rBC sin 90o) = (20 N)(0,2 m)(1) = 4 N.m

Moment of force C :

ΣτC = (FC)(rCC sin 90o) = (10 N)(0)(1) = 0

The resultant of the moment of force :

Στ = Στ1 + Στ2 + Στ3

Στ = -4 + 4 + 0

Στ = 0 N.m

Read :  Boyle's law (constant temperature) - problems and solutions

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 F1 and the axis of rotation is (r1) = 30 cm = 0,3 meters

Distance between F2 and the axis of rotation (r2) = 10 cm = 0,1 meters

Distance between F3 and the axis of rotation (r3) = 20 cm = 0,2 meters

F1 = 10 Newton

F2 = 10 Newton

F3 = 10 Newton

Wanted : Resultant of moment of force about point C.

Solution :

Moment of force 1 :

Στ1 = (F1)(r1 sin 90o) = (10 N)(0,3 m)(1) = -3 N.m

Moment of force 2 :

Στ2 = (F2)(r2 sin 90o) = (10 N)(0,1 m)(1) = 1 N.m

Moment of force 3 :

Στ3 = (F3)(r3 sin 30o) = (10 N)(0,2 m)(0,5) = -1 N.m

The resultant of the moment of force :

Στ = Στ1 + Στ2 + Στ3

Στ = -3 + 1 – 1

Στ = -3 N.m

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

Read :  Motion on the horizontal surface without the friction force – application of Newton's law of motion problems and solutions

5. Three forces F1, F2, and F3 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 53o = 0.8, cos 53o = 0.6, AB = BC = CD = DE = 1 meter)

Known :

The axis of rotation at point C.

Force 1 (F1) = 5 Newton

The distance between the line of action of F1 with the axis of rotation (r1) = 2 meters

Force 2 (F2) = 0.4 Newton

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

Force 3 (F3) = 4.8 Newton

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

Wanted: The moment of force about point C.

Solution :

Moment of force 1 :

τ1 = F1 r sin 53o = (5 N)(2 m)(0,8) = (10)(0,8) N = 8 N

Moment of force 2 :

τ2 = F2 r sin 90o = (0,4 N)(1 m)(1) = -0,4 N

Moment of force 3 :

τ3 = F3 r sin 90o = (4,8 N)(2 m)(1) = -9,6 N

The resultant of the moment of force :

Στ = τ1 – τ2 – τ3 = 8 – 0,4 – 9,6 = 8 – 10 = 2 N.m

Read :  Mechanical energy – problems and solutions

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 (F1) = 6 Newton

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

Force 2 (F2) = 6 Newton

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

Force 3 (F3) = 4 Newton

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 about point C

Solution :

Moment of force 1 :

τ1 = F1 l1 = (6 N)(1 m) = 6 Nm

Moment of force 2 :

τ2 = F2 r2 sin 30o = (6 N)(2 m)(0,5)= 6 Nm

Moment of force 3 :

τ3 = F3 l3 = (4 N)(2 m) = -8 Nm