Moment of force – problems and solutions

Moment of force – problems and solutions

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

Known :

Net force (F_R) = 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

– F_R + F₁ + F₂ – F₃ = 0

– 40 + 10 + F₂ – 20 = 0

– 30 + F₂ – 20 = 0

– 50 + F₂ = 0

F₂ = 50 Newton.

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

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

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

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) = (10 N)(0,4 m)(1) = -4 N.m

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 = Στ₁ + Στ₂ + Στ₃

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 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 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 :

Στ₂ = (F₂)(r₂ 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 :

Στ₃ = (F₃)(r₃ 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 = 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.

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 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 (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 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 :

τ₂ = F₂ 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 :

τ₃ = F₃ 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 = τ₁ – τ₂ – τ₃ = 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 (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

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

Moment of force 2 :

τ₂ = F₂ r₂ 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 :

τ₃ = F₃ l₃ = (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 = τ₁ + τ₂ – τ₃ = 6 + 6 – 8 = 4 N.m

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