Moment of force – problems and solutions

Moment de force – problèmes et solutions

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

Connu :

Net force (F_R) = 40 N Moment of force – problems and solutions 1

Force 1 (F₁) = 10 N

Force (F₃) = 20 N

Voulait: L'intensité de la 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 Newtons.

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 de force :

Στ = 0

10 + 50x – 35 = 0

50x - 25 = 0

50x = 25

x = 25h/50 et XNUMXj/XNUMX

x = 0.5 m

2. Forces of F₁, F₂, F₃, et 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.

Connu :

Forcer F₁= 10 N, the lever arm l₁ = 0 Moment of force – problems and solutions 2

Forcer F₂= 4 N, the lever arm l₂ = 2 mètres

Forcer F₃= 5 N, the lever arm l₃ = 3 mètres

Forcer F₄= 10 N, the lever arm l₄ = 6 mètres

Recherché : the moment of force about point A

Solution:

Moment of force 1 (τ₁) = F₁ l₁ = (10)(0) = 100 000

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.

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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 et 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.

Connu :

The axis rotation at point C. Moment of force – problems and solutions 3

Distance entre F_A et l'axe de rotation (r_AC) = 40 cm = 0,4 mètre

Distance entre F_B et l'axe de rotation (r_BC) = 20 cm = 0.2 mètre

Distance entre F_C et l'axe de rotation (r_CC) = 0 cm

F_A = 10 Newton

F_B = 20 Newton

F_C = 10 Newton

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

Solution:

Moment of force A :

St_A= (F_A)(r_AC péché 90^o) = (10 N)(0,4 m)(1) = -4 Nm

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

Moment of force B :

St_B = (F_B)(r_BC péché 90^o) = (20 N)(0,2 m)(1) = 4 Nm

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

Moment of force C :

St_C= (F_C)(r_CC péché 90^o) = (10 N)(0)(1) = 0

The resultant of the moment of force :

Στ = Στ₁ + Στ₂ + Στ₃

Στ = -4 + 4 + 0

Στ = 0 Nm

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.

Connu :

The axis rotation at point C. Moment of force – problems and solutions 4

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

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Distance entre F₂ et l'axe de rotation (r₂) = 10 cm = 0,1 mètre

Distance entre F₃ et l'axe de rotation (r₃) = 20 cm = 0,2 mètre

F₁ = 10 Newton

F₂ = 10 Newton

F₃ = 10 Newton

Recherché : Resultant of moment of force about point C.

Solution:

Moment of force 1 :

St₁ = (F₁)(r₁ péché 90^o) = (10 N)(0,3 m)(1) = -3 Nm

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

Moment of force 2 :

St₂= (F₂)(r₂ péché 90^o) = (10 N)(0,1 m)(1) = 1 Nm

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

Moment of force 3 :

St₃= (F₃)(r₃ péché 30^o) = (10 N)(0,2 m)(0,5) = -1 Nm

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 Nm

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

5. Three forces F₁, F₂, et 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 mètre)

Connu :

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 mètres

Force 2 (F₂) = 0.4 Newton

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

Force 3 (F₃) = 4.8 Newton

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

Voulait: 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.

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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 – t₂ – t₃ = 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?

Connu :

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 mètres

Force 2 (F₂) = 6 Newton

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

Force 3 (F₃) = 4 Newton

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

Voulait: 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 péché 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 – t₃ = 6 + 6 – 8 = 4 Nm

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