Equilibrium of Forces, Center of Gravity and Moments Practice Questions — JAMB UTME

Question 1Question →

A uniform metre rule of mass $120\text{ g}$ is balanced horizontally on a pivot placed at the $40\text{ cm}$ mark when an unknown mass $m$ is suspended at the $10\text{ cm}$ mark. What is the value of the mass $m$ in grams?

Show answer & explanation

Answer: 40

Answer

The mass

m

required to balance the metre rule horizontally is

40\text{ g}

.

According to the Principle of Moments, a system is in rotational equilibrium when the sum of anticlockwise moments equals the sum of clockwise moments about the pivot. The weight of the

120\text{ g}

uniform metre rule acts at its center of gravity (

50\text{ cm}

mark), which is

10\text{ cm}

to the right of the pivot at

40\text{ cm}

. This creates a clockwise moment of

120\text{ g} \times 10\text{ cm} = 1200\text{ g}\cdot\text{cm}

. The mass

m

is suspended at the

10\text{ cm}

mark, which is

30\text{ cm}

to the left of the pivot, creating an anticlockwise moment of

m \times 30\text{ cm}

. Setting

30m = 1200

yields

m = 40\text{ g}

.

Step-by-Step Solution

1

Determine the position of the center of gravity of the metre rule.

Center of gravity is at the

50\text{ cm}

mark.

A uniform metre rule has its weight concentrated at its geometric midpoint.

2

Calculate perpendicular distances from the pivot at

40\text{ cm}

to all acting forces.

Distance to rule weight =

50\text{ cm} - 40\text{ cm} = 10\text{ cm}

; Distance to mass

m

=

40\text{ cm} - 10\text{ cm} = 30\text{ cm}

.

Moments are calculated by multiplying force (or mass) by perpendicular distance from the turning point.

3

Apply the Principle of Moments for equilibrium.

Sum of anticlockwise moments = Sum of clockwise moments

\implies m \times 30 = 120 \times 10

.

For rotational equilibrium, the total clockwise moment must equal the total anticlockwise moment about the pivot.

4

Solve for the unknown mass

m

.

m = 40\text{ g}

.

Dividing

1200\text{ g}\cdot\text{cm}

by

30\text{ cm}

gives

40\text{ g}

.

Key Concept

Principle of Moments and Center of Gravity of a Uniform Body

Estimated Time:50s

Question 2Question →

A rigid rod $AB$ of length $2.0\text{ m}$ is hinged at end $A$ . A force of $50\text{ N}$ is applied at end $B$ at an angle of $30^\circ$ to the rod. What is the moment of this force about the hinge $A$ ?

25\text{ N m}

50\text{ N m}

86.6\text{ N m}

100\text{ N m}

Show answer & explanation

Answer: $50\text{ N m}$

Answer

The moment of the force about hinge

A

is

50\text{ N m}

.

The moment of a force about a pivot is calculated by multiplying the force magnitude by the perpendicular distance to the pivot line of action (

F \times L \sin\theta

). Substituting

F = 50\text{ N}

,

L = 2.0\text{ m}

, and

\sin(30^\circ) = 0.5

gives

50 \times 2.0 \times 0.5 = 50\text{ N m}

.

Step-by-Step Solution

1

Identify the formula for the moment of a force applied at an angle.

\text{Moment} = F \times d_\perp = F \times L \sin\theta

Moment is defined as the product of the magnitude of the force and the perpendicular distance from the line of action of the force to the pivot point.

2

Substitute the given values into the formula.

\text{Moment} = 50\text{ N} \times 2.0\text{ m} \times \sin(30^\circ)

The force

F = 50\text{ N}

, distance

L = 2.0\text{ m}

, and angle

\theta = 30^\circ

.

3

Calculate the final moment value.

\text{Moment} = 50 \times 2.0 \times 0.5 = 50\text{ N m}

Since

\sin(30^\circ) = 0.5

, the perpendicular distance is

1.0\text{ m}

, giving a moment of

50\text{ N m}

.

Key Concept

Moment of a Force at an Angle

Estimated Time:45s

Question 3Question →

A light rigid bar $PQ$ of length $5.0\text{ m}$ is hinged at end $P$ and held horizontally. An upward force of $40\text{ N}$ is applied at end $Q$ at an angle of $30^\circ$ to the bar. To keep the bar in horizontal equilibrium, a vertical downward force $F$ is applied at a distance of $2.0\text{ m}$ from $P$ . What is the magnitude of the force $F$ ?

20\text{ N}

50\text{ N}

86.6\text{ N}

100\text{ N}

Show answer & explanation

Answer: $50\text{ N}$

Answer

The magnitude of the force

F

required to maintain equilibrium is

50\text{ N}

.

The force of

40\text{ N}

applied at an angle of

30^\circ

to the bar has a perpendicular component of

40\sin 30^\circ = 20\text{ N}

. The counterclockwise moment about pivot

P

is

20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}

. For the bar to remain in equilibrium, the clockwise moment created by

F

must equal

100\text{ N}\cdot\text{m}

, giving

F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m}

, which yields

F = 50\text{ N}

.

Step-by-Step Solution

1

Calculate the perpendicular component of the force applied at end

Q

.

F_{\perp} = 40\text{ N} \times \sin(30^\circ) = 40 \times 0.5 = 20\text{ N}

Only the force component perpendicular to the bar produces a moment about the pivot

P

.

2

Calculate the counterclockwise moment produced by the force at

Q

about pivot

P

.

\text{Moment}_{Q} = 20\text{ N} \times 5.0\text{ m} = 100\text{ N}\cdot\text{m}

Moment is defined as perpendicular force multiplied by distance from the pivot.

3

Apply the Principle of Moments about pivot

P

to find force

F

.

F \times 2.0\text{ m} = 100\text{ N}\cdot\text{m} \implies F = \frac{100}{2.0} = 50\text{ N}

For rotational equilibrium, total clockwise moment about

P

must equal total counterclockwise moment about

P

.

Key Concept

Principle of Moments and Rotational Equilibrium with Forces at an Angle

Question 4Question →

A uniform horizontal wooden beam $AB$ of length $4.0\text{ m}$ and mass $8.0\text{ kg}$ is supported on a pivot placed $1.0\text{ m}$ from end $A$ . A mass of $5.0\text{ kg}$ is suspended from end $B$ . What is the magnitude of the downward vertical force $F$ in newtons that must be applied at end $A$ to keep the beam in horizontal equilibrium? (Take $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 230

Answer

The magnitude of the downward force required at end A to maintain horizontal equilibrium is 230 N.

Taking moments about the pivot, the downward force F at end A produces an anticlockwise moment of F × 1.0 m. This balances the clockwise moments produced by the weight of the beam (80 N × 1.0 m) and the load at end B (50 N × 3.0 m). Equating anticlockwise and clockwise moments gives F × 1.0 = 80 + 150 = 230 N.

Step-by-Step Solution

1

Determine the weights and distance of each force from the pivot point

Weight of beam = 80 N acting at 1.0 m right of pivot; weight at B = 50 N acting at 3.0 m right of pivot; force F acts at 1.0 m left of pivot.

The center of gravity of a uniform 4.0 m beam is at its midpoint (2.0 m from end A).

2

Set up the moment equilibrium equation about the pivot

F × 1.0 = (80 × 1.0) + (50 × 3.0)

For static rotational equilibrium, total anticlockwise moments equal total clockwise moments.

3

Solve for the force magnitude F

F = 230 N

Summing clockwise moments yields 80 + 150 = 230 N m, which divided by 1.0 m gives F = 230 N.

Key Concept

Principle of Moments and Static Equilibrium

Estimated Time:1m 30s

Question 5Question →

A uniform horizontal beam $XY$ of length $3.0\text{ m}$ and weight $80\text{ N}$ is hinged at end $X$ to a vertical wall. It is held in horizontal equilibrium by a light cable attached to end $Y$ that makes an angle of $30^\circ$ with the beam. What is the tension in the cable?

40\text{ N}

80\text{ N}

160\text{ N}

240\text{ N}

Show answer & explanation

Answer: $80\text{ N}$

Answer

The tension in the cable is

80\text{ N}

.

Taking moments about the hinge at end

X

, the clockwise moment due to the weight of the beam acting at its midpoint (

1.5\text{ m}

) is

80\text{ N} \times 1.5\text{ m} = 120\text{ N}\cdot\text{m}

. The counterclockwise moment provided by the cable tension

T

at end

Y

(

3.0\text{ m}

) is

T \sin(30^\circ) \times 3.0\text{ m} = 1.5 T\text{ N}\cdot\text{m}

. Setting clockwise moments equal to counterclockwise moments gives

1.5 T = 120

, yielding

T = 80\text{ N}

.

Step-by-Step Solution

1

Identify the center of gravity and calculate the clockwise moment about the hinge at end

X

.

Since the beam is uniform, its weight of

80\text{ N}

acts at its center of gravity, which is at a distance of

1.5\text{ m}

from

X

. Clockwise moment =

80\text{ N} \times 1.5\text{ m} = 120\text{ N}\cdot\text{m}

.

For a uniform body, weight acts precisely at the midpoint.

2

Determine the counterclockwise moment exerted by the cable tension

T

about point

X

.

The perpendicular component of the tension is

T \sin(30^\circ)

. Counterclockwise moment =

T \sin(30^\circ) \times 3.0\text{ m} = 1.5 T\text{ N}\cdot\text{m}

.

Only the component of force perpendicular to the beam produces a moment about the pivot.

3

Apply the principle of moments for rotational equilibrium to solve for

T

.

1.5 T = 120 \implies T = \frac{120}{1.5} = 80\text{ N}

.

For equilibrium, total counterclockwise moments about any pivot must equal total clockwise moments.

Key Concept

Principle of Moments and Rotational Equilibrium

Estimated Time:1m 30s

Question 6Question →

A uniform horizontal plank $AB$ of length $5.0\text{ m}$ and mass $40\text{ kg}$ rests on two smooth supports, $C$ and $D$ , located $1.0\text{ m}$ from end $A$ and $1.0\text{ m}$ from end $B$ respectively. What is the maximum mass, in kilograms, of an object that can be placed at end $A$ without causing the plank to tilt?

Show answer & explanation

Answer: 60

Answer

The maximum mass that can be placed at end A without tilting the plank is 60 kg.

Just before tilting, the plank rotates around support C, causing the normal reaction at support D to drop to zero. Equating the anticlockwise moment of the added mass at end A about support C (

(m \cdot g) \times 1.0\text{ m}

) to the clockwise moment of the plank's weight about support C (

(40 \cdot g) \times 1.5\text{ m}

) yields

m = 60\text{ kg}

.

Step-by-Step Solution

1

Identify the tipping condition and pivot point

Support C acts as the pivot; the reaction at support D becomes zero (

R_D = 0

).

When extra weight is added at end A, the plank rotates about C and lifts off support D.

2

Calculate perpendicular distances from the pivot C

Distance to added mass

m

=

1.0\text{ m}

; Distance to plank's center of mass =

2.5\text{ m} - 1.0\text{ m} = 1.5\text{ m}

.

The weight of a uniform beam acts at its midpoint (2.5 m from either end).

3

Equate clockwise and counterclockwise moments about C

m \times g \times 1.0\text{ m} = 40\text{ kg} \times g \times 1.5\text{ m}

, giving

m = 60\text{ kg}

.

For the plank to remain balanced just before tipping, total anticlockwise moment must equal total clockwise moment.

Key Concept

Rotational Equilibrium and Tilting of Rigid Bodies

Estimated Time:1m 30s

Question 7Question →

A uniform rigid bar $AB$ of length $2.0\text{ m}$ and weight $120\text{ N}$ is hinged smoothly at end $A$ to a vertical post. The bar is held in equilibrium at an angle of $60^\circ$ above the horizontal by a force $F$ applied at end $B$ acting perpendicular to the bar. What is the magnitude of the force $F$ ?

30\text{ N}

60\text{ N}

120\text{ N}

15\text{ N}

Show answer & explanation

Answer: $30\text{ N}$

Answer

The magnitude of the force required to keep the bar in equilibrium is

30\text{ N}

.

To maintain rotational equilibrium, the clockwise moment created by the weight of the bar about hinge

A

must equal the counterclockwise moment created by force

F

. The weight of

120\text{ N}

acts at the bar's midpoint (

1.0\text{ m}

from

A

), and its perpendicular distance to the vertical line of action is

1.0 \cos(60^\circ) = 0.5\text{ m}

. Thus, the clockwise moment is

120 \times 0.5 = 60\text{ N}\cdot\text{m}

. Since force

F

acts perpendicularly at the end of the

2.0\text{ m}

bar, its moment is

F \times 2.0

. Setting

2.0 F = 60

gives

F = 30\text{ N}

.

Step-by-Step Solution

1

Identify the center of gravity and the position of applied forces.

For a uniform bar of length

L = 2.0\text{ m}

, its weight

W = 120\text{ N}

acts vertically downward at its center of gravity, which is at a distance of

1.0\text{ m}

from hinge

A

.

The weight of a uniform body acts through its midpoint.

2

Determine the perpendicular distance for each force relative to the pivot at

A

.

Perpendicular distance for weight:

d_W = 1.0\text{ m} \times \cos(60^\circ) = 0.5\text{ m}

. Perpendicular distance for force

F

:

d_F = 2.0\text{ m}

(since

F

is perpendicular to the bar).

The moment of a force is defined as the product of the force magnitude and the perpendicular distance from the pivot to the line of action of the force.

3

Apply the Principle of Moments about the pivot

A

.

\sum M_A = 0 \implies F \times 2.0\text{ m} = 120\text{ N} \times 0.5\text{ m} \implies 2.0 F = 60 \implies F = 30\text{ N}

.

For rotational equilibrium, the total counterclockwise moment about any pivot must equal the total clockwise moment.

Key Concept

Principle of Moments and Rotational Equilibrium

Question 8Question →

A non-uniform wooden pole of length $6.0\text{ m}$ and weight $150\text{ N}$ is balanced horizontally on a pivot placed $2.4\text{ m}$ from its heavy end $P$ . The system achieves rotational equilibrium when a load of $50\text{ N}$ is hung directly from end $P$ . What is the distance of the center of gravity of the pole from end $P$ ?

Show answer & explanation

Answer: 3.2

Answer

The distance of the center of gravity of the pole from end

P

is

3.2\text{ m}

.

Taking moments about the pivot at

2.4\text{ m}

from end

P

, the counter-clockwise moment created by the

50\text{ N}

load (

50\text{ N} \times 2.4\text{ m} = 120\text{ N}\cdot\text{m}

) must balance the clockwise moment created by the

150\text{ N}

weight of the pole acting at its center of gravity (

150\text{ N} \times (d - 2.4\text{ m})

). Equating these gives

120 = 150(d - 2.4)

, leading to

d - 2.4 = 0.8\text{ m}

, so

d = 3.2\text{ m}

.

Step-by-Step Solution

1

Identify force positions relative to the pivot

The

50\text{ N}

load is

2.4\text{ m}

to the left of the pivot. The

150\text{ N}

weight acts at the center of gravity, which is

(d - 2.4\text{ m})

to the right of the pivot.

Moments are evaluated relative to the fulcrum to eliminate the unknown normal reaction force at the pivot.

2

Apply the Principle of Moments

Anti-clockwise moment =

50 \times 2.4 = 120\text{ N}\cdot\text{m}

. Clockwise moment =

150 \times (d - 2.4)

. Setting them equal:

120 = 150(d - 2.4)

.

For a body in rotational equilibrium, the total clockwise moment about any pivot equals the total anti-clockwise moment.

3

Solve the equation for distance

d

d - 2.4 = 0.8 \implies d = 3.2\text{ m}

.

Adding the displacement from the pivot (

0.8\text{ m}

) to the pivot position from end

P

(

2.4\text{ m}

) yields the position of the center of gravity from end

P

.

Key Concept

Rotational equilibrium and Principle of Moments for non-uniform rigid bodies

Estimated Time:1m 30s

Question 9Question →

A uniform horizontal rod $AB$ of length $2.0\text{ m}$ and mass $6.0\text{ kg}$ is suspended horizontally by two light vertical strings attached at end $A$ and at a point $C$ located $0.5\text{ m}$ from end $B$ . Taking $g = 10\text{ m/s}^2$ , what is the tension in the string at point $C$ in Newtons?

Show answer & explanation

Answer: 40

Answer

The tension in the string at point C is 40 N.

The weight of the rod (

60\text{ N}

) acts at its midpoint (

1.0\text{ m}

from end

A

). Point

C

is located

1.5\text{ m}

from end

A

. Taking moments about end

A

gives

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

, which evaluates to

T_C = 40\text{ N}

.

Step-by-Step Solution

1

Calculate total weight and identify the center of gravity position

Weight

W = 60\text{ N}

acting at

1.0\text{ m}

from end

A

For a uniform rod, the weight acts vertically downwards at its geometric center.

2

Set up the moment equilibrium equation taking end A as pivot

60\text{ N} \times 1.0\text{ m} = T_C \times 1.5\text{ m}

Taking moments about point

A

eliminates the force at

A

and equates clockwise moment from weight to counter-clockwise moment from tension at

C

.

3

Solve for the tension force at point C

T_C = 40\text{ N}

Dividing the total moment of

60\text{ N}\cdot\text{m}

by the moment arm of

1.5\text{ m}

yields

40\text{ N}

.

Key Concept

Principle of Moments and Rotational Equilibrium

Estimated Time:1m 30s

Question 10Question →

A uniform beam $PQ$ of length $4.0\text{ m}$ and mass $20\text{ kg}$ is supported horizontally on a pivot at end $P$ and by a vertical wire attached at end $Q$ . A load of mass $30\text{ kg}$ is placed on the beam at a distance of $1.0\text{ m}$ from $P$ . Taking the acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the tension in the vertical wire attached at $Q$ in newtons?

Show answer & explanation

Answer: 175

Answer

The tension in the vertical wire attached at end

Q

is

175\text{ N}

.

By applying the principle of moments about the pivot at

P

, the sum of downward clockwise moments produced by the

30\text{ kg}

load (

300\text{ N} \times 1.0\text{ m} = 300\text{ N}\cdot\text{m}

) and the beam's center of gravity (

200\text{ N} \times 2.0\text{ m} = 400\text{ N}\cdot\text{m}

) equals

700\text{ N}\cdot\text{m}

. Equating this to the counterclockwise moment of the tension force (

T \times 4.0\text{ m}

) gives

T = 175\text{ N}

.

Step-by-Step Solution

1

Determine forces and their perpendicular distances from the pivot at

P

.

The load exerts a downward force of

300\text{ N}

at

1.0\text{ m}

from

P

. The uniform beam's weight of

200\text{ N}

acts at its midpoint (

2.0\text{ m}

from

P

). The vertical tension

T

acts upward at

Q

(

4.0\text{ m}

from

P

).

Before applying the principle of moments, all force magnitudes and their distance arms relative to the pivot point must be identified.

2

Equate total clockwise moments to total counterclockwise moments about

P

.

(300\text{ N} \times 1.0\text{ m}) + (200\text{ N} \times 2.0\text{ m}) = T \times 4.0\text{ m}

For rotational equilibrium, the sum of clockwise moments about any pivot must equal the sum of counterclockwise moments about that same pivot.

3

Calculate the value of the tension force

T

.

T = \frac{700}{4.0} = 175\text{ N}

Simplifying the moment equation gives the magnitude of the upward supporting force.

Key Concept

Principle of Moments and Rotational Equilibrium

Question 11Question →

A light rigid lever of length $2.5\text{ m}$ is pivoted horizontally at one end $O$ . A downward vertical weight of $60\text{ N}$ is hung from the lever at a distance of $1.5\text{ m}$ from $O$ . An upward force $F$ inclined at an angle of $30^\circ$ to the lever is applied at the free end. What is the magnitude of the force $F$ required to maintain horizontal equilibrium?

36\text{ N}

72\text{ N}

104\text{ N}

180\text{ N}

Show answer & explanation

Answer: $72\text{ N}$

Answer

The magnitude of the force

F

required to maintain equilibrium is

72\text{ N}

.

According to the Principle of Moments, for a body in rotational equilibrium, the total clockwise moment about a pivot equals the total counterclockwise moment. The load creates a clockwise moment of

60\text{ N} \times 1.5\text{ m} = 90\text{ N}\cdot\text{m}

. The force

F

applied at

30^\circ

has a perpendicular component of

F \sin 30^\circ = 0.5F

. The counterclockwise moment is

0.5F \times 2.5\text{ m} = 1.25F

. Setting

1.25F = 90\text{ N}\cdot\text{m}

gives

F = 72\text{ N}

.

Step-by-Step Solution

1

Calculate the clockwise moment about the pivot

O

caused by the hanging load.

\text{Clockwise Moment} = 60\text{ N} \times 1.5\text{ m} = 90\text{ N}\cdot\text{m}

The force of

60\text{ N}

acts perpendicularly at a distance of

1.5\text{ m}

from the pivot.

2

Express the counterclockwise moment about the pivot

O

exerted by the force

F

.

\text{Counterclockwise Moment} = F \sin(30^\circ) \times 2.5\text{ m} = 1.25 F\text{ N}\cdot\text{m}

Only the perpendicular component of the force,

F \sin(30^\circ)

, contributes to the moment about pivot

O

.

3

Equate the clockwise and counterclockwise moments according to the Principle of Moments.

1.25 F = 90 \implies F = \frac{90}{1.25} = 72\text{ N}

For rotational equilibrium, the sum of clockwise moments must equal the sum of counterclockwise moments.

Key Concept

Principle of Moments and Perpendicular Force Components

Question 12Question →

A uniform horizontal beam $AB$ of length $4.0\text{ m}$ and mass $10\text{ kg}$ is hinged smoothly to a vertical wall at end $A$ . It is held horizontally in static equilibrium by a light cable attached to end $B$ and anchored to the wall above $A$ , making an angle of $30^\circ$ with the beam. A mass of $5\text{ kg}$ is suspended from the beam at a distance of $3.0\text{ m}$ from hinge $A$ . Taking acceleration due to gravity $g = 10\text{ m/s}^2$ , what is the tension in the cable?

175\text{ N}

87.5\text{ N}

275\text{ N}

17.5\text{ N}

Show answer & explanation

Answer: $175\text{ N}$

Answer

The tension in the cable is

175\text{ N}

.

Applying the principle of moments about the hinge at end A, the clockwise moments due to the beam's center of mass (

100\text{ N}

at

2.0\text{ m}

) and the suspended load (

50\text{ N}

at

3.0\text{ m}

) are balanced by the counterclockwise moment of the cable tension (

T \sin 30^\circ

at

4.0\text{ m}

). Solving

(100 \times 2.0) + (50 \times 3.0) = 2.0 T

yields

T = 175\text{ N}

.

Step-by-Step Solution

1

Calculate the downward gravitational forces (weights) acting on the system.

Weight of beam

W_{\text{beam}} = m_{\text{beam}} g = 10\text{ kg} \times 10\text{ m/s}^2 = 100\text{ N}

acting at

2.0\text{ m}

from

A

. Weight of load

W_{\text{load}} = m_{\text{load}} g = 5\text{ kg} \times 10\text{ m/s}^2 = 50\text{ N}

acting at

3.0\text{ m}

from

A

.

Forces causing clockwise moments must be expressed in force units (newtons) and located at their respective lines of action.

2

Formulate the equilibrium condition using the Principle of Moments about hinge

A

.

\sum \tau_A = 0 \implies (W_{\text{beam}} \times 2.0\text{ m}) + (W_{\text{load}} \times 3.0\text{ m}) = T \sin(30^\circ) \times 4.0\text{ m}

The hinge

A

eliminates reaction forces at the hinge from the moment equation.

3

Substitute numerical values and solve for tension

T

.

(100 \times 2.0) + (50 \times 3.0) = T \times 0.5 \times 4.0 \implies 200 + 150 = 2.0 T \implies 350 = 2.0 T \implies T = 175\text{ N}$.

Perpendicular distance from

A

to line of action of tension is

4.0 \sin 30^\circ = 2.0\text{ m}

.

Key Concept

Equilibrium of rigid bodies and Principle of Moments under non-perpendicular forces

Estimated Time:2m 0s

Question 13Question →

A uniform horizontal wooden rod $XY$ of length $3.0\text{ m}$ and weight $50\text{ N}$ rests horizontally on two smooth supports located at $X$ (the left end) and at a point $Z$ which is $0.6\text{ m}$ from end $Y$ . A block of weight $120\text{ N}$ is placed on the rod at a distance of $0.9\text{ m}$ from end $X$ . What is the magnitude of the upward reaction force, in newtons, at support $Z$ ?

Show answer & explanation

Answer: 76.25

Answer

The magnitude of the upward reaction force at support

Z

is

76.25\text{ N}

.

Taking moments about support

X

, the total clockwise moment is the sum of the moment due to the load (

120\text{ N} \times 0.9\text{ m} = 108\text{ N}\cdot\text{m}

) and the weight of the rod (

50\text{ N} \times 1.5\text{ m} = 75\text{ N}\cdot\text{m}

), giving

183\text{ N}\cdot\text{m}

. Equating this to the counterclockwise moment of the reaction force at

Z

(

R_Z \times 2.4\text{ m}

) yields

R_Z = \frac{183}{2.4} = 76.25\text{ N}

.

Step-by-Step Solution

1

Identify the perpendicular distance of each force and support from pivot point

X

.

Center of gravity position

x_{cg} = 1.5\text{ m}

, load position

x_{L} = 0.9\text{ m}

, and support

Z

position

x_{Z} = 3.0 - 0.6 = 2.4\text{ m}

.

Taking moments about

X

requires knowing the exact moment arm for each force from

X

.

2

Set up the equation for rotational equilibrium about point

X

.

Total clockwise moment =

(120 \times 0.9) + (50 \times 1.5) = 183\text{ N}\cdot\text{m}

; Total counterclockwise moment =

R_Z \times 2.4

.

Choosing pivot

X

eliminates the unknown reaction force

R_X

because its distance from

X

is zero.

3

Equate clockwise moments to counterclockwise moments and solve for

R_Z

.

R_Z = \frac{183}{2.4} = 76.25\text{ N}

.

For a body in rotational equilibrium, the algebraic sum of moments about any point must equal zero.

Key Concept

Principle of Moments and Rotational Equilibrium

Question 14Question →

A light, rigid horizontal bar $AB$ of length $1.5\text{ m}$ is smoothly pivoted at end $A$ . A vertical downward load of $40\text{ N}$ is hung from end $B$ . The bar is kept in horizontal equilibrium by a light string attached at point $C$ , located $1.0\text{ m}$ from $A$ . The string exerts a tension force $T$ pulling upwards at an angle of $30^\circ$ relative to the horizontal bar. What is the magnitude of the tension $T$ in newtons?

Show answer & explanation

Answer: 120

Answer

The magnitude of the tension

T

in the string is

120\text{ N}

.

For the bar to maintain rotational equilibrium, the clockwise moment about pivot

A

must equal the counterclockwise moment about

A

. The

40\text{ N}

load exerts a clockwise moment of

40\text{ N} \times 1.5\text{ m} = 60\text{ N}\cdot\text{m}

. The string tension

T

exerts a counterclockwise moment given by its vertical component multiplied by the distance from the pivot:

(T \sin 30^\circ) \times 1.0\text{ m} = 0.5T \text{ N}\cdot\text{m}

. Equating the two moments gives

0.5T = 60\text{ N}\cdot\text{m}

, yielding

T = 120\text{ N}

.

Step-by-Step Solution

1

Calculate the clockwise moment about the pivot at end

A

τ_{\text{clockwise}} = 40\text{ N} \times 1.5\text{ m} = 60\text{ N}\cdot\text{m}

The weight at

B

acts vertically downward at a perpendicular distance of

1.5\text{ m}

from pivot

A

.

2

Determine the perpendicular component of tension

T

relative to the bar

F_{\perp} = T \sin 30^\circ = 0.5T

Only the component of force perpendicular to the bar produces a moment about the pivot.

3

Set up the counterclockwise moment about pivot

A

τ_{\text{counterclockwise}} = (0.5T) \times 1.0\text{ m} = 0.5T \text{ N}\cdot\text{m}

The string is attached at point

C

, which is

1.0\text{ m}

away from pivot

A

.

4

Apply the principle of moments for rotational equilibrium and solve for

T

0.5T = 60 \implies T = 120\text{ N}

For rotational equilibrium, total clockwise moments must equal total counterclockwise moments about any pivot.

Key Concept

Principle of moments and rotational equilibrium for forces acting at non-perpendicular angles.

Estimated Time:1m 30s

Question 15Question →

A uniform rigid beam $MN$ of length $2.0\text{ m}$ and weight $80\text{ N}$ is smoothly pivoted at end $M$ . A vertical load of $120\text{ N}$ is hung at a distance of $1.5\text{ m}$ from $M$ . The beam is held horizontally in equilibrium by a cable attached at end $N$ pulling upward at an angle of $30^\circ$ to the horizontal beam. What is the magnitude of the tension in the cable?

100\text{ N}

130\text{ N}

260\text{ N}

340\text{ N}

Show answer & explanation

Answer: $260\text{ N}$

Answer

The magnitude of the tension in the cable is

260\text{ N}

.

For rotational equilibrium about pivot

M

, the sum of clockwise moments must equal the counterclockwise moment. Clockwise moment from the beam's weight and load is

(80\text{ N} \times 1.0\text{ m}) + (120\text{ N} \times 1.5\text{ m}) = 260\text{ N}\cdot\text{m}

. Counterclockwise moment from the cable tension is

T \times 2.0\text{ m} \times \sin(30^\circ) = 1.0 T\text{ N}\cdot\text{m}

. Equating the two gives

T = 260\text{ N}

.

Step-by-Step Solution

1

Identify the positions and lines of action of all forces relative to pivot

M

.

The weight of the uniform beam (

80\text{ N}

) acts at its center of gravity (

1.0\text{ m}

from

M

). The suspended load (

120\text{ N}

) acts at

1.5\text{ m}

from

M

. Cable tension

T

acts at

2.0\text{ m}

from

M

at an angle of

30^\circ

to the beam.

Rotational equilibrium requires evaluating moments created by all forces about the pivot point.

2

Calculate the sum of clockwise moments about pivot

M

.

\sum \tau_{\text{clockwise}} = (80\text{ N} \times 1.0\text{ m}) + (120\text{ N} \times 1.5\text{ m}) = 80\text{ N}\cdot\text{m} + 180\text{ N}\cdot\text{m} = 260\text{ N}\cdot\text{m}

.

Both downward forces exert clockwise turning effects about pivot

M

.

3

Determine the counterclockwise moment exerted by the inclined cable tension

T

.

\tau_{\text{counterclockwise}} = T \times d \sin\theta = T \times 2.0\text{ m} \times \sin(30^\circ) = 1.0 T\text{ N}\cdot\text{m}.

Only the component of tension perpendicular to the beam (

T \sin 30^\circ

) produces a moment about the pivot.

4

Apply the Principle of Moments to calculate tension

T

.

1.0 T = 260 \implies T = 260\text{ N}.

For the beam to remain horizontally balanced, total clockwise moment must equal total counterclockwise moment.

Key Concept

Principle of Moments and Rotational Equilibrium with Inclined Forces

Question 16Question →

A light rigid rod $OP$ of length $1.2\text{ m}$ is pivoted smoothly at end $O$ . A vertical downward load of $30\text{ N}$ is suspended from end $P$ . An upward force $F$ is applied at the midpoint of the rod at an angle of $30^\circ$ to the horizontal rod to keep it in horizontal equilibrium. What is the magnitude of the force $F$ ?

60\text{ N}

120\text{ N}

240\text{ N}

40\text{ N}

Show answer & explanation

Answer: $120\text{ N}$

Answer

The magnitude of the force

F

is

120\text{ N}

.

The correct answer is

120\text{ N}

. For rotational equilibrium about the pivot, the clockwise moment created by the suspended load (

30\text{ N} \times 1.2\text{ m} = 36\text{ N}\cdot\text{m}

) must equal the counterclockwise moment generated by force

F

. Because force

F

acts at an angle of

30^\circ

at the midpoint (

0.6\text{ m}

), its effective perpendicular component is

F \sin(30^\circ) = 0.5F

. Equating the moments gives

0.5F \times 0.6\text{ m} = 36\text{ N}\cdot\text{m}

, which yields

0.3F = 36

, resulting in

F = 120\text{ N}

.

Step-by-Step Solution

1

Calculate the clockwise moment about the pivot

O

due to the load at end

P

.

\tau_{\text{clockwise}} = 30\text{ N} \times 1.2\text{ m} = 36\text{ N}\cdot\text{m}

The force of

30\text{ N}

acts vertically downwards at a perpendicular distance of

1.2\text{ m}

from the pivot.

2

Determine the perpendicular distance (or perpendicular component of force) for

F

applied at the midpoint.

Midpoint distance =

\frac{1.2\text{ m}}{2} = 0.6\text{ m}

; Perpendicular force component =

F \sin(30^\circ) = 0.5F

Only the component perpendicular to the line of action contributes to the moment about pivot

O

.

3

Set up the counterclockwise moment expression about pivot

O

.

\tau_{\text{counterclockwise}} = F \sin(30^\circ) \times 0.6\text{ m} = 0.3F\text{ N}\cdot\text{m}

The moment is the product of the perpendicular force component and the distance from the pivot.

4

Equate clockwise and counterclockwise moments to solve for

F

.

0.3F = 36 \implies F = \frac{36}{0.3} = 120\text{ N}

According to the principle of moments, total clockwise moments must equal total counterclockwise moments for rotational equilibrium.

Key Concept

Principle of Moments and Rotational Equilibrium

Estimated Time:1m 15s

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