Load Distribution and Balancing Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB)

Question 1Question →

Two workers are carrying an engine block weighing $300$ pounds using a rigid, lightweight $10$ -foot pole that rests on their shoulders. If the engine block is hung exactly $2$ feet away from the first worker and $8$ feet away from the second worker, how much of the weight is the first worker supporting?

240

pounds

150

pounds

60

pounds

300

pounds

Show answer & explanation

Answer: $240$ pounds

Answer

The first worker supports

240

pounds.

In a load distribution system, a support carries a fraction of the total weight that is inversely proportional to its distance from the load. To find the load on the first worker, you can use the second worker's shoulder as a pivot point. The total length of the pole is

10

feet. The downward torque from the engine block is

300 \text{ pounds} \times 8 \text{ feet} = 2400 \text{ ft-lbs}

. The upward torque from the first worker must equal this, so

F_1 \times 10 \text{ feet} = 2400 \text{ ft-lbs}

. Solving this gives

F_1 = 240

pounds. Intuitively, because the weight is much closer to the first worker, they bear the vast majority of the load.

Step-by-Step Solution

1

Identify the total length of the support and the distances to the load.

Total pole length =

10

feet. Distance from the first worker to the load =

2

feet. Distance from the second worker to the load =

8

feet.

Establishing the lever arm distances is necessary to calculate the torques balancing the system.

2

Set up a torque equilibrium equation using the second worker as the pivot point (fulcrum).

F_1 \times 10 = 300 \times 8

In static equilibrium, the upward torque created by the first worker (

F_1

) must equal the downward torque created by the engine block relative to the second worker.

3

Solve the equation for the force exerted by the first worker (

F_1

).

10F_1 = 2400 \Rightarrow F_1 = 240

pounds.

Isolating the variable gives the exact load supported by the first worker.

Key Concept

Load Distribution and Equilibrium

Question 2Question →

A $120$ -pound crate rests on a sturdy $12$ -foot plank spanning two sawhorses placed at opposite ends. The center of the crate is located $3$ feet from the left sawhorse. Assuming the weight of the plank is negligible, what is the downward force, in pounds, exerted by the crate on the left sawhorse?

Show answer & explanation

Answer: 90

Answer

The downward force exerted on the left sawhorse is

90

pounds.

Because the crate is positioned

3

feet from the left sawhorse and

9

feet from the right sawhorse on a

12

-foot span, the left sawhorse must support

\frac{9}{12}

(or

75\%

) of the crate's weight. Multiplying

120

pounds by

0.75

yields exactly

90

pounds.

Step-by-Step Solution

1

Calculate the distance from the crate to the right sawhorse.

12 - 3 = 9

feet.

The weight supported by one end of a beam is proportional to the distance from the load to the opposite end.

2

Determine the fraction of the total weight carried by the left sawhorse.

\frac{9}{12}

, which simplifies to

\frac{3}{4}

or

0.75

.

This fraction represents the ratio of the opposite distance to the total span, dictating how much load is distributed to the left support.

3

Calculate the downward force on the left sawhorse.

120 \times 0.75 = 90

pounds.

Multiplying the total crate weight by the calculated proportion gives the specific load carried by that support.

Key Concept

Load Distribution and Balancing

Estimated Time:45s

Question 3Question →

A $100$ -pound pump is placed on a $10$ -foot scaffold board that spans between two structural beams, Beam X and Beam Y. The board's own weight is negligible. If the pump is located $2$ feet from Beam X, how much of the pump's weight is supported by Beam X?

20

pounds

50

pounds

80

pounds

100

pounds

Show answer & explanation

Answer: $80$ pounds

Answer

The correct answer is

80

pounds.

Because the pump is situated much closer to Beam X, that beam must carry a proportionally larger share of the weight. This can be solved by taking the torque around Beam Y: the total board length (

10

feet) multiplied by the upward force at Beam X must equal the pump's weight (

100

pounds) multiplied by its distance from Beam Y (

8

feet). Solving

(F_X \times 10 = 100 \times 8)

results in a load of

80

pounds on Beam X.

Step-by-Step Solution

1

Identify the distances of the load from both supports.

The pump is

2

feet from Beam X. Since the board is

10

feet long total, the pump is

8

feet from Beam Y (

10 - 2 = 8

).

To calculate the load distribution, the distance to the opposite support must be determined.

2

Apply the principle of moments (torque) using Beam Y as the pivot point.

Torque equation: Force at Beam X

\times

Total distance

=

Pump weight

\times

Distance to Beam Y. Therefore,

F_X \times 10 = 100 \times 8

.

The sum of torques around any pivot point must be zero for the structural system to remain in static equilibrium.

3

Solve the equation for the force on Beam X.

10 \times F_X = 800

, which simplifies to

F_X = 80

pounds.

This calculation isolates

F_X

to provide the exact weight supported by Beam X.

Key Concept

Load Distribution and Static Equilibrium

Question 4Question →

A construction crew is moving several identical uniform steel I-beams. Each beam is $12$ feet long and weighs $600$ pounds. Two workers will carry each beam by supporting it from beneath at different points along its length.

Rank the following carrying configurations based on the amount of weight supported by the worker positioned on the LEFT side, from the GREATEST load to the LEAST load.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The configuration with the greatest load on the left worker is when the left worker is 3 feet inward (400 lbs), followed by 2 feet inward (360 lbs), then both workers at extreme ends (300 lbs), and finally the least load occurs when the right worker is moved inward (200 lbs).

As a support point moves closer to a uniform object's center of gravity, it takes on a proportionally larger share of the total weight. Moving the opposite support closer to the center transfers weight away from the first support.

Step-by-Step Solution

1

Determine the location of the center of gravity (CG).

For a uniform

12

-foot beam, the CG is exactly in the middle at the

6

-foot mark.

The weight of a uniform object can be treated as a single downward point force acting at its center.

2

Establish the formula for calculating load distribution.

Load on Left Worker = Total Weight

\times

(Distance from Right Worker to CG) / (Total Distance Between Workers)

This formula balances the torques around the right worker, which acts as a fulcrum when determining the left worker's lifting force.

3

Calculate the left worker's load for each configuration.

Item 1:

600 \times (6 / 12) = 300

lbs.
Item 2:

600 \times (3 / 9) = 200

lbs.
Item 3:

600 \times (6 / 10) = 360

lbs.
Item 4:

600 \times (6 / 9) = 400

lbs.

Applying the formula from step 2 translates the distances into exact pounds supported.

4

Order the results from greatest to least.

400

lbs (Item 4),

360

lbs (Item 3),

300

lbs (Item 1),

200

lbs (Item 2).

The question requests ranking from greatest load supported by the left worker to the least.

Key Concept

Load Distribution and Balancing

Estimated Time:2m 0s

Question 5Question →

Three identical $10$ -foot planks are each placed across two sawhorses, one sawhorse at the left end (Support X) and one at the right end (Support Y). A single $60$ -pound bag of cement is placed on each plank at a different location. Rank the scenarios below based on the amount of downward force carried by Support X, from greatest to least.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The bag closest to Support X places the greatest load on it, followed by the centered bag, while the bag closest to Support Y places the least load on Support X.

Support X must carry the largest portion of the load when the weight is closest to it. As the weight moves further away across the plank, the amount of force supported by X decreases.

Step-by-Step Solution

1

Understand the principle of load distribution.

The closer a weight is to a given support, the more of that weight the support must carry.

This is a basic rule of mechanical balance and levers.

2

Evaluate the load on Support X for the bag placed

3

feet from it.

This bag is very close to Support X, so Support X carries more than half the weight.

The distance to X is smaller than the distance to Y.

3

Evaluate the load on Support X for the centered bag.

This bag is exactly

5

feet from both supports, so Support X carries exactly half the weight.

A centered load is shared evenly.

4

Evaluate the load on Support X for the bag placed

2

feet from Support Y.

This bag is

8

feet away from Support X, so Support Y carries almost all the weight, leaving very little for Support X.

The distance to X is much larger than the distance to Y.

Key Concept

The proportion of a load carried by a support increases as the center of gravity moves closer to that support.

Question 6Question →

A heavy generator is suspended horizontally by two vertical cables, Cable X on the left and Cable Y on the right. The center of gravity of the generator is located closer to Cable Y. Which of the following describes the load distribution between the two cables?

Cable X carries more of the load than Cable Y.

Cable Y carries more of the load than Cable X.

Cable X and Cable Y carry equal amounts of the load.

The distribution depends entirely on the generator's total mass, not its center of gravity.

Show answer & explanation

Answer: Cable Y carries more of the load than Cable X.

Answer

Cable Y carries more of the load than Cable X.

In mechanical comprehension, when a load is supported by multiple points, the support closer to the object's center of gravity must carry a proportionally larger share of the weight to maintain static equilibrium. Since the generator's center of gravity is closer to Cable Y, Cable Y bears the greater load.

Step-by-Step Solution

1

Identify the location of the supports and the center of gravity.

Cable X is on the left, Cable Y is on the right, and the center of gravity is closer to Cable Y.

Understanding the geometry of the setup is the first step in determining load distribution.

2

Apply the principle of load distribution and equilibrium.

The support closer to the center of gravity must provide a larger upward reaction force to balance the weight.

The center of gravity acts as the specific point where the entire downward force of the object's mass is concentrated.

3

Determine which cable carries the greater load.

Cable Y carries a greater portion of the generator's weight.

Because Cable Y is physically closer to the concentration of the downward force, it must support more of the load than Cable X.

Key Concept

In a system with multiple supports, the support closest to the object's center of gravity carries the greatest proportion of the load.

Question 7Question →

A uniform wooden beam is $10$ feet long and weighs $50$ pounds. It rests horizontally on two sawhorses. Sawhorse A is placed $4$ feet from the left end, and Sawhorse B is placed $9$ feet from the left end. A $50$ -pound bag of cement is placed at the extreme right end of the beam ( $10$ feet from the left end). Assuming the system is in static equilibrium, what is the upward reaction force exerted by Sawhorse B?

30

pounds

50

pounds

60

pounds

70

pounds

Show answer & explanation

Answer: $70$ pounds

Answer

The upward reaction force exerted by Sawhorse B is

70

pounds.

For the system to be in static equilibrium, the moments around any pivot point must sum to zero. Choosing Sawhorse A as the pivot: the beam's center of gravity (

5

ft mark) is

1

ft right of the pivot, creating

50 \times 1 = 50

ft-lbs of clockwise torque. The bag (

10

ft mark) is

6

ft right of the pivot, creating

50 \times 6 = 300

ft-lbs of clockwise torque. To balance the total

350

ft-lbs of clockwise torque, Sawhorse B (located

5

ft right of the pivot) must provide

350 / 5 = 70

pounds of upward force.

Step-by-Step Solution

1

Identify the pivot point and the locations of all downward forces.

Using Sawhorse A (at

4

ft) as the pivot. The beam's weight (

50

lbs) acts at its center (

5

ft). The cement bag (

50

lbs) acts at the right end (

10

ft).

Choosing one support as the pivot eliminates its unknown force from the equilibrium equation.

2

Calculate the lever arm distances from the pivot for each force.

Beam weight lever arm:

5 - 4 = 1

ft. Cement bag lever arm:

10 - 4 = 6

ft. Sawhorse B lever arm:

9 - 4 = 5

ft.

Torque is determined by multiplying a force by its perpendicular distance from the pivot point.

3

Calculate the total clockwise torque acting on the beam.

Beam torque:

50 \text{ lbs} \times 1 \text{ ft} = 50 \text{ ft-lbs}

. Cement bag torque:

50 \text{ lbs} \times 6 \text{ ft} = 300 \text{ ft-lbs}

. Total clockwise torque =

350 \text{ ft-lbs}

.

Both the beam's weight and the cement bag push downward on the right side of the pivot, creating clockwise rotation.

4

Set the clockwise torque equal to the counterclockwise torque and solve for Sawhorse B's force.

F_B \times 5 \text{ ft} = 350 \text{ ft-lbs}

, which means

F_B = 350 / 5 = 70

pounds.

In static equilibrium, the torques must perfectly balance to prevent the beam from rotating.

Key Concept

Rotational Equilibrium and Center of Gravity

Alternative Method

You can verify this by calculating moments around Sawhorse B instead. The beam's center (

5

ft) is

4

ft to the left (CCW torque:

50 \times 4 = 200

). The cement bag (

10

ft) is

1

ft to the right (CW torque:

50 \times 1 = 50

). Sawhorse A is

5

ft to the left (CW torque:

F_A \times 5

). Balancing torques:

200 = 50 + 5(F_A) \rightarrow F_A = 30

lbs. Since total downward force is

100

lbs,

F_B

must be

100 - 30 = 70

lbs.

Estimated Time:2m 0s

Question 8Question →

A footbridge spans a $20$ -foot wide creek, resting on a concrete pillar at each end (Left Pillar and Right Pillar). A person weighing $160$ pounds stands on the bridge, exactly $5$ feet from the Right Pillar. If we ignore the weight of the bridge itself, how many pounds of the person's weight are supported by the Left Pillar?

Show answer & explanation

Answer: 40

Answer

40

The weight supported by the Left Pillar is inversely proportional to the load's distance from it. Since the person is

15

feet from the Left Pillar and

5

feet from the Right Pillar on a

20

-foot bridge, they are

3

times closer to the Right Pillar. Therefore, the Right Pillar carries

3/4

of the weight (

120

lbs) and the Left Pillar carries

1/4

of the weight (

40

lbs).

Step-by-Step Solution

1

Determine the relevant distances along the bridge.

Total span =

20

ft. Distance to Right Pillar =

5

ft. Distance to Left Pillar =

15

ft.

To calculate the load on one support, we need the total distance and the distance to the opposite support.

2

Set up the load distribution calculation for the Left Pillar.

Load on Left Pillar = Total Weight

\times

(Distance to Right Pillar / Total Span)

The proportion of weight supported by one end of a beam is equal to the distance from the load to the opposite end, divided by the total length.

3

Calculate the supported weight using the values.

160 \times (5 / 20) = 160 \times (1 / 4) = 40

pounds.

Solving the equation gives the exact weight carried by the left support.

Key Concept

Load Distribution and Balancing

Alternative Method

You can also use the concept of torque (moments) in static equilibrium. Taking the Right Pillar as the pivot point: (Force of Left Pillar

\times

20

ft) = (

160

lbs

\times

5

ft). Therefore,

20 \times F_{left} = 800

, which means

F_{left} = 40

lbs.

Estimated Time:45s

Question 9Question →

A single-axle utility trailer has its axle located $8$ feet behind the hitch. The empty trailer weighs $600$ pounds, and its center of gravity is $5$ feet behind the hitch. A machine weighing $1,200$ pounds is loaded onto the trailer so that its center of gravity is $6$ feet behind the hitch. When the trailer is parked horizontally, what is the downward force, in pounds, exerted on the hitch?

Show answer & explanation

Answer: 525

Answer

525

The correct answer is derived by treating the trailer as a lever in static equilibrium. Taking the sum of torques around the axle, the total rotational force from the empty trailer and the loaded machine is

4200

lb-ft. Dividing this by the

8

-foot lever arm to the hitch yields exactly

525

pounds of force.

Step-by-Step Solution

1

Identify a pivot point and determine all relevant distances.

Using the axle as the pivot point. Distance from axle to trailer CG is 3 ft (

8 - 5

). Distance from axle to machine CG is 2 ft (

8 - 6

). Distance from axle to hitch is 8 ft.

Establishing a pivot point allows us to use the principle of moments (torques) to find the unknown reaction force.

2

Calculate the total counter-clockwise torque around the axle.

Trailer torque =

600 \text{ lbs} \times 3 \text{ ft} = 1800 \text{ lb-ft}

. Machine torque =

1200 \text{ lbs} \times 2 \text{ ft} = 2400 \text{ lb-ft}

. Total torque =

4200 \text{ lb-ft}

.

The weight of the trailer and the machine both push down, attempting to rotate the system around the axle.

3

Set up the static equilibrium equation.

4200 \text{ lb-ft} = F_{hitch} \times 8 \text{ ft}

.

For the trailer to remain stationary, the clockwise torque provided by the hitch must perfectly balance the total counter-clockwise torque.

4

Solve for the force at the hitch (

F_{hitch}

).

F_{hitch} = \frac{4200}{8} = 525 \text{ lbs}

.

Dividing the total torque by the distance to the hitch gives the required balancing force.

Key Concept

Static Equilibrium and Torque Distribution

Alternative Method

You can also solve this by setting the hitch as the pivot point. The downward torque would be

(600 \times 5) + (1200 \times 6) = 3000 + 7200 = 10200

lb-ft. The upward force from the axle balances this:

F_{axle} \times 8 = 10200

, so

F_{axle} = 1275

lbs. Since the total weight is

1800

lbs (

600 + 1200

), the remaining force supported by the hitch is

1800 - 1275 = 525

lbs.

Estimated Time:2m 0s

Question 10Question →

Two technicians, Technician X and Technician Y, are carrying a heavy transmission using a horizontal lifting bar. Technician X holds the left end of the bar and Technician Y holds the right end. The transmission is secured to the bar much closer to Technician X than to Technician Y. Which of the following statements correctly describes how the weight is distributed between them?

Technician X supports a greater portion of the weight than Technician Y.

Technician Y supports a greater portion of the weight than Technician X.

Both technicians support exactly equal portions of the weight.

Technician Y supports a greater portion of the weight due to the mechanical advantage of the longer bar section.

Show answer & explanation

Answer: Technician X supports a greater portion of the weight than Technician Y.

Answer

Technician X supports a greater portion of the weight than Technician Y.

In load balancing, the support that is physically closer to the center of gravity of the load carries a proportionally larger share of the weight. Because the transmission is secured closer to Technician X, Technician X must exert more upward force to maintain equilibrium.

Step-by-Step Solution

1

Identify the locations of the two supports and the applied load.

The supports are at each end of the bar (Technician X and Technician Y), and the load (the transmission) is located closer to Technician X.

Understanding the physical arrangement is necessary to analyze the forces.

2

Apply the general rule of load distribution for offset weights.

When a weight is placed between two supports but off-center, the support closest to the weight carries a proportionally larger share of the total load.

This is derived from the principles of static equilibrium and rotational moments; a shorter lever arm from the load requires a greater supporting force.

3

Conclude which technician bears more weight based on their proximity to the load.

Since the transmission is closer to Technician X, Technician X bears a greater portion of the load.

The weight distribution is inversely proportional to the distance from the load to the support.

Key Concept

In a system with two supports, an offset load distributes its weight unevenly, with the support closest to the load's center of gravity bearing the larger portion of the weight.

Question 11Question →

A uniform wooden plank is $16$ feet long and weighs $120$ pounds. It rests horizontally on two supports. The left support is positioned at the extreme left end of the plank, and the right support is positioned $12$ feet away from the left end. If a $60$ -pound weight is placed on the extreme right end of the plank, how much upward force must the right support provide to maintain equilibrium?

90

pounds

120

pounds

160

pounds

180

pounds

Show answer & explanation

Answer: $160$ pounds

Answer

160

pounds

The system is in equilibrium, meaning the torques must balance. Taking the left support as the pivot, the downward forces create a clockwise torque: the plank's weight (

120

lbs) acts at its center (

8

ft away) creating

960 \text{ lb-ft}

of torque, and the box (

60

lbs) acts at the end (

16

ft away) creating another

960 \text{ lb-ft}

. The total downward torque is

1920 \text{ lb-ft}

. The right support is located

12

feet away and must push upward to create an equal counter-clockwise torque. Therefore,

12 \times \text{Force} = 1920

, which means the required force is

160

pounds.

Step-by-Step Solution

1

Set the left support as the pivot point (fulcrum).

All distances for torque calculations will be measured from the

0

-foot mark.

Choosing one of the unknown supports as the pivot eliminates its force from the torque equation, allowing us to solve directly for the other support.

2

Calculate the total clockwise torque acting on the plank.

Plank torque =

120 \text{ lbs} \times 8 \text{ ft} = 960 \text{ lb-ft}

. Weight torque =

60 \text{ lbs} \times 16 \text{ ft} = 960 \text{ lb-ft}

. Total clockwise torque =

1920 \text{ lb-ft}

.

The uniform plank's weight acts at its center of gravity (

8

feet from the left), and the external weight acts at the extreme right end (

16

feet from the left). Both create a downward, clockwise rotation around the left support.

3

Equate the clockwise torque to the counter-clockwise torque to solve for the right support's force.

F_{right} \times 12 \text{ ft} = 1920 \text{ lb-ft}

. Dividing by

12

gives

F_{right} = 160 \text{ lbs}

.

For the system to remain in static equilibrium, the upward (counter-clockwise) torque provided by the right support must exactly balance the downward (clockwise) torques.

Key Concept

Static Equilibrium and Principle of Moments

Question 12Question →

A uniform steel bar is $10$ feet long and weighs $200$ pounds. It is placed on a pivot located exactly $2$ feet from its left end. To keep the bar perfectly horizontal, a downward force must be applied to the extreme left end. What is the magnitude of this downward force?

100

pounds

120

pounds

300

pounds

500

pounds

Show answer & explanation

Answer: $300$ pounds

Answer

The required downward force is

300

pounds.

To maintain a horizontal position, the torques around the pivot must balance perfectly. Because the bar is uniform, its entire

200

-pound weight acts at its center of gravity, which is

5

feet from the end. The pivot is located

2

feet from the left end, meaning the center of gravity is exactly

3

feet to the right of the pivot (

5 - 2 = 3

). This creates a clockwise torque of

600

lb-ft (

200

lbs

\times

3

ft). To balance this rotation, a downward force applied at the left end (which is

2

feet from the pivot) must create an equal

600

lb-ft counter-clockwise torque. Dividing the required

600

lb-ft by the

2

-foot distance gives a required downward force of

300

pounds.

Step-by-Step Solution

1

Determine the location of the center of gravity (CG).

The CG is exactly in the middle of the

10

-foot bar, which is

5

feet from either end.

A uniform object's weight is evenly distributed, so its total weight acts geometrically at its center.

2

Calculate the distances from the pivot point.

The left end is

2

feet from the pivot. The CG is

3

feet from the pivot (

5

feet -

2

feet).

Torque is calculated using the distance from the exact pivot (fulcrum) to the point where the force acts.

3

Calculate the clockwise torque created by the bar's weight.

200

lbs

\times

3

ft =

600

lb-ft.

The weight of the bar pulls down on the right side of the pivot, creating a clockwise rotational tendency.

4

Calculate the required counter-clockwise force at the left end.

Force

\times

2

ft =

600

lb-ft, which means Force =

300

lbs.

For the bar to remain horizontal (in equilibrium), the counter-clockwise torque must exactly balance the clockwise torque.

Key Concept

Principle of Moments (Torque Equilibrium)

Question 13Question →

A uniform $10$ -foot wooden beam weighing $50$ pounds is placed on a flat surface. In four different scenarios, exactly $150$ pounds of additional weight is placed on the beam in various configurations. Arrange the scenarios in order of the beam's overall balance point (center of gravity), measured from the LEFT end of the beam, from the GREATEST distance to the LEAST distance.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from greatest to least distance from the left end is: the two

75

-pound weights (furthest right), the

50

/

100

-pound weights, the

100

/

50

-pound weights, and finally the single

150

-pound weight (closest to the left).

Because the total system weight is identical (

200

lbs) in every scenario, the relative position of the center of gravity is determined purely by the sum of the moments (weight

\times

distance) of the added weights from the left end. Calculating these moments yields

1050

,

900

,

600

, and

450

ft-lbs respectively. A larger moment means the center of gravity is pulled further to the right, which translates to a greater overall distance from the left end.

Step-by-Step Solution

1

Identify the components that determine the overall center of gravity (balance point).

The center of gravity's distance from an origin (the left end) is found by dividing the total turning moment (torque) by the total weight of the system.

This is the fundamental principle of moments for finding the balance point of a loaded beam.

2

Observe the constants across all four scenarios.

In every scenario, the total system weight is identical at

200

lbs (

50

lb beam +

150

lbs of added weight), and the beam's own moment from the left end remains constant.

Because the total weight denominator is identical across all choices, the relative position of the balance point is determined solely by the sum of the moments of the added weights.

3

Calculate the turning moment (Weight

\times

Distance) of the added weights relative to the left end for each scenario.

Two

75

-lb weights:

(75 \times 4) + (75 \times 10) = 1050

ft-lbs.

50

-lb and

100

-lb weights:

(50 \times 2) + (100 \times 8) = 900

ft-lbs.

100

-lb and

50

-lb weights:

(100 \times 2) + (50 \times 8) = 600

ft-lbs.
Single

150

-lb weight:

150 \times 3 = 450

ft-lbs.

This calculation provides a numerical value directly proportional to how far right the weight is biased.

4

Order the scenarios by their added moments from greatest to least.

The order is

1050 > 900 > 600 > 450

. Therefore, the balance point distances follow the exact same order.

A greater turning moment from the left end means the center of gravity is located further to the right, which translates to a greater distance from the left end.

Key Concept

The center of gravity of a system shifts proportionally toward the location with the greatest turning moment (Force

\times

Distance). When total mass is constant, ranking moments ranks the balance points.

Estimated Time:1m 30s

Question 14Question →

A uniform window-washing scaffold is $20$ feet long and weighs $180$ pounds. It is supported horizontally by two vertical cables, one attached at the extreme left end and one at the extreme right end. A worker weighing $220$ pounds stands exactly $5$ feet from the left cable. If the system is in static equilibrium, what is the tension, in pounds, in the right cable?

Show answer & explanation

Answer: 145

Answer

145

The system is in equilibrium, so the sum of the moments about the left support must be zero. The downward weights create a clockwise moment of

2900 \text{ lb-ft}

(

180 \text{ lbs} \times 10 \text{ ft}

plus

220 \text{ lbs} \times 5 \text{ ft}

). To balance this, the right cable, located

20

feet away, must provide an equal counter-clockwise moment. Dividing

2900

by

20

yields a tension of

145

pounds.

Step-by-Step Solution

1

Select a pivot point to calculate moments (torques).

The left end of the scaffold is chosen as the pivot point.

Choosing the left end eliminates the unknown tension of the left cable from the equation, allowing direct calculation of the right cable's tension.

2

Calculate the total clockwise moment caused by the downward weights.

Scaffold moment =

180 \text{ lbs} \times 10 \text{ ft} = 1800 \text{ lb-ft}

. Worker moment =

220 \text{ lbs} \times 5 \text{ ft} = 1100 \text{ lb-ft}

. Total clockwise moment =

2900 \text{ lb-ft}

.

The scaffold's weight acts at its center (

10

ft), and the worker's weight acts at

5

ft from the pivot. Moment equals force multiplied by distance.

3

Equate the clockwise moment to the counter-clockwise moment and solve.

T_R \times 20 \text{ ft} = 2900 \text{ lb-ft}

, which gives

T_R = 145 \text{ lbs}

.

For the scaffold to be in static equilibrium, the sum of all moments about any pivot point must be zero.

Key Concept

Rotational equilibrium and calculation of moments (torque) to find support reaction forces.

Question 15Question →

A commercial box truck has a wheelbase (the distance between the front and rear axles) of $15$ feet. The empty truck weighs $6,000$ pounds, with its center of gravity located $6$ feet behind the front axle. A $3,000$ -pound industrial generator is loaded onto the truck and positioned directly above the rear axle. What is the total load supported by the rear axle in this configuration?

4,500

pounds

5,400

pounds

6,000

pounds

6,600

pounds

Show answer & explanation

Answer: $5,400$ pounds

Answer

5,400

pounds

To find the load on the rear axle, we must evaluate the contributions of the empty truck and the generator separately. For the empty truck, the center of gravity is

6

feet from the front axle, meaning it is closer to the front. Taking moments around the front axle, the rear axle supports

6/15

of the

6,000

-pound truck, which equals

2,400

pounds. Because the

3,000

-pound generator is placed directly over the rear axle, its entire weight is added to that support. Adding both contributions together yields

2,400 + 3,000 = 5,400

pounds.

Step-by-Step Solution

1

Calculate the portion of the empty truck's weight supported by the rear axle.

Using the front axle as a pivot, the moment equation is: Rear Axle Force ×

15

ft =

6,000

lbs ×

6

ft. Solving for the force gives

36,000 / 15 = 2,400

lbs.

The principle of moments dictates that the reaction force at a support is proportional to the distance of the load from the opposite support.

2

Determine the load added by the generator to the rear axle.

Since the

3,000

-lb generator is placed directly over the rear axle,

100\%

of its weight rests on that axle (

3,000

lbs).

Loads placed directly on top of a support do not transfer any moment to the other supports.

3

Calculate the total weight supported by the rear axle.

2,400\text{ lbs} + 3,000\text{ lbs} = 5,400\text{ lbs}

.

The total load on a support is the sum of all individual load contributions.

Key Concept

Principle of Moments and Offset Load Distribution

Estimated Time:1m 15s

Question 16Question →

A uniform $12$ -foot diving board weighing $50$ pounds is attached to a rear tie-down anchor at one end. The board rests on a fulcrum support located exactly $3$ feet from the rear anchor. A $100$ -pound swimmer stands at the extreme opposite end of the board. What is the required downward holding force (tension) at the rear tie-down anchor to keep the board horizontally balanced?

150

pounds

300

pounds

350

pounds

500

pounds

Show answer & explanation

Answer: $350$ pounds

Answer

350

pounds

To find the balancing force, we must ensure all rotational forces (torques) around the fulcrum cancel out. The swimmer (

100

lbs at a

9

-foot distance from the fulcrum) creates

900

lb-ft of torque. The board's center of gravity (

50

lbs at a

3

-foot distance from the fulcrum) creates

150

lb-ft of torque. The total clockwise torque is

1050

lb-ft. Because the rear anchor is

3

feet from the fulcrum on the opposite side, it must pull down with a force of

1050 \div 3 = 350

pounds to maintain equilibrium.

Step-by-Step Solution

1

Identify the pivot point (fulcrum) and calculate the distances of all forces from this point.

The fulcrum is at

3

ft. The rear anchor is

3

ft away. The board's center of gravity is at

6

ft (

3

ft from fulcrum). The swimmer is at

12

ft (

9

ft from fulcrum).

Torque is calculated by multiplying force by the distance to the fulcrum, not to the edge of the object.

2

Calculate the total clockwise torque caused by the board and the swimmer.

Board torque:

50 \text{ lbs} \times 3 \text{ ft} = 150 \text{ lb-ft}

. Swimmer torque:

100 \text{ lbs} \times 9 \text{ ft} = 900 \text{ lb-ft}

. Total clockwise torque =

1050 \text{ lb-ft}

.

These forces push down on the right side of the fulcrum, acting to rotate the front of the diving board downward.

3

Determine the downward force required at the rear anchor to balance the system.

The rear anchor must provide an equal counter-clockwise torque.

\text{Force} \times 3 \text{ ft} = 1050 \text{ lb-ft}

.

\text{Force} = \frac{1050}{3} = 350 \text{ lbs}

.

For static equilibrium, the opposing torques around the fulcrum must be equal.

Key Concept

Static Equilibrium and Torque

Question 17Question →

A heavy jet engine is suspended horizontally from a $12$ -foot rigid lifting beam. The lifting beam is supported by two overhead hoists: Hoist 1 at the left end and Hoist 2 at the right end. Rank the following attachment points for the jet engine based on the tension load experienced by Hoist 1, from GREATEST to LEAST.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from greatest to least load on Hoist 1 is:

3

feet from Hoist 1,

6

feet from Hoist 1,

9

feet from Hoist 1, and

12

feet from Hoist 1.

In a simple beam supported at both ends, the load distributed to each support is directly proportional to how close the weight is to that support. The closer the jet engine is to Hoist 1, the more weight Hoist 1 must support. Therefore, arranging the attachment points from the shortest distance (

3

feet) to the longest distance (

12

feet) correctly orders the load on Hoist 1 from greatest to least.

Step-by-Step Solution

1

Identify the relationship between load position and support reaction forces in a static beam system.

The closer a load is to a specific support point, the greater the proportion of the weight that support must carry.

This is a fundamental principle of lever mechanics and load distribution.

2

Determine the relative distances of each attachment point from Hoist 1.

The distances are

3

feet,

6

feet,

9

feet, and

12

feet.

Establishing the distances allows for direct comparison of the load distribution.

3

Order the distances from shortest to longest to rank the load from greatest to least.

Shortest distance (

3

ft) creates the greatest load on Hoist 1, followed by

6

ft,

9

ft, and the longest distance (

12

ft) creates the least load.

Because the load on a support is inversely proportional to the load's distance from that support.

Key Concept

The proportion of a load carried by a support is inversely related to the distance between the center of gravity of the load and that support.

Estimated Time:1m 0s

Question 18Question →

An irregularly shaped $300$ -pound machine casing is $10$ feet long. For maintenance, it is suspended horizontally by two vertical chains, one attached at the extreme left end and one at the extreme right end. The casing's center of gravity is located $2$ feet from the left end.

How much weight is supported by the right chain?

60

pounds

75

pounds

150

pounds

240

pounds

Show answer & explanation

Answer: $60$ pounds

Answer

60

pounds

Because the center of gravity is offset, the weight is not distributed equally. By using the left end as a pivot point, we can balance the torques. The weight of the casing creates a downward torque of

300 \text{ lbs} \times 2 \text{ ft} = 600 \text{ ft-lbs}

. The right chain, pulling upward at a distance of

10

feet, must create an equal opposite torque. Therefore,

F_{right} \times 10 \text{ ft} = 600 \text{ ft-lbs}

, which means the right chain supports

60

pounds.

Step-by-Step Solution

1

Choose a pivot point to set up a torque equation.

Using the left end (where the left chain is attached) as the pivot eliminates the left chain's unknown force from the calculation.

In static equilibrium, the sum of torques around any point must be zero. Choosing one support as the pivot makes it easier to solve for the other support.

2

Calculate the downward torque generated by the casing's weight.

The downward torque is the weight (

300

pounds) multiplied by its distance from the left pivot (

2

feet):

300 \times 2 = 600

foot-pounds.

Torque is Force multiplied by Distance from the fulcrum.

3

Set the downward torque equal to the upward torque generated by the right chain and solve.

The upward torque is the unknown force of the right chain (

F_{right}

) multiplied by its distance from the pivot (

10

feet). So,

F_{right} \times 10 = 600

. Dividing both sides by

10

gives

F_{right} = 60

pounds.

Upward torque must equal downward torque for the casing to remain balanced horizontally.

Key Concept

Load Distribution and Balancing

Estimated Time:1m 0s

Question 19Question →

During a routine weight and balance check, a small aircraft is parked horizontally on weighing scales. The scale under the nose wheel reads $500$ pounds, and the combined reading from the scales under the two main wheels is $1,500$ pounds. The nose wheel is located exactly $100$ inches forward of the main landing gear axis. How many inches forward of the main landing gear axis is the aircraft's center of gravity?

Show answer & explanation

Answer: 25

Answer

25

In static equilibrium, the sum of the moments around any point must be zero. By using the main landing gear axis as the reference point, the moment created by the total weight of the aircraft (

2,000

pounds) acting at the center of gravity must equal the moment created by the nose wheel's load. The nose wheel is

100

inches away and supports

500

pounds, creating a moment of

50,000

pound-inches. Dividing this total moment by the total weight (

50,000 \div 2,000

) gives a center of gravity located

25

inches forward of the main gear.

Step-by-Step Solution

1

Determine the total weight of the aircraft.

500 + 1,500 = 2,000

pounds

The total weight is needed to find the proportion of the load carried by each support.

2

Set up the balance equation (sum of moments) around the main landing gear axis.

(Total\ Weight) \times (CG\ Distance) = (Nose\ Wheel\ Weight) \times (Distance\ to\ Nose\ Wheel)

In static equilibrium, the moment generated by the total weight at the CG equals the sum of the moments of individual components.

3

Substitute the known values into the equation and solve for the CG distance.

2,000 \times d = 500 \times 100 \Rightarrow 2,000d = 50,000 \Rightarrow d = 25

inches

Solving the algebraic equation isolates the distance from the main gear to the center of gravity.

Key Concept

Center of Gravity and Static Equilibrium

Alternative Method

Instead of using the main gear as the pivot, you can use the nose wheel as the pivot. The moment of the main gear is

1,500 \times 100 = 150,000

. Dividing by the total weight (

2,000

) gives

75

inches from the nose wheel. Subtracting this from the total wheelbase (

100 - 75

) gives

25

inches from the main gear.

Estimated Time:1m 0s

Question 20Question →

A single-axle utility trailer has a total loaded weight of $2,000$ pounds. The loaded trailer's center of gravity is located $3$ feet forward of the axle. If the trailer hitch is located $12$ feet forward of the axle, what is the downward force exerted on the hitch when the trailer is horizontally parked?

500

pounds

1,000

pounds

1,500

pounds

8,000

pounds

Show answer & explanation

Answer: $500$ pounds

Answer

500

pounds

In static equilibrium, the sum of torques around any pivot point must be zero. By treating the axle as the pivot, the weight of the trailer (

2,000

pounds) acts at a distance of

3

feet from the pivot, creating a torque of

6,000

ft-lbs. To balance this, the upward force from the hitch (located

12

feet away) must equal

6,000

ft-lbs divided by

12

feet, which yields

500

pounds.

Step-by-Step Solution

1

Identify the pivot point and the forces acting on the system.

The axle acts as the pivot point (fulcrum). The two forces creating torque around the axle are the trailer's weight (

2,000

lbs) pushing down at the center of gravity, and the hitch support pushing up.

Setting the axle as the pivot eliminates its unknown support force from the initial torque equation.

2

Calculate the torque (moment) created by the trailer's weight around the axle.

Torque = Force × Distance =

2,000\text{ lbs} \times 3\text{ ft} = 6,000\text{ lb-ft}

.

This is the rotational force that the hitch must counterbalance to keep the trailer horizontal.

3

Set up the equilibrium equation to solve for the force at the hitch.

F_{\text{hitch}} \times 12\text{ ft} = 6,000\text{ lb-ft}

.

In static equilibrium, the clockwise torque must equal the counterclockwise torque.

4

Solve for the hitch force.

F_{\text{hitch}} = 6,000 \div 12 = 500\text{ pounds}

.

Dividing the required balancing torque by the lever arm distance gives the necessary support force.

Key Concept

Static Equilibrium and Torque (Moments)

Load Distribution and Balancing

Answer

Step-by-Step Solution

Key Concept

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