An irregularly shaped 300-pound machine casing is 10 feet long. For maintenance, it is sus… — Armed Services Vocational Aptitude Battery (ASVAB)

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?

$60$ poundsAnswer
B
$75$ pounds
C
$150$ pounds
D
$240$ 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

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.

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.

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

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