A uniform window-washing scaffold is 20 feet long and weighs 180 pounds. It is supported h… — Armed Services Vocational Aptitude Battery (ASVAB)

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?

Cevap: 145 pounds

Cevap

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

20

yields a tension of

145

pounds.

Adım Adım Çözüm

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.

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.

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.

Anahtar Kavram

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

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