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Difficulty: HardLevers and Fulcrums

A steam boiler uses a horizontal safety valve lever hinged at one end (the fulcrum). The uniform steel lever is 50 inches50\text{ inches} long and weighs 20 lbs20\text{ lbs}. A pressure valve is attached to the lever 10 inches10\text{ inches} from the fulcrum and exerts an upward force of 150 lbs150\text{ lbs}. To keep the valve closed, what is the minimum downward-acting weight that must be hung at the very end of the lever (50 inches50\text{ inches} from the fulcrum)?

  1. 20 lbs20\text{ lbs}Answer
  2. B
    30 lbs30\text{ lbs}
  3. C
    90 lbs90\text{ lbs}
  4. D
    750 lbs750\text{ lbs}

Answer

The minimum weight that must be hung at the end is 20 lbs20\text{ lbs}.
The system is in equilibrium when the total upward torque equals the total downward torque. The upward torque from the valve (1500 in-lbs1500\text{ in-lbs}) must be balanced by the downward torque of the lever's center of gravity (500 in-lbs500\text{ in-lbs}) plus the torque of the applied weight at the end (20 lbs×50 inches=1000 in-lbs20\text{ lbs} \times 50\text{ inches} = 1000\text{ in-lbs}).

Step-by-Step Solution

1
Identify the opposing torques acting around the fulcrum.
Upward torque is generated by the steam valve. Downward torque is generated by the lever's own weight plus the applied weight at the end.
For the lever to remain balanced (closed), the total clockwise torque must equal the total counter-clockwise torque.
2
Calculate the upward torque from the steam valve.
150 lbs×10 inches=1500 in-lbs150\text{ lbs} \times 10\text{ inches} = 1500\text{ in-lbs} (upward).
Torque equals force multiplied by the perpendicular distance from the fulcrum.
3
Calculate the downward torque contributed by the uniform lever's own weight.
The center of gravity is at 25 inches25\text{ inches} (half of 50 inches50\text{ inches}). Torque = 20 lbs×25 inches=500 in-lbs20\text{ lbs} \times 25\text{ inches} = 500\text{ in-lbs} (downward).
A uniform object's weight acts through its geometric center.
4
Determine the remaining downward torque required and solve for the unknown weight.
1500 in-lbs−500 in-lbs=1000 in-lbs1500\text{ in-lbs} - 500\text{ in-lbs} = 1000\text{ in-lbs} of downward torque needed. W×50 inches=1000 in-lbsW \times 50\text{ inches} = 1000\text{ in-lbs}, so W=20 lbsW = 20\text{ lbs}.
The weight at the end must provide exactly the remaining torque to maintain equilibrium.

Key Concept

Torque Equilibrium and Lever Mechanics

Alternative Method

Instead of calculating full torques, recognize that an idealized weightless lever would require 30 lbs30\text{ lbs} of force at the end (150÷5150 \div 5). The lever's own weight (20 lbs20\text{ lbs}) acts at its midpoint, which effectively provides 10 lbs10\text{ lbs} of downward force translated to the very end (20÷220 \div 2). Thus, the required external weight is 30−10=20 lbs30 - 10 = 20\text{ lbs}.
Estimated Time:1m 30s
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Topics

Levers and Fulcrums

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