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Similarly, an inch-pound (or pound-inch) is the torque of one pound of force applied to one inch of distance from the pivot, and is equal to 1 ⁄ 12 lbf⋅ft (0.1129848 N⋅m). It is commonly used on torque wrenches and torque screwdrivers for setting specific fastener tension.
The foot-pound force (symbol: ft⋅lbf, [1] ft⋅lb f, [2] or ft⋅lb [3]) is a unit of work or energy in the engineering and gravitational systems in United States customary and imperial units of measure. It is the energy transferred upon applying a force of one pound-force (lbf) through a linear displacement of one foot.
The foot (standard symbol: ft) [1] [2] is a unit of length in the British imperial and United States customary systems of measurement. The prime symbol, ′, is commonly used to represent the foot. [3] In both customary and imperial units, one foot comprises 12 inches, and one yard comprises three feet.
The pound-force is the product of one avoirdupois pound (exactly 0.45359237 kg) and the standard acceleration due to gravity, approximately 32.174049 ft/s 2 (9.80665 m/s 2). [ 5 ] [ 6 ] [ 7 ] The standard values of acceleration of the standard gravitational field ( g n ) and the international avoirdupois pound (lb) result in a pound-force equal ...
One slug is a mass equal to 32.17405 lb (14.59390 kg) based on standard gravity, the international foot, and the avoirdupois pound. [3] In other words, at the Earth's surface (in standard gravity), an object with a mass of 1 slug weighs approximately 32.17405 lbf or 143.1173 N. [4] [5]
Since a pound of force (pound force) accelerates a pound of mass at 32.174 049 ft/s 2 (9.80665 m/s 2; the acceleration of gravity, g), we can scale down the unit of force to compensate, giving us one that accelerates 1 pound mass at 1 ft/s 2 rather than at 32.174 049 ft/s 2; and that is the poundal, which is approximately 1 ⁄ 32 pound force.
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The conversion for the poundal is given by 1 pdl = 1 lb·ft/s 2 = 0.138 254 954 376 N (precisely). [ 1 ] To convert between the absolute and gravitational FPS systems one needs to fix the standard acceleration g which relates the pound to the pound-force.