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The pound-force is equal to the gravitational force exerted on a mass of one avoirdupois pound on the surface of Earth.Since the 18th century, the unit has been used in low-precision measurements, for which small changes in Earth's gravity (which varies from equator to pole by up to half a percent) can safely be neglected.
For example, if a person places a force of 10 N at the terminal end of a wrench that is 0.5 m long (or a force of 10 N acting 0.5 m from the twist point of a wrench of any length), the torque will be 5 N⋅m – assuming that the person moves the wrench by applying force in the plane of movement and perpendicular to the wrench.
Conversions between units in the metric system are defined by their prefixes (for example, 1 kilogram = 1000 grams, 1 milligram = 0.001 grams) and are thus not listed in this article. Exceptions are made if the unit is commonly known by another name (for example, 1 micron = 10 −6 metre).
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.
In engineering and physics, g c is a unit conversion factor used to convert mass to force or vice versa. [1] It is defined as = In unit systems where force is a derived unit, like in SI units, g c is equal to 1.
With compressive force counted as negative tensile force, the rate of change of the tensile force in the direction of the g-force, per unit mass (the change between parts of the object such that the slice of the object between them has unit mass), is equal to the g-force plus the non-gravitational external forces on the slice, if any (counted ...
The inverse relationship between force per unit current and of a linear motor has been demonstrated. To translate this model to a rotating motor, one can simply attribute an arbitrary diameter to the motor armature e.g. 2 m and assume for simplicity that all force is applied at the outer perimeter of the rotor, giving 1 m of leverage.
The SI unit of force is the newton (symbol N), which is the force required to accelerate a one kilogram mass at a rate of one meter per second squared, or kg·m·s −2.The corresponding CGS unit is the dyne, the force required to accelerate a one gram mass by one centimeter per second squared, or g·cm·s −2. A newton is thus equal to ...