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Obviously, a rotating frame of reference is a case of a non-inertial frame. Thus the particle in addition to the real force is acted upon by a fictitious force...The particle will move according to Newton's second law of motion if the total force acting on it is taken as the sum of the real and fictitious forces.
Power in mechanical systems is the combination of forces and movement. In particular, power is the product of a force on an object and the object's velocity, or the product of a torque on a shaft and the shaft's angular velocity. Mechanical power is also described as the time derivative of work.
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.
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 ...
In this inertial frame, the concept of centrifugal force is not required as all motion can be properly described using only real forces and Newton's laws of motion. In a frame of reference rotating with the stone around the same axis as the stone, the stone is stationary. However, the force applied by the string is still acting on the stone.
The force on the mass is equal to the vector sum of the spring force and the kinetic frictional force. When the velocity changes sign (at the maximum and minimum displacements), the magnitude of the force on the mass changes by twice the magnitude of the frictional force, because the spring force is continuous and the frictional force reverses ...
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.
This force does zero work because it is perpendicular to the velocity of the ball. The magnetic force on a charged particle is F = qv × B, where q is the charge, v is the velocity of the particle, and B is the magnetic field. The result of a cross product is always perpendicular to both of the original vectors, so F ⊥ v.