✔ 最佳答案
An object's weight, henceforth called "actual weight", is the downward force exerted upon it by a gravitational field. By contrast, an object's apparent weight is the (usually upward) force (the normal force, or reaction force), typically transmitted through the ground, that opposes the (usually downward) acceleration of a supported object, preventing it from falling.
An object's apparent weight is equal to its actual weight, unless:
The object has an acceleration, as in a lift, a rocket, or a rollercoaster.
Some force other than the earth's gravity and the normal force is acting on the object. This may, for example, be buoyancy, centrifugal force due to the Earth's rotation, magnetic force, or the gravitational force of another astronomical body.
Suppose that Alice has a mass of 65 kilograms and is standing stationary on the floor. Gravity is pulling her downward with a force of:
Fgrav = mg
= 65 kg × 9.81 m/s^2
= 637.65 N (newtons)
where m is mass and g is the acceleration due to gravity. By definition, Fgrav, the downward force of gravity, is Alice's actual weight. (Note that the force of gravity varies slightly over the surface of the earth, and 9.81 m/s^2 is only an approximate value. See standard gravity, Physical geodesy, Gravity anomaly and Gravity for further information.)
However, since Alice is at rest, the net force acting on her must be zero (otherwise Alice would be accelerating, according to Newton's second law). Since the net force is zero, the upward force exerted by the floor must exactly balance the downward force of gravity, meaning that Alice's actual weight and apparent weight are the same. (Here we are ignoring some minor effects such as buoyancy and centrifugal force, discussed later.)
If an object is accelerating upwards or downwards then its apparent weight respectively increases or decreases. Consider Alice again, but now in a lift accelerating downwards at 3 m/s^2 (metres per second per second). Let Fgravity be the downwards force on Alice due to gravity, Fnormal be the normal (upwards) force exerted by the floor of the lift, and Fnet be the net force on Alice.
It is essential to be clear about the signs of these forces and accelerations. Let us take downward forces and accelerations to be positive and upward forces and accelerations to be negative. This means that Fgravity is positive, Fnormal is negative, and Alice's acceleration is positive.
We know that Alice is accelerating at +3 m/s^2, and we know that her mass is 65 kg. Applying an equation we get from Newton's second law, we have that
Fnet = ma
= 65 kg × 3 m/s^2
= 195 N
However, this net force is the sum of the force due to gravity and the normal force. Thus,
Fgravity + Fnormal = Fnet
= 195 N
We already know from the earlier example that Fgravity = 637 N. Therefore
637 N + Fnormal = 195 N
so Fnormal = −442 N
Thus the normal force is 442 N (upwards), so Alice's apparent weight is 442 N, and it feels to Alice as if her weight has decreased by about 30% — even though her actual weight (the force exerted on her by gravity) remains unchanged. If Alice were standing on a weighing scale then the weight registered would also be 442 N. This is because a scale does not measure an object's actual weight, but rather measures the force that it exerts on the scale.
The workings above consolidate into the neat formula
Fnormal = m(a − g)
It is important to understand that it is acceleration, not velocity, that causes changes in apparent weight. In a lift travelling upwards or downards at any constant speed – however great – Alice's apparent weight will be the same as if the lift were at rest.
