Real and Apparent Weight (F.Sc – Physics – Chapter 5.11)
We often hear that objects appear to be weightless in a spaceship circling round the Earth. In order to examine the effect in some detail, let us first define, what do we mean by the weight?
The real weight of an object is the gravitational pull of the Earth on the object. Similarly the weight of an object on the surface of the moon is taken to be the gravitational pull of the Moon on the object.
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The moment you switch on your mobile phone, your location can be tracked immediately by global positioning system. |
Generally the weight of an object is measured by a spring balance. The force exerted by the object on the scale is equal to gravity on the object, i.e., the weight of the object. This is not always true, as will be explained a little later, so we call the reading of the scale as apparent weight.
To illustrate this point, let us consider the apparent weight of an object of mass m, suspended by a string and spring balance, in a lift as shown in Fig. 5.17 (a).
When the lift is at rest, Newton’s second law tells us that the acceleration of the object is zero; the resultant force on it is also zero. If w is the gravitational force acting on it and T is the tension in the string then we have,
T – W = ma
As a = 0
Hence, T = W ……….. (5.26)
This situation will remain as long as a = 0. The scale thus shows the real weight of the object. The weight of the object seems to a person in the lift to vary, depending on its motion.
When the lift is moving upwards with acceleration a, then
T – W = ma
Or T = W + ma ……….. (5.27)
The object will then weigh more than its real weight by an amount ma.
Now suppose, the lift and hence, the object is moving downwards with an acceleration a (Fig. 5.17 b),
then we have
W – T = ma
Which shows that
T = w – ma …………… (5.28)
The tension in the string, which is the scale reading, is less than w by an amount ma. To a person in the accelerating lift, the object appears to weigh less than w.
Its apparent weight is then (w – ma).
Let us now consider that the lift is falling freely under gravity. Then a=g, and hence
T = w – mg
As the weight w of the body is equal to mg so
T = mg – mg = 0
The apparent weight of the object will be shown by the scale to be zero.
It is understood from these considerations that apparent weight of the object is not equal to its true weight in an accelerating system. It is equal and opposite of the force required stopping it from falling in that frame of reference.
Do You Know? |
Your apparent weight differs from your true weight when the velocity of the elevator changes at the start and end of a ride, not during the rest of the ride when that velocity is constant. |