-
Laws of Motion
Laws of Motion
APPARENT WEIGHT OF A MAN IN A LIFT/ELEVATOR AND ROCKET MOTION
Apparent weight of a man in a lift:
Suppose a person of mass m is standing on a weighing machine placed in an elevator / lift. The actual weight of the person = mg. This acts on the weighing machine which offers a reaction R given by the reading of the weighing machine. This reaction exerted by the surface of contact on the person is apparent weight of the person. We shall discuss how R is related to mg in the following different situations:
(i)When the elevator is at rest Acceleration of the person = 0
\(\therefore\) Net force on the person = 0
\(\therefore\) R-mg=0
or R = mg (FBD)
i.e apparent weight is equal to the actual weight of the person, -
Laws of Motion
ii) When the elevator is moving uniformly inthe upward or downward direction.
Again, In uniform motion, acceleration of the person=0. Proceeding as in case (i) R=mg
i.e., apparent weight = actual weight of the person -
Laws of Motion
(iii) When the elevator is moving upwards with an acceleration "a" or moving downwards with retardation a
R1 - mg = ma
R1 = mg + ma=m(g+a)
Thus, R1 > mg
Hence apparent weight of the person becomes more than the actual weight, when the elevator is accelerating upwards. Fractional change / Increase
in apparent weight \( \frac{{R_1 - mg}} {{mg}} = \frac{a} {g}\% \) Increase in apparent weight
\( \frac{{R_1 - mg}} {{mg}} \times 100 = \frac{a} {g} \times 100 \) -
Laws of Motion
(iv) When the elevator is moving downwards with an acceleration a or moving upwards with retardation a
\( mg - R_2 = ma \)
\( R_2 = mg - ma = m(g - a) \)
Thus R2<mg
Hence apparent weight of the person becomes less than the actual weight when the elevator is accelerating downwards.
Fractional change in apparent weight
\( \frac{{R_2 - mg}} {{mg}} = - \frac{a} {g} \)
% change in apparent weight = \( - \frac{a} {g} \times 100 \) -
Laws of Motion
(v) If the lift falls freely under gravity, a = g
\( mg - R_3 = mg \)
\( \therefore R_3 = m(g - g) = 0 \)
i.e apparent weight of the body becomes zero or the body becomes weightless
Note that weightlessness is felt only because the force of reaction between the person and the plane with which he is in contact vanishes.
Note : When downward acceleration is greater than g. ie a >g, then R=m(g-a) becomes negative. In that event, the person will loose contact from the floor of the lift and after some time the ceiling of the lift will hit head.
The tension in the cable used to carry the lift up or down is equal to the apparent weight of the lift with passengers, which is the system.
(vi) Tension in the Cable:
(i)If the lift is accelerating upwards or retarding downwards.
T-Mg = Ma
T=M (g+a)
M= Mass of the lift + passengers
Note: If breaking force of cable is given then minimum distance in which lift can be stopped can be obtained from
\( T_{breaking} = M(g + a_{\max } );S_{\min } = \frac{{u^2 }} {{2a_{max} }} \)
The minimum time in which the lift can be stopped is \( t_{\text{min}} = \frac{u} {{a_{max} }} \)
u = initial velocity of the lift
ii) If the lift is moving downward, with acceleration a or moving upward with retardation’ a’
Mg-T=Ma
T=M(g- a)
Note : A body is kept on the floor of a lift at rest. The lift starts descending at acceleration a
i) If a > g the displacement of the body in time t is \( \frac{1} {2}gt^2 \)
ii) If a <g the displacement of the body in time t is \( \frac{1} {2}at^2 \)