A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and minimum weights recorded is 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find a) the true weight of the person and b) the magnitude of the acceleration. Take g= 9.9 m/s2.
Answers
m1 given = 72kg
m2 given= 60kg
Therefore m1+m2/2 = 72+60/2 = 132/2 = 66kg
b) acceleration
66(g-a)= 60g
(g-a)= 60/66g
(g-a) = 10/11g
take 11 to that side
11g-11a= 10g
11g-10g= 11a
g= 11a
therefore, a= g/11
as g is given = 9.9m/s^2
thus, 9.9/11 = 0.9m/s^2
Answer:
The true weight of the person is 653.4 N
The magnitude of the acceleration is 0.9m/s²
Explanation:
Given weight = Wmax= 72kg
wmin= 60kg
Maximum weight will be shown when the elevator accelerates upwards.
Let N be the normal reaction on the person by the weighing machine.
So, from the FBD [free-body diagram of the person],
N=mg+ma …(1)
This is maximum weight, N = 72 × 9.9 N
When decelerating upwards, minimum weight will be recorded.
N’=mg+m(-a) …(2)
This is minimum weight, N‘ = 60 × 9.9 N
solving equations (1) and (2), we have:
2 mg = 1306.8
m=1306.8/2×9.9
=66 kg
So, the true mass of the man is 66 kg.
And true weight = 66 x9.9 = 653.4 N
(b) Using equation (1) to find the acceleration, we get:
mg + ma = 72 × 9.9
a=72×9.9-66×9.9/66
=9.9×6/66
=9.9/1
a=0.9 m/s²
∴the true weight of the person is 653.4 N
The magnitude of the acceleration is 0.9m/s²
