Download Aptitude Crack
img not found img not found

Applied Mechanics - Engineering

Q1:

A spring scale in a stationary lift shows a reading of 60 kg for a man standing on it. If the lift starts descending at an acceleration of g/5, the scale reading would be

A 48 kg

B 60 kg

C 72 kg

D none of these.

ANS:A - 48 kg

When lift moving up m(g+a)=force.
Scale reading =m(g+a)/g=10(g+g/3)/g.
When lift moving down.
Scale reading =m(g*g/5)/g.
=60(9.81-(9.81/5))/9.81.
= 48.256 ~ 48 kg. As R = ma_mg
R = m(g_a)
a = g/5
By putting the values we get;
R = 60(g-g/5).
R = 60(4g/5),
R = 48kg.