Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 2 (Q.No. 26)
26.
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
Discussion:
6 comments Page 1 of 1.
Suresh said:
4 years ago
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.
R = m(g_a)
a = g/5
By putting the values we get;
R = 60(g-g/5).
R = 60(4g/5),
R = 48kg.
(1)
Sujoy said:
5 years ago
If lift was moving descending or downward,
T=W(1-f/g) where T=tension, W = weight, f= acceleration, g=gravity.
So, T = 60(1-g/5/g).
T=(60*4/5) = 48 kg.
T=W(1-f/g) where T=tension, W = weight, f= acceleration, g=gravity.
So, T = 60(1-g/5/g).
T=(60*4/5) = 48 kg.
Vidyashri Hiremath said:
7 years ago
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.
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.
Sufazo said:
9 years ago
Here, we can use G-g/5.
Ehsan said:
9 years ago
Nicely explained @Asif.
Asif Ali said:
9 years ago
At g = 60kg.
At g/5 => 60 - 60/5 = 60 - 12 = 48kg.
At g/5 => 60 - 60/5 = 60 - 12 = 48kg.
(1)
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