Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 21)
21.
If the tension in a cable supporting a lift moving upwards is twice the tension when the lift is movng downwards, the acceleration of the lift, is
Discussion:
6 comments Page 1 of 1.
Remya Raman said:
4 years ago
Upward tension = m(g+a)
here given downward tension = 2t = m(g-a).
t = 2m(g-a) = m(g+a)
= 2mg + ma = mg - ma
= mg = 3ma.
g = 3a.
a = g/3.
here given downward tension = 2t = m(g-a).
t = 2m(g-a) = m(g+a)
= 2mg + ma = mg - ma
= mg = 3ma.
g = 3a.
a = g/3.
(6)
Suneel kurmi said:
8 years ago
Tension in the cable.
Moving downward T= m(g-a).
Upward T=m(g+a).
B option is correct.
Moving downward T= m(g-a).
Upward T=m(g+a).
B option is correct.
(1)
Kiran said:
8 years ago
Yes, your explanation is correct @Amr.
Amr said:
1 decade ago
According to motion equation sum of forces = mass*acceleration (a).
In case going down force up is tension t and down is weight mass*g.
So m*g-t = m*a.
So t = m*g-m*a....(1).
In case of going up force up is 2*t and force down is m*g.
So 2*t-m*g = m*a.
So from 1 t = m*g-m*a then 2*m*g-2*m*a-m*g = m*a.
Then m*g-2*m*a = m*a.
So m*g = 3*m*a. So g = 3 a.
So a = g/3.
In case going down force up is tension t and down is weight mass*g.
So m*g-t = m*a.
So t = m*g-m*a....(1).
In case of going up force up is 2*t and force down is m*g.
So 2*t-m*g = m*a.
So from 1 t = m*g-m*a then 2*m*g-2*m*a-m*g = m*a.
Then m*g-2*m*a = m*a.
So m*g = 3*m*a. So g = 3 a.
So a = g/3.
Jay bermudo said:
1 decade ago
If you look closely, there are three cables in the problem. It says the word "lift" three times so the cable is three. So gravity divided by three is the answer.
M Zafar said:
1 decade ago
Any one explain how?
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