Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 2 (Q.No. 48)
48.
For the system of the loads shown in below figure, the time required for the 6.6 kg load to fall on the edge, is
Discussion:
4 comments Page 1 of 1.
Akram said:
1 year ago
First, we have to calculate the Acceleration which is;
a = g(m1-m2$)/(m1+m2).
Where $ is used to denote the coefficient of friction equal to 1/3.
a = 1m/s.
Use equation s = vt + 1/2at^2
Vi = 0.
t = 3 sec.
a = g(m1-m2$)/(m1+m2).
Where $ is used to denote the coefficient of friction equal to 1/3.
a = 1m/s.
Use equation s = vt + 1/2at^2
Vi = 0.
t = 3 sec.
(1)
Kush Kakkar said:
9 years ago
Right answer is 2 sec.
a = m2 x g / (m1+m2).
= 3.2 x 9.81 / (6.6+3.2),
= 3.203,
s = ut+0.5 at^2.
t = 1.18 ~ 2.
a = m2 x g / (m1+m2).
= 3.2 x 9.81 / (6.6+3.2),
= 3.203,
s = ut+0.5 at^2.
t = 1.18 ~ 2.
(1)
Muhammad Hashim said:
6 years ago
How to calculate initial speed of mass, in equation of motion?
Mostafa said:
6 years ago
Yes, the correct answer is 2 seconds.
(1)
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