Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 4 (Q.No. 29)
29.
Two loads of 50 kg and 75 kg are hung at the ends of a rope passing over a smooth pulley shown in below figure. The tension in the string is :
Discussion:
6 comments Page 1 of 1.
Bavin said:
4 years ago
Every explanations are helpful. Thank you all....
SHUBHAM SIDDHARTH NAYAK said:
5 years ago
Alternative method of solving:
50 kg moves upward hence
T = 50(g+a).
75 kg moves downwards hence
T = 75(g-a).
By solving we get;
a = g/5=2 m/sec2.
T = 50(10+2) = 600N = 60 kg.
50 kg moves upward hence
T = 50(g+a).
75 kg moves downwards hence
T = 75(g-a).
By solving we get;
a = g/5=2 m/sec2.
T = 50(10+2) = 600N = 60 kg.
(2)
Guriya said:
5 years ago
2m1*m2/m1+m2.
m1=50.
m2=75.
2(50)(75)/50+75.
=> 60.
m1=50.
m2=75.
2(50)(75)/50+75.
=> 60.
(1)
Nom said:
6 years ago
Right. Thanks @Pbhavik.
Pbhavik35 said:
6 years ago
2*(m1)*(m2)*g ÷ (m1+m2).
m1= 75 m2= 50 g=9.81.
Ans= 588.6N,
Convert 588.6/9.81 = 60.
m1= 75 m2= 50 g=9.81.
Ans= 588.6N,
Convert 588.6/9.81 = 60.
Rafikul said:
7 years ago
Please explain the answer in detail.
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