Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 1 (Q.No. 43)
43.
A string of length 90 cm is fastened to two points A and B at the same level 60 cm apart. A ring weighing 120 g is slided on the string. A horizontal force P is applied to the ring such that it is in equilibrium vertically below B. The value of P is :
Discussion:
5 comments Page 1 of 1.
Akki said:
1 year ago
Right method, Thanks @Bosco.
Tee said:
2 years ago
Right method, Thanks @Bosco.
Bosco said:
7 years ago
Taking Moments at point B.
M1 * d1 = M2 * d2
m1 = 120g, d1 = 60cm d2 = 90cm.
Lengh of string. Hence subtitute 120g * 60cm = m2 * 90.
By solving m2 = 120 * 60/90 = > m = 80g. Correct answer.
M1 * d1 = M2 * d2
m1 = 120g, d1 = 60cm d2 = 90cm.
Lengh of string. Hence subtitute 120g * 60cm = m2 * 90.
By solving m2 = 120 * 60/90 = > m = 80g. Correct answer.
Dhyank said:
8 years ago
Why ? please explain.
Abdilahi khalif H.mohamoud said:
8 years ago
120*60/90=80.
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