Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 6 (Q.No. 17)
17.
Two blocks A and B of masses 150 kg and 50 kg respectively are connected by means of a string as shown in the below figure. The tension in all the three strings __________ be same.
Discussion:
13 comments Page 1 of 2.
Joew said:
5 years ago
T can be equal for balanced pulley. Nowhere it's mentioned.
Vijay said:
6 years ago
If it will at rest then T will be equal.
Supratip said:
6 years ago
@Subhendu.
If there are more than 2 pully, will the formula work?
If there are more than 2 pully, will the formula work?
SUBHENDU MUKHERJEE said:
6 years ago
Use the formula, 2M1M2/M1+M2= 2 * 150 * 50/150 + 50 = 100 * 150/200 = 75.
So, the tension in 1st string from the right side is 75. Which is the same in the other two as 150/2 =75.
So 75 n tension in each string.
So, the tension in 1st string from the right side is 75. Which is the same in the other two as 150/2 =75.
So 75 n tension in each string.
(1)
Sukhdeep Singh said:
7 years ago
The system is not in equilibrium at this point of time. We will not apply an extra 25 kg of effort at 50KG block end. To make the system in equibrium. Both Ta and Tb will be 75 Kg then. This type of system is 2nd pulley system, where the mechanical advantage will be 2 and the effort will reduce to half to become 75 Kg.
Sajin s said:
7 years ago
Here, Tension[T] Is equal.
Prasanth b said:
7 years ago
If Tb=50Kg and Ta=75Kg, how come that all T's are the same?
Justin James said:
7 years ago
By using pulley the weight reduce to half and the same on B equalizes the weight.
Shibu s said:
7 years ago
Please explain the answer.
Vivek said:
7 years ago
If Tb=50Kg and Ta=75Kg, how come that all T's are the same?
How can it same? It is different tension.
How can it same? It is different tension.
(1)
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