Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 17)
17.
If the masses of both the bodies, as shown in the below figure, are reduced to 50 percent, then tension in the string will be
Discussion:
16 comments Page 1 of 2.
Vip said:
7 years ago
Here, F or T is directly proportional to mass M, when M is reduced 50 % then Force tension is also reduced 50 % means half.
(4)
Arogya said:
1 decade ago
T = (m1*m2)/(m1+m2).
When m value is reduced to half.
New tension may be assumed as T1.
Now T1 = ((m1/2)*(m2/2))/((m1/2)+(m2/2)).
T1 = (2/4)(m1*m2)/(m1+m2) = (1/2)(m1*m2)/(m1+m2) = T/2.
Therefore new tension also reduced to half.
When m value is reduced to half.
New tension may be assumed as T1.
Now T1 = ((m1/2)*(m2/2))/((m1/2)+(m2/2)).
T1 = (2/4)(m1*m2)/(m1+m2) = (1/2)(m1*m2)/(m1+m2) = T/2.
Therefore new tension also reduced to half.
(3)
Paul said:
7 years ago
2T= (m'g+m"g).
Now 50% reduced from both:
2T* = .5(m'g+m"g) = .5X2T.
T* = .5T.
Now 50% reduced from both:
2T* = .5(m'g+m"g) = .5X2T.
T* = .5T.
(2)
Abhijit Sasmal said:
9 years ago
Tension in both side of the string will be equal because tension is directly proportional to the body mass. If reduce 50% in mass then tension will be half and equal to both side.
(1)
Imran Faizi said:
9 years ago
Tension is directly proportional to the mass of the body. If there is a certain amount of reduction in mass of the bodies then Tension (T) will also be reduced by the same amount. So it will be half.
(1)
GOURAV said:
8 years ago
Well explained @Arogya.
(1)
Salim Ansari said:
5 years ago
T = mg + ma.
(1)
Vinay BEL said:
5 years ago
Since T= 2(m1*m2/m1+m2) g.
If we put the value of mass half.
So, the answer is [A].
If we put the value of mass half.
So, the answer is [A].
(1)
Nivesh said:
1 decade ago
Since T= (m1*m2/m1+m2) g.
So, tension will be halved (T/2) when both m1 & m2 are halved.
So, tension will be halved (T/2) when both m1 & m2 are halved.
Anil kumar Dhaked said:
1 decade ago
T = (m1*m2/m1+m2).g
So,Tension force will be 50 percent of T(T*50/100)when both m1 or m2 are 50% that means halves.
So,Tension force will be 50 percent of T(T*50/100)when both m1 or m2 are 50% that means halves.
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