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 2 of 2.
Kaushik Das said:
1 decade ago
m1a-T = T-m2a.
Therefore T = (m1+m2)a/2 = X(say).
Now if masses are halved, then tension = X/2 = T/2.
Therefore T = (m1+m2)a/2 = X(say).
Now if masses are halved, then tension = X/2 = T/2.
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.
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)
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)
GOURAV said:
8 years ago
Well explained @Arogya.
(1)
Salim Ansari said:
5 years ago
T = mg + ma.
(1)
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