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.
PRIYA said:
1 decade ago
We know that,
F=ma.
Therefore T=ma.
T is directly proportional to m.
Thus m is reduced to half hence T is also reduced to half.
F=ma.
Therefore T=ma.
T is directly proportional to m.
Thus m is reduced to half hence T is also reduced to half.
Dileepkumar said:
1 decade ago
Tension is nothing but with small difference is equal to the force which is equal to the product of mass and acceleration. The tension is directly proportional to the mass.
Malik said:
1 decade ago
If body is in equilibrium than it means weight = tension; as weight reduces to 50% of original. So it is obvious that tension becomes half.
Meriolen said:
1 decade ago
Looking at the figure if mass is reduced equally on both sides irrespective of percentage, the tension reduces by the same amount. Simple logic.
Sharda said:
1 decade ago
From the formula.
T=Mg.
We know simply tension (T) is directly proportional to the mass of body (M). And hence 50 percent reduction in the mass reduces the same amount of tension in the string. Hence tension will be half.
T=Mg.
We know simply tension (T) is directly proportional to the mass of body (M). And hence 50 percent reduction in the mass reduces the same amount of tension in the string. Hence tension will be half.
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