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.
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)
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.
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.
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.
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.
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.
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