Question
Question: Two blocks of mass \({M_1} = 20kg\) and \({M_2} = 12kg\) are connected by a metal rod of mass \(8kg\...
Two blocks of mass M1=20kg and M2=12kg are connected by a metal rod of mass 8kg. The system is pulled vertically up by applying a force of 480N as shown. The tension at the midpoint of the rod is
(A) 144
(B) 96
(C) 240
(D) 192
Solution
We will first find the net acceleration of the entire system by applying Newton’s 2nd Law on the whole rigid body. To find tension at the midpoint of the rod, we divide the whole system into two parts, splitting the rod, so that our first system becomes M1 with half the rod, and the second system is the other half of the rod attached to M2. Next we use Newton’s 2nd Law to balance forces on the systems and find equations with tension at the mid-point of the rod T as a variable. Using the previously calculated value of net acceleration of the system, we find the tension at the midpoint of the rod T.
Step by step solution: We know the acceleration of the system is a, where it can be found by applying Newton’s 2nd Law on the system,
480N−(M1+8+M2)g=(M1+8+M2)a
⇒480N=(M1+8+M2)(a+g)
⇒480=(20+8+12)(a+10)
⇒40480=(a+10)
⇒a=2m/s2
There, if the whole body is accelerating at a=2m/s2, then the upper half will also be accelerating at the same value of a
Let us assume the Tension as T at the midpoint of the rod.
The net force on the first system is 480N−T−(M1+4)g=(M1+4)a
Since the entire body is moving as a whole rigid mass with the same acceleration, here we will use the previously calculated value of a=2m/s2,
Therefore, since the upper half of the rod is of mass 4kg, the equation becomes,
⇒480−T−(M1+4)g=(M1+4)2
⇒480−T−(20+4)(10)=(20+4)2
⇒480−(20+4)12=T
⇒T=192N
Hence option (D) is correct
Note: Since the entire structure is a rigid body, the net acceleration of the system will be assumed to be uniform and the same both when the entire structure is being considered as a whole and also when the upper half of the rod and M1 is considered.