Question
Two blocks of mass $${M_1} = 20\,kg$$ and $${M_2} = 12\,kg$$ are connected by a metal rod of mass $$8\,kg.$$ The system is pulled vertically up by applying a force of $$480\,N$$ as shown. The tension at the mid-point of the rod is :
Two blocks of mass $${M_1} = 20\,kg$$ and $${M_2} = 12\,kg$$ are connected by a metal rod of mass $$8\,kg.$$ The system is pulled vertically up by applying a force of $$480\,N$$ as shown. The tension at the mid-point of the rod is :
A.
$$144\,N$$
B.
$$96\,N$$
C.
$$240\,N$$
D.
$$192\,N$$
Answer :
$$192\,N$$
Solution :
Acceleration produced in upward direction
$$\eqalign{ & a = \frac{F}{{{M_1} + {M_2} + {\text{Mass}}\,{\text{of}}\,{\text{metal}}\,{\text{rod}}}} \cr & = \frac{{480}}{{20 + 12 + 8}} = 12\,m{s^{ - 2}} \cr} $$
Tension at the mid point
$$\eqalign{ & T = \left( {{M_2} + \frac{{{\text{Mass}}\,{\text{of}}\,{\text{rod}}}}{2}} \right)a \cr & = \left( {12 + 4} \right) \times 12 = 192\,N \cr} $$
Acceleration produced in upward direction
$$\eqalign{ & a = \frac{F}{{{M_1} + {M_2} + {\text{Mass}}\,{\text{of}}\,{\text{metal}}\,{\text{rod}}}} \cr & = \frac{{480}}{{20 + 12 + 8}} = 12\,m{s^{ - 2}} \cr} $$
Tension at the mid point
$$\eqalign{ & T = \left( {{M_2} + \frac{{{\text{Mass}}\,{\text{of}}\,{\text{rod}}}}{2}} \right)a \cr & = \left( {12 + 4} \right) \times 12 = 192\,N \cr} $$
