Question

a rod of negligible mass having length l=2m is pivoted at its centre and two masses of m1=6 kg and m2=3 kg are hung from the ends. find tension in the supports to the blocks of mass 3 kg and 6kg

Answer 1

Given in the following question:

Length of the rod $\\ = 2m$

Mass of first block (m₁) $\\ = 6 kg$

Mass of second block(m₁) $\\ = 3 kg$

The tension due to first block (T₁) =

From equilibrium,

 $\\ T_{1} = m_{1} \times g\\\\ T_{1} = 6 \times 10\\\\ T_{1} = 6 0$

The tension due to second block (T₂) =

From equilibrium,

 $\\ T_{2} = m_{1} \times g\\\\ T_{2} = 3 \times 10\\\\ T_{2} = 3 0$

Hence, the tension would be 30 N and 60 N.