Question

b) Three blocks are connected as shown on a horizontal frictionless table, and pulled to the right
with a force of T3 = 60N

If M1 = 10 kg, M2 = 20 kg, M3 = 30kg.
prove T1/T2 = 1/3​

Answer 1

☆Concept:

▪︎the whole system will move together, so all the blocks will have acceleration = a

▪︎Fnet=Fdriving-Fopposing=ma

☆Solution:

Fnet=Fdriving-Fopposing=ma

For block $\sf{ m_{2}}$.

• $\sf{ T_{2} - T_{1} = m_{2}a}$

•$\sf{ T_{2} - T_{1} = 20a}$------(1)

For block $\sf{ m_{1}}$.

• $\sf{ T_{1} - 0 = m_{1}a}$

•$\sf{ T_{1} = 10a}$--------(2)

》Putting $\ T_{1}$ in equation (1)

▪︎$\sf{ T_{2} - T_{1} = 20a}$

▪︎$\sf{ T_{2} = 20a + T_{1}}$

▪︎$\sf{ T_{2} = 20a+ 10a}$

▪︎$\sf{ T_{2} = 30a}$

》$\bf{ T_{1} = 10a}$

》$\bf{ T_{2} = 30a}$

》$\bf{ \dfrac{ T_{1}}{ T_{2}} = \dfrac{1}{3}}$

Hence proved!