Figure 5.33 shows an arrangement in which four disks are suspended by cords. The longer, top cord loops over a frictionless pulley and pulls with a force of magnitude 98 N on the wall to which it is attached. The tensions in the three shorter cords are${\mathit{T}}_{\mathbf{1}}$$\mathbf{=}\mathbf{58}\mathbf{.}\mathbf{8}\mathbf{N}\mathbf{,}{\mathit{T}}_{\mathbf{2}}\mathbf{=}\mathbf{49}\mathbf{.}\mathbf{0}\mathbf{N}$and ${\mathit{T}}_{\mathbf{3}\mathbf{}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{N}$. (a) What is the mass of disk A (b) What is the mass of disk B, (c) What is the mass of disk C, and (d) What is the mass of disk D?

1. The magnitude of force is, $F_{\mathrm{Pull}}=98\mathrm{N}$.
2. The tensions in three strings is,$T_1$$=58.8\mathrm{N}$,$T_2=49.0\mathrm{N}$,$T_3=9.8\mathrm{N}$ .
3. The acceleration due to gravity is,$g=9.8\mathrm{m}/{\mathrm{s}}^2$ .

The tension will be equal to the weight of the mass attached to that string. With this, we can find the masses of all disks. Since the system is in equilibrium, the net force on all the disks is zero.

Formula:

Weight of the block,$W=mg$ (i)

Where m is the mass of disk, g is the acceleration due to gravity.

Tension in the string balances the weight of the block, $T-w=0$ (ii)

As T₃ has attached to only disk D, so from equation (ii), we get

$T_3=m_{D}g$

Substitute the values in the above equation.

$9.8\mathrm{N}=m_D\times 9.8\mathrm{m}/{\mathrm{s}}^2$

$m_D=1.0\mathrm{kg}$$m_D=1.0\mathrm{kg}$

Hence, the value of mass of disk D is$1.0\mathrm{kg}$

As, T₂ has attached to disk C and D, so from equation (ii), we get

$$\begin{gathered} T_2=\left(m_c+m_D\right)g \\ 49\mathrm{N}=\left({\mathrm{m}}_{\mathrm{c}}+1\mathrm{kg}\right)\times 9.8\mathrm{m}/{\mathrm{s}}^2\left(\because m_{\mathit{D}}=1.0\mathrm{kg}\right) \\ \frac{49}{9.8}=m_c+1 \\ {m}_c+1=5 \\ {m}_c=4.0\mathrm{kg} \end{gathered}$$

As T₁ has attached to disk B, C and D, so from equation (ii), we get

$$\begin{gathered} T_1=\left(m_B+m_C+m_D\right)g \\ 58.8\mathrm{N}=\left({\mathrm{m}}_{\mathrm{B}}+4\mathrm{kg}+1\mathrm{kg}\right)9.8\mathrm{m}/{\mathrm{s}}^2\left(\because m_D=1.0\mathrm{kg}\&m_c=4.0\mathrm{kg}\right) \\ \left(m_B\mathit{+}\mathit{5}\right)=6 \\ {\mathrm{m}}_{\mathrm{B}}=1.0\mathrm{kg} \end{gathered}$$

Hence, the value of mass of disk B is$1.0\mathrm{kg}$

As the pulling force given to us is 98 N. So that would be equal to the sum of weights of all disks.

$$\begin{gathered} F_{pull}=\left(m_A+m_B+m_C+m_D\right)g \\ 98\mathrm{N}=\left(m_A+1\mathrm{kg}+4\mathrm{kg}+1\mathrm{kg}\right)\times 9.8\mathrm{m}/{\mathrm{s}}^2\left(\because m_B=1.0\mathrm{kg},m_D=1.0\mathrm{kg}\&{\mathrm{m}}_{\mathrm{C}}=4.0\mathrm{kg}\right) \\ (m_A+6)=10 \\ {m}_A=4.0\mathrm{kg} \\ \mathrm{The}\mathrm{mass}\mathrm{of}\mathrm{disk}\mathrm{A}\mathrm{is}4.0\mathrm{kg} \end{gathered}$$