91Ó°ÊÓ

• Superhero’s mass = 90.0 kg.
• Trusty Sidekick’s mass = 55.0 kg.

• The mass of the rope is negligible.

Here, T’ is the tension in the upper rope above Superhero and T is the tension in the rope between the Superhero and Trusty Sidekick.

In the upper circle of the diagram, take the summation of forces in the vertical direction and equate it to zero as:

$\begin{array}{c}∑F_y=0\\ \mathbf{T}\mathbf{\text{'}}\mathbf{−}\mathbf{T}\mathbf{−}{\mathbf{w}}_{\mathbf{S}}\mathbf{=}\mathbf{0}\\ \mathbf{T}\mathbf{\text{'}}\mathbf{=}\mathbf{T}\mathbf{+}{\mathbf{w}}_{\mathbf{S}}\end{array}$

Here, w_S is the weight of the superhero.

Since the tension in the rope above Trusty Sidekick is due to its weight only, therefore

$\begin{array}{c}T\text{'}=w_T+w_S\\ T\text{'}=m_{T}g+m_{S}g\end{array}$

Substitute 90 kg for m_S and 55 kg for m_T in the above expression, and we get,

$\begin{array}{c}T\text{'}=90\text{ k²µ}×9.8{\text{ m/²õ}}^{\text{2}}+55\text{ k²µ}×9.8{\text{ m/²õ}}^{\text{2}}\\ =882\text{ k²µ}â‹\dots {\text{m/s}}^2+539\text{ k²µ}â‹\dots {\text{m/s}}^{\text{2}}\\ =\left(882+539\right)\text{ N}\\ =1421\text{ N}\end{array}$

Hence, the tension in the rope above the Superhero is 1421 N.

In the lower circle of the diagram, take the summation of forces in the vertical direction and equate it to zero as:

$\begin{array}{c}\mathbf{T}\mathbf{−}{\mathbf{w}}_{\mathbf{T}}\mathbf{=}\mathbf{0}\\ \mathbf{T}\mathbf{=}{\mathbf{m}}_{\mathbf{T}}\mathbf{g}\end{array}$

Substitute 55 kg for m_Tin the above expression, and we get,

$\begin{array}{c}T=55\text{ k²µ}×{\text{9.8m/s}}^2\\ =539\text{ k²µ}â‹\dots {\text{m/s}}^2\\ =539\text{ N}\end{array}$

Hence, the tension in the rope in the lower circle is 539 N.