91Ó°ÊÓ

Given Data:

The mass of the wrecking ball is $M=75\mathrm{kg}.$

The mass of the chain is $m=26\mathrm{kg}$.

The maximum tension in the chain is found by considering the mass of the chain with the attached wrecking ball in the heavy-duty chain, and the minimum tension by considering the mass of the wrecking ball only. The mass of the heavy-duty chain varies according to the length of the chain, so the three fourth of the weight is considered for the tension at three-fourths from the bottom.

The maximum tension in the chain is calculated by using the vertical equilibrium of the wrecking ball as:

${\mathit{T}}_{\mathbf{1}}\mathbf{=}\left(m+M\right)\mathit{g}$

Here $g$is the gravitational acceleration, and its value is $9.8\mathrm{m}/{\mathrm{s}}^2$, $m$is the mass of a heavy-duty chain and $M$is the mass of the wrecking ball.

Substitute all the values in the above equation, and we get,

$$\begin{gathered} T_1=\left(26\mathrm{kg}+75\mathrm{kg}\right)×\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ {T}_1=990\mathrm{N} \end{gathered}$$

The minimum tension in the chain is given as:

$T_2=Mg$

Substitute all the values in the above equation, and we get,

role="math" localid="1663745837985" $$\begin{gathered} T_2=\left(75\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ {T}_2=735\mathrm{N} \end{gathered}$$

Therefore, the maximum and minimum tension in the chain is 990 N and 735 N .

The tension in the chain at three fourth from the bottom is calculated as:

$$\begin{gathered} T=Mg+\frac{3}{4}mg \\ T=\left(M+\frac{3}{4}m\right)g \end{gathered}$$

Substitute all the values in the above equation, and we get,

$$\begin{gathered} T=\left(75\mathrm{kg}+\frac{3}{4}\left(26\mathrm{kg}\right)\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ T=926\mathrm{N} \end{gathered}$$

Therefore, the tension in the chain at three fourth from the bottom is 926 N .