Q3E - University Physics with Modern Physics

Question

 A 75.0-kg wrecking ball hangs from a uniform, heavy-duty chain of mass 26.0 kg. 

(a) Find the maximum and minimum tensions in the chain. 

(b) What is the tension at a point three-fourths of the way up from the bottom of the chain?

Step-by-Step Solution

Answer

(a) The maximum and minimum tension in the chain is 990N and 735N.

(b) The tension in the chain at three fourth from the bottom is 926N.

Step 1: Tension in the chain

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.

Step 2: Determine the maximum and minimum tensions in the chain (a)

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,

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

The minimum tension in the chain is given as:

 $T_2=Mg$ 

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

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

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

Step 3: Determine the tension in the chain at three fourth from the bottom (b)

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 .