Q60P - University Physics with Modern Physics

Question

An adventurous archaeologist crosses between two rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope (Fig. P5.60). The rope will break if the tension in it exceeds $\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathit{N}$ , and our hero’s mass is 90.0kg . (a) If the angle $\mathit{\theta }$ is $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{°}$ , what is the tension in the rope? (b) What is the smallest value  can have if the rope is not to break?

                                 

Step-by-Step Solution

Answer

(a) The tension in the rope is 2541.78N .

(b) The smallest value $\theta$ can have if the rope is not to break is $0.974°$ .

Step 1: Identification of the given data

The given data is listed below as:

 

• The mass of the hero is, m = 90.0 kg
• The maximum tension the rope can contain is, $T=2.50\times {10}^{4}N$

Step 2: Significance of the tension

Tension is described as the pulling force that is transmitted in an axial direction through a cable or string. Moreover, tension is also the pair of some action-reaction forces that acts on each end of the pulley.

Step 3: (a) Determination of the tension of the rope

$\theta$The equation of the tension of the rope is expressed as:

 

$$\begin{gathered} 2T \sin \theta =m \\ T=\frac{mg}{2 \sin \theta } \end{gathered}$$

 

Here, T is the tension of the rope, m is the mass of the hero, g is the acceleration due to gravity and  is the angle subtended by the rope with the horizontal.

 

Substitute 90.0 kg for m , $9.8m/s^2$ for g and $10.0°$ for $\theta$ in the above equation.

 

$$\begin{gathered} T=\frac{\left(90.0kg\right)\left(9.8 m/s^2\right)}{2 \sin \left(10.0°\right)} \\ =\frac{882kg.m/s^2}{0.347} \\ =2541.78kg.m/s^2\times \left(\frac{1N}{1 kg.m/s^2}\right) \\ =2541.78 N \end{gathered}$$

 

Thus, the tension in the rope is 2541.78 N .

Step 4: (b) Determination of the smallest value of θ

The equation of the smallest value of the angle can be expressed as:

 

 $$\begin{gathered} 2T_1\sin \theta =mg \\ \theta =arc\sin \left(\frac{mg}{2T_1}\right) \end{gathered}$$

 

Here, $T_1$ is the maximum tension the rope can contain, m is the mass of the hero, g is the acceleration due to gravity and $\theta$ is the angle subtended by the rope with the horizontal.

 

Substitute $90.0kg$for m , $9.8m/s^2$ for g and $2.50\times {10}^{4}N$ for $T_1$ in the above equation.

 

$$\begin{gathered} \theta =arc\sin \left(\frac{\left(90.0kg\right)\left(9.8m/s^2\right)}{2\left(2.50\times {10}^{4}N\right)}\right) \\ =arc\sin \left(\frac{\left(882kg.m/s^2\right)}{\left(50000 N\right)}\right) \\ =arc\sin \left(0.017kg.m/s^2/N\times \frac{1N}{1kg.m/s^2}\right) \\ =arc\sin \left(0.017\right) \end{gathered}$$

 

Hence, further as:

 

$$\begin{gathered} \theta =arc\sin \left(0,017\right) \\ =0.974° \end{gathered}$$ 

 

Thus, the smallest value $\theta$ can have if the rope is not to break is $0.974°$ .