Question

Bob traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25 m away, Fig. 4-65. Assume the rope can provide a tension force up to 29 KN before breaking, and use a “safety factor” of 10 (that is, the rope should only be required to undergo a tension force of 2.9 kN). (a) If Bob’s mass is 72.0 kg, determine the distance x that the rope must sag at a point halfway across if it is to be within its recommended safety range. (b) If the rope sags by only one-fourth the distance found in (a), determine the tension force in the rope. Will the rope break?

Figure 4-65 Problem 74

Short answer

(a) The rope must sag by a distance of 2.35 m at the mid-point.

(b) The tension force in the rope will be 7.55 kN, and the rope will not break.

Step-by-step solution

Step 1. Understanding the concept of a free body diagram

A free body diagram is a representation of all the forces acting on the body. By drawing a free body diagram of an object, it becomes easy to analyze its motion.

Step 2. Given data and assumptions

The mass of Bob,$m=72.0\mathrm{kg}$

The tension force at which the rope will break,$T_{\mathrm{critical}}=29\mathrm{kN}$

The advised maximum tension limit (safety factor),$T_{\mathrm{limit}}=2.9\mathrm{kN}$

The distance between both trees,$l=25\mathrm{m}$

Step 3. FBD of the point where Bob hangs

The FBD of the point where Bob hangs himself can be drawn as:

In the above figure, T represents the tension in the rope.

Step 4. Applying Newton’s second law on the point where Bob hangs

As the system is in equilibrium, the net force acting on the point where Bob hangs must be zero. Thus, applying Newton’s law second law in the vertical direction will give you:

$mg=2T\sin \theta$

Using the definition of the sine function, you can write the above equation as:

role="math" localid="1645678904919" $mg=2T\frac{x}{\sqrt{x^2+{\left(\frac{l}{2}\right)}^2}}\dots \left(i\right)$

Step 5. Calculating the value of x when T≤Tlimit

Let us take the limiting case when $T=T_{\mathrm{limit}}$. Substituting this value in equation (i), you get:

$72\times 9.8=2\times 2.9\times {10}^3\frac{x}{\sqrt{x^2+{\left(\frac{25}{2}\right)}^2}}$

Simplifying the above equation, you get:

$\frac{72\times 9.8}{2\times 2.9\times {10}^3}=\frac{x}{\sqrt{x^2+{\left(\frac{25}{2}\right)}^2}}$

Squaring both sides, you get:

$x^2=0.015\left(x^2+{\left(\frac{25}{2}\right)}^2\right)$

Solve the above equation for x.

$x=2.35\mathrm{m}$

Thus, the rope must sag by a distance of 2.35 m at the mid-point for the tension force to not exceed the limit of 2.9 kN.

Step 6. Calculating the value of tension when the rope sags by x4

Substitute $\frac{x}{4}$instead of x in equation (i).

$72\times 9.8=2T\frac{\left(\frac{2.34}{4}\right)}{\sqrt{{\left(\frac{2.34}{4}\right)}^2+{\left(\frac{25}{2}\right)}^2}}$

Solving the above equation for T, you get:

$\begin{array}{c}T=7546.71\mathrm{N}\\ =7.55\mathrm{kN}\end{array}$

As $T<T_{\mathrm{critical}},$the rope will not break.