91影视

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.

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,$l=25\mathrm{m}$

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

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

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

$mg=2T\sin 胃$

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

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

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

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

Simplifying the above equation, you get:

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

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.

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

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

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.