91影视

When a force is exerted on a body by another body, the second body exerts a similar force in a different direction on the first object.

Given data:

The mass of each climber is $m=75\mathrm{kg}$.

The angle of inclination is $胃=31掳$.

Draw a free-body diagram.

Here, $F_{T_3}$is the tension force for the third climber, $F_{T_2}$is the tension force for the second climber, and $F_{T_1}$is the tension force for the first climber.

Applying the equilibrium condition along the horizontal direction for the third climber,

$\begin{array}{c}{F}_{T_3}-mg\sin 胃=0\\ F_{T_3}-\left(75\mathrm{kg}\right)\left(9.8\mathrm{m}{/\mathrm{s}}^2\right)\sin \left(31掳\right)=0\\ F_{T_3}=378.5\mathrm{N}\end{array}$.

Thus, the tension in the rope between the last two climbers is $378.5\mathrm{N}$.

In the equilibrium condition, the force $F_{T_2}$is equal to the force $F_{T_3}$. Therefore,

$\begin{array}{c}{F}_{T_2}=F_{T_3}\\ F_{T_2}=378.5\mathrm{N}\end{array}$

From the free-body diagram, the force exerted between the first two climbers can be calculated as

$\begin{array}{c}{F}_{T_1}-F_{T_2}-mg\sin 胃=0\\ F_{T_1}=F_{T_2}+mg\sin 胃\\ F_{T_1}=\left(378.5\mathrm{N}\right)+\left(75\mathrm{kg}\right)\left(9.8\mathrm{m}{/\mathrm{s}}^2\right)\sin \left(31掳\right)\\ F_{T_1}=757\mathrm{N}\end{array}$

Thus, the tension in the rope between the first and second climbers is $757\mathrm{N}$.