Question

a mountain climber, in the process of crossing between two cliffs by a rope, pauses to rest. she weighs 545 n. as the drawing shows, she is closer to the left cliff than to the right cliff, with the result that the tensions in the left and right sides of the rope are not the same. find the tensions in the rope to the left and to the right of the mountain climber.

$t_{l}=$
$t_{r}=$

Explanation:

Step1: Set up equilibrium equations

The climber is in equilibrium. In the vertical - direction, the sum of the vertical components of the tensions equals the weight of the climber. In the horizontal - direction, the sum of the horizontal components of the tensions is zero. Let $T_L$ be the tension in the left - hand side of the rope and $T_R$ be the tension in the right - hand side of the rope.
The vertical - force equation is $T_L\sin(65^{\circ})+T_R\sin(80^{\circ}) = 545$ N. The horizontal - force equation is $T_L\cos(65^{\circ})=T_R\cos(80^{\circ})$, so $T_L=\frac{\cos(80^{\circ})}{\cos(65^{\circ})}T_R$.

Step2: Substitute $T_L$ into the vertical - force equation

Substitute $T_L=\frac{\cos(80^{\circ})}{\cos(65^{\circ})}T_R$ into $T_L\sin(65^{\circ})+T_R\sin(80^{\circ}) = 545$ N.
We get $\frac{\cos(80^{\circ})}{\cos(65^{\circ})}T_R\sin(65^{\circ})+T_R\sin(80^{\circ}) = 545$ N.
Using the trigonometric identity $\frac{\cos(80^{\circ})\sin(65^{\circ})}{\cos(65^{\circ})}+ \sin(80^{\circ})=\tan(80^{\circ})\cos(80^{\circ})+\sin(80^{\circ})$.
$\frac{\cos(80^{\circ})\sin(65^{\circ})}{\cos(65^{\circ})}+ \sin(80^{\circ})=\frac{\cos(80^{\circ})\sin(65^{\circ})+\sin(80^{\circ})\cos(65^{\circ})}{\cos(65^{\circ})}=\frac{\sin(80^{\circ} + 65^{\circ})}{\cos(65^{\circ})}=\frac{\sin(145^{\circ})}{\cos(65^{\circ})}$.
$T_R=\frac{545}{\frac{\cos(80^{\circ})\sin(65^{\circ})}{\cos(65^{\circ})}+\sin(80^{\circ})}=\frac{545\cos(65^{\circ})}{\cos(80^{\circ})\sin(65^{\circ})+\sin(80^{\circ})\cos(65^{\circ})}=\frac{545\cos(65^{\circ})}{\sin(145^{\circ})}$.
$\cos(65^{\circ})\approx0.423$, $\sin(145^{\circ})\approx0.574$, $T_R=\frac{545\times0.423}{0.574}\approx400$ N.

Step3: Calculate $T_L$

Since $T_L=\frac{\cos(80^{\circ})}{\cos(65^{\circ})}T_R$, $\cos(80^{\circ})\approx0.174$, $\cos(65^{\circ})\approx0.423$, $T_L=\frac{0.174}{0.423}\times400\approx165$ N.

Answer:

$T_L = 165$ N
$T_R = 400$ N