Question

Three mountain climbers who are roped together in a line are ascending an icefield inclined at 29° to the horizontal (Fig. 4–67). The last climber slips, pulling the second climber off his feet. The first climber is able to hold them both. If each climber has a mass of 75 kg, calculate the tension in each of the two sections of rope between the three climbers. Ignore friction between the ice and the fallen climbers.
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Accepted Answer

Identify the forces acting on each climber. For each climber, the forces include their weight (gravitational force), the tension in the rope, and the normal force from the ice surface. Since the ice is frictionless, there is no frictional force to consider. Break the forces into components parallel and perpendicular to the incline. The weight of each climber can be expressed as $$ F_g = m \cdot g $$, where $$ m  is $$the mass of the climber and $$ g  is $$the acceleration due to gravity. The component of the weight parallel to the incline is $$ F_{g, 	ext{parallel}} = m \cdot g \cdot \sin(	heta) $$, and the component perpendicular to the incline is $$ F_{g, 	ext{perpendicular}} = m \cdot g \cdot \cos(	heta) . $$Analyze the forces acting on the last climber (Climber 3). The tension in the rope connecting Climber 3 to Climber 2 must balance the parallel component of Climber 3's weight. Thus, $$ T_2 = m \cdot g \cdot \sin(	heta) . $$Analyze the forces acting on the second climber (Climber 2). Climber 2 is being pulled by Climber 3's weight and is also pulling Climber 1. The tension in the rope connecting Climber 2 to Climber 1 must balance the combined parallel components of the weights of Climbers 2 and 3. Thus, $$ T_1 = 2 \cdot m \cdot g \cdot \sin(	heta) . $$Substitute the known values into the equations. Use $$ m = 75 \ 	ext{kg} $$, $$ g = 9.8 \ 	ext{m/s}^2 $$, and $$ 	heta = 29^\circ  to $$calculate the tensions $$ T_1 $$ and $$ T_2 . $$Ensure that the trigonometric functions are applied correctly to find the numerical values of the tensions.

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