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Chapter 5: Problem 21

Two unfortunate climbers, roped together, are sliding freely down an icy mountainside. The upper climber (mass \(75 \mathrm{kg}\) ) is on a slope at \(12^{\circ}\) to the horizontal, but the lower climber (mass \(63 \mathrm{kg})\) has gone over the edge to a steeper slope at \(38^{\circ},(\mathrm{a})\) Assuming frictionless ice and a massless rope, what's the acceleration of the pair? (b) The upper climber manages to stop the slide with an ice ax. After the climbers have come to a complete stop, what force must the ax exert against the ice?

Short Answer

Expert verified
The acceleration of the pair is found using the total force divided by the total mass. For the force exerted by the ice ax, it should be equal and opposite to the combined force along the slope that is causing the climbers to slip.

Step by step solution

01

STEP 1: Identify the forces acting on both climbers

Let's consider the force due to gravity for both climbers. The force due to gravity can be decomposed along the slope and perpendicular to it. The force along the slope, that's causing climbers to slip, is \(F_{gs} = mg \cdot sin(\Theta)\) where m is mass of climber, g is acceleration due to gravity and \(\Theta\) is angle of slope with horizontal. The forces of both climbers will add up because lower climber is on steeper slope.
02

STEP 2: Calculate the total force and acceleration of the climbers

The net force causing the climbers to slip down is sum of forces due to gravity along the slope. The net force, using Newton's second law, is \(F = m_1 * g * sin(\Theta_1) + m_2 * g * sin(\Theta_2)\) where \(m_1, \Theta_1)\) are mass and angle of slope with horizontal of top climber and \(m_2, \Theta_2)\) are mass and angle for lower climber. The acceleration 'a' of the climbers together is given by \(a = F / (m_1 + m_2)\).
03

STEP 3: Determine the force exerted by the ice ax

To stop the climbing pair, the ice ax must provide a force equal to the sliding force but in opposite direction. So, the force 'F_ax' that the ax has to exert against the ice is \(F_ax = m_1 * g * sin(\Theta_1) + m_2 * g * sin(\Theta_2)\).

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