Two unfortunate climbers, roped together, are sliding freely down an icy mountai
ID: 1595145 • Letter: T
Question
Two unfortunate climbers, roped together, are sliding freely down an icy mountainside. The upper climber (mass 75 kg) is on a slope at 12 degree to the horizontal, but the lower climber (mass 62 kg) has gone over the edge to a steeper slope at 38 degree. Draw a free body diagram of each climber and clearly specify your choice of coordinate system and unit vector directions. Assuming frictionless ice and a massless rope, what is the acceleration of the pair? 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?Explanation / Answer
The climber are connected by the rope. So they experience the same acceleration. Set up separate force balance
for each climber introducing the (unknown) tension in the rope. It pulls up the lower climber and drags down the
upper climber in the corresponding balances. Then eliminate the tension and solve for the acceleration:
Let T be the tension in the rope.
the force balances are:
for the lower climber:
m_l·a = m_l·g·sin(38°) - T
T = m_l·g·sin(36°) - m_l·a
for the upper climber
m_u · a = m_u·g·sin(12°) + T
T = m_u·a - m_u·g·sin(12°)
Equate the two equations for T and solve for a:
m_l·g·sin(38°) - m_l·a = m_u·a - m_u·g·sin(12°)
a = g · (m_l·sin(38°) + m_u·sin(12°) ) / (m_l + m_u)
a = 9.81m/s^2· (62kg · sin(38°) + 75kg · sin(12°) ) / (62kg + 75kg)
a = 3.85 m/s^2
The upper climber made the axe exert a force equal to
F = ma
F = 75kg(3.85)N
F = 288.7 N
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