Answer:
a) The acceleration of the pair is 9.81 m/s2.
b) The force exerted by the ax against the ice is equal to the sum of the weights of the two climbers, 145 kg, multiplied by the acceleration of the pair, 9.81 m/s2. Therefore, the ax must exert a force of 1425.45 N against the ice.
Hi, I've been stuck on this for days! I'd love if I can get some help.
Two unfortunate climbers, roped together, are sliding freely down an icy mountainside. The upper climber (mass 80kg ) is on a slope at 16∘ to the horizontal, but the lower climber (mass 65kg ) has gone over the edge to a steeper slope at 32∘.
a)Assuming frictionless ice and a massless rope, what is 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?
What I've done:
a)Let T be tension
a: acceleration
First Climber
F=W+Tsin(16)
F=ma
W+Tsin(16)=ma
Tsin(16)=80a-(80*9.81)....eqn1
Second climber
F=W-Tsin(32)
F=ma
W-Tsin(32)=ma
rearranging:
Tsin(32)=(65*9.81)-65a....eqn2
eqn1/eqn2:(algebraic rearranging)
where a=9.81? (not sure if correct)
b)I'm not too sure how to go about this.
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