What is the acceleration and tension?

Question

You and a friend are glacier climbing. Your friend falls off an icy cliff! Luckily, the two of you are attached by a rope and your friend does not plummet into the abyss. Unluckily, you are attached by a rope and you start sliding off of the cliff! Assuming there is no friction between the rope and the icy cliff, determine the acceleration of you and your friend and the tension T in the rope. The tension and acceleration should be ONLY in terms of your mass (m1), your friend’s mass (m2) and acceleration due to gravity (g). But don’t worry, you have a pick axe and the outlook is not so grim (but the rescuing part is not included in this problem). Draw 2 free body diagrams.

So far I have a= (Fnorm+Fmg+FT(tenstion))/(m1+m2)

And tension, T= (m1m2)/(g(m1+m2))

Is that right or wrong?

Answer 1

I have no idea who posed this question, but it does critically depend on the characteristics of the rope. Let's assume that the ice is frictionless, so that this is a system in gravitatonal free-fall (with reduced g on account of slope). Then the center of gravity of the system falls at a steady acceleration by this reduced g. What happens to the climbers relative to their center of mass depends on the rope. Let's say the rope is infinitely non-stretchable. Then the falling climber will impart his momentum onto the resting climber exactly the same as if the resting climber was struck by his partner, and what happens does depend on the relative masses m1 & m2, as well as the speed the falling climber was going when the rope became taut. The tension on the rope would be the impulse function. On the other hand, if the rope was stretchy, then you have a complicated dynamic system which is solved by a differential equation, and the tension of the rope would gyrate. The acceleration of each of the climbers would be g' - T(t) / m1 and g' + T(t) / m2, at least at the beginning if climber 2 is still above climber 1, g' being the reduced acceleration due to gravity, and T(t) being the tension of the rope as a function of time. Note that the 2-climber system is falling at constant acceleration g', whatever is happening between the two of them cannot affect this constant acceleration of the center of gravity.