# Homework Help: Helping thief escape

1. Feb 26, 2016

### brycenrg

1. The problem statement, all variables and given/known data
A 75 kg thief wants to escape a building. He has a rope that supports 58kg how can he escape?

2. Relevant equations
f=ma

3. The attempt at a solution
75kg*g = W
58kg*g = T
T - W = 75kg*a
a = (t-w)/m
If he applies a acceleration upwards at 2.2 m/s^2 he would survive.
When i do the calculation I get a negative number, which would mean he would be applying a acceleration downwards which doesnt make sense. Why is that?
In my equation T is positive

2. Feb 26, 2016

### Staff: Mentor

Why would he want to climb upward to get out of a building? Shouldn't he lower himself down from the building?

3. Feb 26, 2016

### brycenrg

That makes sense he has to accell downwards to go. But what it means is if he accels downwards more that 2.2 then the rope wouldnt support him right?

4. Feb 26, 2016

### Staff: Mentor

I think you are confused. If he was just hanging from the rope, with no accleration, the rope would break, right? If he slid down the rope, just barely hanging onto it (which would burn his hands), he would be acclerating at just under $9.8 \frac m {sec^2}$. Would the rope break then?

5. Feb 26, 2016

### brycenrg

I am, I want to say if he used friction to slow him down to just = to T and under that force he would be ok. Which is under 2.2m/s^2.
But it seems like your saying he has to be going faster than gravity, which doesnt make sense to me

6. Feb 26, 2016

### Staff: Mentor

Yes, it doesn't make sense, and that isn't what I was saying, either.

The acceleration due to gravity is acting downward. What is the tension on the rope if his downward acceleration, relative to the fixed rope, is less than 2.2 m/sec^2? What's the tension on the rope if his downward acceleration, again relative to the fixed rope, is more than 2.2 m/sec^2? You're getting bogged down in the minutiae of the problem and are missing the bigger picture.