# Homework Help: Desperatly in need of HW help!

1. Dec 5, 2005

### avb203796

A rock-climbing group scales a cliff. One of the climbers has a mass of 80 kg, and the rope has a breaking tension that is 10% more than the tension of the rope when the climber is at a dead hang. What is the maximum upward accelaration at which the climber may be pulled up the cliff without the rope breaking? If he descends the cliff at an acceleration of 1.0m/s^2 what will be the tension of the rope?

2. Dec 5, 2005

### Astronuc

Staff Emeritus
Well in a gravitational field, the weight is given by mg, where m is the mass and g is the local acceleration of gravity.

When a rope is pulled with an acceleration, then the tension in the rope = m (g+a) where a is the acceleration imposed by the rope.

If the climber accelerates downward or decelerates, his effective weight is m(g-a).

3. Dec 6, 2005

### andrevdh

Tension in rope

The climber experiences an upwards force, T. The tension in the rope is the same magnitude but in the opposite opposite direction since they form an action-reaction pair - that is the rope experiences a similar force downwards as a result of supporting the climber. By taking the upwards direction as positive and using Newton's second law we have that the resultant force R that the climber experiences is given by:
$$R=T-W$$
therefore
$$ma=T-mg$$
in this equation you can still insert a sign for the acceleration of the climber depending on if he is going downwards or upwards.

Last edited: Nov 29, 2006
4. Dec 6, 2005

### avb203796

Ok so then the tension in the rope when the climber is at a dead hang would equal the cimbers weight because there would be no acceleration right? so therefore fo rhtis scenario the tension in the rope would be = mg=(80kg)(9.80m/s^2)=784kg*m/s^2
so then the breaking tension of the rope would be = 110%(784)=862.4?
Is this correct?

Last edited: Dec 6, 2005
5. Dec 6, 2005

### avb203796

Assuming what I have above is correct you would then set the equation T=m(g+a)=864.2 and then you would plug in and solve for "a" so you would get 1.2225m/s^2 for the maximum upward acceleration at whoch the climber may be pulled up without the rope breaking correct?

6. Dec 6, 2005

### avb203796

and finally you find the tension of the rope if the climber descends downward at an acceleration of 1.0m/s^2 by using the equation T= m(g-a) right? so therefore you would have T=m(g-a)=80kg(9.80m/s^2-1.0m/s^2)= 704 kg*m/s^2. Is this all correct?

7. Dec 6, 2005

### Astronuc

Staff Emeritus
Correct.

Check your numbers. The climber has mass 80 kg.

Also since Tbreak = 1.1 mg, then Tbreak/m = 1.1g = g+amax.

Correct.

8. Dec 6, 2005

### avb203796

The maximum upward accelaration would actually be 0.98 not 1.2225 right?

9. Dec 6, 2005

### Astronuc

Staff Emeritus
Correct.

And for future reference refer to andrevdh's post. It is a very nice description of the general problem.

10. Dec 6, 2005

### avb203796

Great thank you!