# Calculate Force Generated for Climbing Fall Factors

• redial
In summary, the conversation discusses a formula for calculating the force generated when an object of weight x is connected to a rope with a 5% stretch and free falls a length of the rope l (or half the length of the rope). The conversation also mentions the importance of knowing the amount of stretch in the rope per unit of force, as well as the time of deceleration and potential injury to the object. The final part of the conversation focuses on finding the force generated when a 100kg load is dropped from 1m above the anchor point, with the rope stretching 15% overall. The conversation ends with a discussion on the correct equation to use for calculating the maximum force of the rope's elasticity and the importance of knowing
redial
Can anyone help me with a formula to calculate the force generated, when and object weighing "x", connected to a rope with 5% , free falls the length of the rope "l", or half the length of the rope? I am looking for fall factors with respect to climbing. This may be fairly simple?

what's the "with 5%" about? the amount that the rope stretches per unit of force (like Hooke's law) is a needed parameter. then we can figger out what the time of decelleration is when the rope is pulled tight at the bottom of the fall. if the rope were steel cable that might not be expected to stretch measureably, the "object" hanging on it might break a few bones at the bottom of the fall.

Assume that I have a 1m rope attached to the load at one end and, to a fixed anchor at the other. And the stretch in the rope will be 15% overall including knots.

Does that help?

redial said:
Can anyone help me with a formula to calculate the force generated, when and object weighing "x", connected to a rope with 5% , free falls the length of the rope "l", or half the length of the rope? I am looking for fall factors with respect to climbing. This may be fairly simple?

>connected to a rope with 5%
What's that? Friction coefficient \mu=0.05?
>free falls the length of the rope "l", or half the length of the rope?
This doesn't make sense. For what you need the rope's length?
Climbing: You need overcome your weight (assume zero your acceleration) and the friction coefficient between your hands and the rope must be sufficient for no-slipping. (and, of course, the rope must supports your weight!)

Alright... Rope tied to a 100kg load and to an anchor point. Drop the load from the same height as the anchor, thus the fall is 1m, the rope stretches 15%. Also drop the load from 1m above the anchor, thus the fall is 2 m.

My question is How much force is generated?

Rope = 1m

Drop distance = 1m

Is it simply mass x gravity x length?

redial said:
Assume that I have a 1m rope attached to the load at one end and, to a fixed anchor at the other. And the stretch in the rope will be 15% overall including knots.

Does that help?

Aah, okey! But why you didn't say that it is an elastic pendulum?
So you have this pendulum starting from the rest in horizontal position (string tense) and because the gravity action it falls to the vertical position and strectching 15%? Please confirm.

Because I'm not a scientist... lol. just kidding.

Not exactly. the load drops vertically, coming to a fairly abrupt stop at the rope. THe construction of the rope allows for 15% stretch.

After thinking about the equation given in the link, it seems to be in error.

Instead, using the principle of conservation of mechanical energy we should have,

$$\Delta K + \Delta U_g + \Delta U_{rope} = 0$$

$$\Delta K = 0$$ since the object is stationary at the initial and final positions.

Now let H be the distance from the anchor point to the object at the initial position, d be the distance the object falls, and L be the total length of the rope.

$$\Delta U_g = mg \Delta y = mg(-2H - d)$$

and

$$\Delta U_{rope} = \frac{1}{2} kd^2$$

Now substituting these into the first equation we get,

$$\frac{1}{2} kd^2 - mgd - 2mgH = 0$$

$$d = \frac{mg + \sqrt{(mg)^2 + 4kmgH}}{k}$$

Note that only the principal root matters here.

Now since the system will follow Hooke's Law we have,

$$F_{max} = kd$$

Plugging in d to that gives,

$$F_{max} = mg + \sqrt{(mg)^2 + 4kmgH}$$

Which is almost the same as what the link gave, except this is correct.

The "springiness" factor can be described as $$k = \frac{ e_{rope}}{L}$$

Hence...

$$F_{max} = mg + \sqrt{(mg)^2 + 2e_{rope}mg \frac{2H}{L}}$$

(2H is the distance the object falls before the rope begins to stretch)

Hope that helps.

CS

BTW

$$e_{rope}$$ is the elasticity of the rope. I looked up a book value for it and a typical one is ~20 kN.

CS

redial said:
Because I'm not a scientist... lol. just kidding.

Not exactly. the load drops vertically, coming to a fairly abrupt stop at the rope. THe construction of the rope allows for 15% stretch.

Finally I could understand your problem, redial.
Let H<L be the distance between the fixed end point of the rope, of length L, and the weight tied by the another rope's end (so the rope is slack). The weight is released from the rest and falls freely a height L-H to the natural (not deformed) length L of the rope. The weight drops an additional height d during the deformation of the rope. Thus the total height that the weight drops is L-H+d. The corresponding potential energy is stored by the rope on the form of potential energy of deformation (assuming elastic this process), that is
U_d=1/2.k.d^2,
where d=0.15L is given. By conservation,
U_w=U_d,
or
mg(L-H+d)=1/2.k.d^2.
Notice that this equation is used to find k (the elastisc constant). Once you get k, the force due to deformation of the rope is
F_d=k.d.
(this is the force required to deform the rope 15% of its length; the additional potential energy due to height d, is also stored by the rope as energy of deformation).

perucho said:
Finally I could understand your problem, redial.
Let H<L be the distance between the fixed end point of the rope, of length L, and the weight tied by the another rope's end (so the rope is slack). The weight is released from the rest and falls freely a height L-H to the natural (not deformed) length L of the rope. The weight drops an additional height d during the deformation of the rope. Thus the total height that the weight drops is L-H+d. The corresponding potential energy is stored by the rope on the form of potential energy of deformation (assuming elastic this process), that is
U_d=1/2.k.d^2,
where d=0.15L is given. By conservation,
U_w=U_d,
or
mg(L-H+d)=1/2.k.d^2.
Notice that this equation is used to find k (the elastisc constant). Once you get k, the force due to deformation of the rope is
F_d=k.d.
(this is the force required to deform the rope 15% of its length; the additional potential energy due to height d, is also stored by the rope as energy of deformation).

You can raise the doubt whether the force is k.d or mg+k.d. The latter one is simply FALSE! Let's make a free body diagram of the weight at an arbitrary rope's deformation x (this one is taken positive downwards from the rope's bottom end at the natural length L of the rope). So upwards is the force F_x exerted by the rope on the weight and downwards is the proper weight mg. So by the Newton's second law
mg -F_x=m.d^2 x/dt^2.
Therefore
F_x=mg-m.d^2 x/dt^2.
Observe that the acceleration is negative (the weight stops!) and F_x takes into account the weight action and the dynamics (inerce) one as well. But by the Hooke's law
F_x=k.x

## 1. What is a fall factor?

The fall factor is a measure of the severity of a fall in rock climbing. It is calculated by dividing the length of the fall by the length of rope available to absorb the fall. A fall factor of 1 or higher is considered a high-impact fall.

## 2. How do you calculate the force generated for a climbing fall?

The force generated for a climbing fall can be calculated using the following formula: force = mass x acceleration. The mass is the weight of the climber, and the acceleration is the fall factor multiplied by the acceleration due to gravity (9.8 m/s^2).

## 3. Can the force generated for a climbing fall be reduced?

Yes, the force generated for a climbing fall can be reduced by using proper equipment and techniques. This includes using dynamic ropes, placing protection along the climbing route, and using proper belay techniques to minimize the length of a fall.

## 4. How does the fall factor affect the force generated for a climbing fall?

The fall factor has a direct impact on the force generated for a climbing fall. As the fall factor increases, the force generated also increases. This is because a higher fall factor means there is less rope available to absorb the energy of the fall, resulting in a greater impact force on the climber.

## 5. Why is it important to calculate the force generated for a climbing fall?

Calculating the force generated for a climbing fall is important because it helps climbers understand the potential impact of a fall and take appropriate safety measures. It also allows for the selection of appropriate equipment, such as ropes and protection, to ensure the safety of the climber.

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