Deceleration of a falling load of a rope

In summary, the homework statement is trying to calculate the force an object experiences when falling due to gravity that is caught by a rope. The equations that were used were F=ma, v=0, u=1.57m/s, x=l*σ/E, and σ=F/A. In order to calculate the deceleration, x was needed. To calculate the stress on the rope, the equation for young's modulus was needed, but the problem was that the force was not known. A steel rope behaves just as really strong spring, provided you do not strain it beyond its yield point. So using E=0.5kx2 and relating k back to the Young's modulus
  • #1
tommy060289
22
0
Hey everyone

Homework Statement


Im trying to calculate the force encountered by an object falling due to gravity that is caught by a rope.

Obviously F=ma and since mass is constant I need to find the rate of deceleration. I've tried to using the equation v^2 = u^2 +2ax as v = 0 and u =1.57 m/s so I need to find x to calculate the deceleration.

In order to calculate s I was going to use the equation for young's modulus which is σ/ε and calculate the strain, As E = 200 GPa and I just need to find the stress on the rope, but here in lies my problem, as I can not figure out how to do this. I know the CSA as r = 0.003 m but I don't know what to use for my force value as surely the force is what I am trying to work out in the first place, and I don't think using mass * gravity is sufficient as the force must be greater as we have a velocity

Homework Equations



F = m*a

v^2 = u^2 + 2ax

x = l*σ/E

σ =F/A

The Attempt at a Solution



My attempt is described above but I just can't figure out what to use for F
 
Physics news on Phys.org
  • #2
tommy060289 said:
Hey everyone

Homework Statement


Im trying to calculate the force encountered by an object falling due to gravity that is caught by a rope.

Obviously F=ma and since mass is constant I need to find the rate of deceleration. I've tried to using the equation v^2 = u^2 +2ax as v = 0 and u =1.57 m/s so I need to find x to calculate the deceleration.

In order to calculate s I was going to use the equation for young's modulus which is σ/ε and calculate the strain, As E = 200 GPa and I just need to find the stress on the rope, but here in lies my problem, as I can not figure out how to do this. I know the CSA as r = 0.003 m but I don't know what to use for my force value as surely the force is what I am trying to work out in the first place, and I don't think using mass * gravity is sufficient as the force must be greater as we have a velocity

Homework Equations



F = m*a

v^2 = u^2 + 2ax

x = l*σ/E

σ =F/A

The Attempt at a Solution



My attempt is described above but I just can't figure out what to use for F

That equation you have quoted applies to constant acceleration.

imagine that instead of a rope, you had a spring. As the spring stretches under the load, the force gets stronger and stronger, meaning higher and higher acceleration, so a formula for a constant acceleration situation is hardly appropriate.

The formula could be used to find the average force - the constant force that would achieve the same event [stopping the falling mass].

You could just as easily use work energy considerations however.

Gravity has done work accelerating the mass as it falls, then the rope does work stopping the falling mass.
If the object fell 3 m, then its weight force times 3m is the amount of work done gaining energy [3mg].
If the object then stopped in 3 cm, clearly the force would have to be 100 times the weight force to do the same amount of work. That force however would not be constant, so 100mg is only the average force [100mg x 0.03 = 3mg; total work the same].
 
  • #3
But I'm still struggling going about calculating the distance in which the object is stopped. Is there anyway to calculate this. I could use young's modulus but again, the problem lies in calculating the stress?
 
  • #4
tommy060289 said:
But I'm still struggling going about calculating the distance in which the object is stopped. Is there anyway to calculate this. I could use young's modulus but again, the problem lies in calculating the stress?

Even a steel rope behaves just as really strong spring - provided you do not strain it beyond its yield point. so make use of E = 0.5kx2 and relate k back to the Young's modulus

You are not going to get a number answer, unless you know the mass, how far it fell and young's modulus of the catching rope/wire, and how thick and long it was to begin with.
Of course you could just use M for mass, h for how far it fell, g for gravity, γ for Young's modulus, etc. and derive an algebraic solution.
 
  • #5
thanks peter, ill try that!
 

1. What is the deceleration of a falling load of a rope?

The deceleration of a falling load of a rope refers to the rate at which the speed of the load decreases as it falls due to the friction and resistance of the rope.

2. How is the deceleration of a falling load of a rope calculated?

The deceleration of a falling load of a rope can be calculated by dividing the change in speed by the time it takes for the load to fall. This can be represented by the formula: a = (vf - vi) / t, where a is the deceleration, vf is the final velocity, vi is the initial velocity, and t is the time.

3. What factors affect the deceleration of a falling load of a rope?

The deceleration of a falling load of a rope can be affected by various factors such as the weight and size of the load, the type and condition of the rope, the angle at which the rope is being pulled, and the presence of any external forces such as wind resistance.

4. How does the length of the rope impact the deceleration of a falling load?

The length of the rope can impact the deceleration of a falling load by increasing the amount of friction and resistance that the load experiences. A longer rope means more surface area for the load to rub against, resulting in a slower deceleration.

5. Can the deceleration of a falling load of a rope be controlled?

Yes, the deceleration of a falling load of a rope can be controlled by adjusting various factors such as the type and condition of the rope, the angle at which the rope is being pulled, and the presence of any external forces. Properly maintaining and using the rope can also help to control the deceleration of the load.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
755
  • Introductory Physics Homework Help
Replies
4
Views
232
  • Introductory Physics Homework Help
Replies
6
Views
705
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
2
Replies
39
Views
2K
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
619
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
3K
Back
Top