Why do wrecking balls amplify damage done?

  • Thread starter Thread starter tahayassen
  • Start date Start date
  • Tags Tags
    Balls Damage
AI Thread Summary
Wrecking balls amplify damage due to the combination of kinetic and gravitational potential energy generated during their swing, making them more effective than direct mechanical action. For smaller buildings, direct mechanical methods can suffice, but larger structures require the significant force that a swinging wrecking ball or explosives can provide. A calculation showed that a 1-ton ball moving at 15 ft per second can generate about 90,000 lbs of force when stopping within an inch, highlighting the concentrated impact. This force is challenging to replicate with machines at greater heights. The discussion clarifies the physics behind the effectiveness of wrecking balls in demolition.
tahayassen
Messages
269
Reaction score
1

Homework Statement



A wrecking ball is commonly used to destroy old buildings. The energy comes from a motor pulling a large pendulum back, which then allows it to swing against a wall. Why not just use the motor to do the same work on the wall?

2. The attempt at a solution

The kinetic energy AND the gravitational potential energy is used to break the wall with a pendulum whereas a motor will only have kinetic energy.
 
Physics news on Phys.org
For small buildings direct mechanical action is used, see,

http://www.youtube.com/watch?v=I9FP7Y2Ig1Y&feature=related

For large and stronger structures use the ball or explosives, see,

http://www.youtube.com/watch?v=SJkdYuzViqo&feature=related

Calculate the peak force generated when a 1 ton ball moving at 15 ft per second stops within 1 inch. If I did it right I get about 90,000 lbs. force concentrated over a relatively small area. It would be difficult to apply such a force many stories up with machines we have.
 
Spinnor said:
For small buildings direct mechanical action is used, see,

http://www.youtube.com/watch?v=I9FP7Y2Ig1Y&feature=related

For large and stronger structures use the ball or explosives, see,

http://www.youtube.com/watch?v=SJkdYuzViqo&feature=related

Calculate the peak force generated when a 1 ton ball moving at 15 ft per second stops within 1 inch. If I did it right I get about 90,000 lbs. force concentrated over a relatively small area. It would be difficult to apply such a force many stories up with machines we have.

I thought force is mass times acceleration - not mass times speed.

edit: Never mind. I get it now. Thanks.
 
Last edited:
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top