Tag: ramps

I teach a class on safety helmets (hard hats) and had a question about one of their specifications. The manufacturer rates their crown impact energy level at 40 foot-pounds. Would this be equivalent to taking an object that weighs 20 pounds and dropping it 2 feet onto a hard hat? – AH

Assuming that the wearer doesn’t let the helmet move and that the object that hits the helmet is rigid, my answer is approximately yes. If a 20-pound rigid object hits the hat from a height of 2 feet, that object will transfer just over 40 foot-pounds of energy to the helmet in the process of coming to a complete stop. The “just over” has to do with the object’s continued downward motion as it dents the hat and the resulting release of additional gravitational potential energy. Also, the need for a rigid dropped object lies in a softer object’s ability to absorb part of the impact energy itself; a dropped 20-pound sack of flour will cause less damage than a dropped 20-pound anvil.

However, the true meaning of the “40 foot-pound” specification is that the safety helmet is capable of absorbing 40 foot-pounds of energy during an impact on its crown. This energy is transferred to the helmet by doing work on it: by pushing its crown downward as the crown dents downward. The product of the downward force on the crown times the distance the crown moves downward gives the total work done on the helmet and this product must not exceed 40 foot-pounds or the helmet may fail to protect the wearer. Since the denting force typically changes as the helmet dents, this varying force must be accounted for in calculating the total work done on the helmet. While I’m not particularly familiar with safety helmets, I know that bicycle helmets don’t promise to be useable after absorbing their rated energies. Bicycle helmets contain energy-absorbing foam that crushes permanently during severe impacts so that they can’t be used again. Some safety helmets may behave similarly.

Finally, an object dropped from a certain height acquires an energy of motion (kinetic energy) equal to its weight times the height from which it was dropped. As long as that dropped object isn’t too heavy and the helmet it hits dents without moving overall, the object’s entire kinetic energy will be transferred to the helmet. That means that a 20-pound object dropped from 2 feet on the helmet will deposit 40 found-pounds of energy in the helmet. But if the wearer lets the helmet move downward overall, some of the falling object’s energy will go into the wearer rather than the helmet and the helmet will tolerate the impact easily. On the other hand, if the dropped object is too heavy, the extra gravitational potential energy released as it dents the helmet downward will increase the energy transferred to the helmet. Thus a 4000-pound object dropped just 1/100th of a foot will transfer much more than 40 foot-pounds of energy to the helmet.

My teacher said that if you lift a 5 pound sack, you are doing work but if you carry the sack, you aren’t doing any work. Why is that?

When you lift the sack, you are pushing it upward (to support its weight) and it is moving upward. Since the force you exert on the sack and the distance it is traveling are in the same direction, you are doing work on the sack. As a result, the sack’s energy is increasing, as evidenced by the fact that it is becoming more and more dangerous to a dog sitting beneath it.

But when you carry the sack horizontally at a steady pace, the upward force you exert on the sack and the horizontal distance it travels are at right angles to one another. You don’t do any work on the sack in that case. The evidence here is that the sack doesn’t become any more dangerous; its speed doesn’t increase and neither does its altitude. It just shifts from one place to an equivalent one to its side.

Please explain the forces that allow one team to win a Tug-O-War contest. — ES

If we neglect the mass of the rope, the two teams always exert equal forces on one another. That’s simply an example of Newton’s third law—for every force team A exerts on team B, there is an equal but oppositely directed force exerted by team B on team A. While it might seem that these two forces on the two teams should always balance in some way so that the teams never move, that isn’t the case. Each team remains still or accelerates in response to the total forces on that team alone, and not on the teams as a pair. When you consider the acceleration of team A, you must ignore all the forces on team B, even though one of those forces on team B is caused by team A. There are two important forces on team A: (1) the pull from team B and (2) a force of friction from the ground. That force of friction approximately cancels the pull from the team B because the two forces are in opposite horizontal directions. As long as the two forces truly cancel, team A won’t accelerate. But if team A doesn’t obtain enough friction from the ground, it will begin to accelerate toward team B. The winning team is the one that obtains more friction from the ground than it needs and accelerates away from the other team. The losing team is the one that obtains too little friction from the ground and accelerates toward the other team.

I don’t understand work done without any acceleration. Since F=ma and a=0, F=0 and thus W=0.

You are merging two equations out of context. The force you exert on an object can be non-zero without causing that object to accelerate. For example, if someone else is pushing back on the object, the object may not accelerate. If the object moves away from you as you push on it, then you’ll be doing work on the object even though it’s not accelerating. The only context in which you can merge those two equations (Force=mass x acceleration and Work=Force x distance) is when you are exerting the only force on the object. In that case, your force is the one that determines the object’s acceleration and your force is the one involved in doing work. In that special case, if the object doesn’t accelerate, then you do no work because you exert no force on the object! If someone else is pushing the object, then the force causing it to accelerate is the net force and not just your force on the object. As you can see, there are many forces around and you have to be careful tacking formulae together without thinking carefully about the context in which they exist.

What is thermal energy?

While we ordinarily associate energy with an object’s overall movement or position or shape, the individual atoms and molecules within the object can also have their own separate portions of energy. Thermal energy is the energy associated with the motions and positions of the individual atoms within the object. While an object may be sitting still, its atoms and molecules are always jittering about, so they have kinetic energies. When they push against one another during a bounce, they also have potential energies. These internal energies, while hard to see, are thermal energy.

When a rubber ball bounces or rebounds, does the weight of the ball determine how many times it bounces?

Each time the ball bounces, it rises to a height that is a certain fraction of its height before that bounce. The ratio of these two heights is the fraction of the ball’s energy that is stored and returned during the bounce. A very elastic ball will return about 90% of its energy after a bounce, returning to 90% of its original height after a bounce. A relatively non-elastic ball may only return about 20% of its energy and bounce to only 20% of its original height. It is this energy efficiency that determines how many times a ball bounces. The missing energy is usually converted into thermal energy within the ball’s internal structure.

When an egg falls and hits the table, the table pushes up on it, doesn’t it? The same with a bouncing ball?

Yes, when a falling object hits a table, the table pushes up on the falling object. What happens from then on depends on the object’s characteristics. The egg shatters as the table pushes on it and the ball bounces back upward.

You discussed how an egg doesn’t bounce because it doesn’t have time and instead it breaks. Why, then, does a mouse ball (in a computer mouse) or a bowling ball not bounce? It doesn’t break, so why doesn’t the support force make it bounce back upward. Does this relate to elasticity?

Actually, both a mouse ball and a bowling ball will bounce somewhat if you drop them on a suitably hard surface. It does have to do with elasticity. During the impact, the ball’s surface dents and the force that dents the ball does work on the ball—the force on the ball’s surface is inward and the ball’s surface moves inward. Energy is thus being invested in the ball’s surface. What the ball does with this energy depends on the ball. If the ball is an egg, the denting shatters the egg and the energy is wasted in the process of scrambling the egg’s innards. But in virtually any normal ball, some or most of the work done on the ball’s surface is stored in the elastic forces within the ball—this elastic potential energy, like all potential energies, is stored in forces. This stored energy allows the surface to undent and do work on other things in the process. During the rebound, the ball’s surface undents. Although it’s a little tricky to follow the exact flow of energy during the rebound, the elastic potential energy in the dented ball becomes kinetic energy in the rebounding ball. But even the best balls waste some of the energy involved in denting their surfaces. That’s why balls never bounce perfectly and never return to their original heights when dropped on a hard, stationary surface. Some balls are better than others at storing and returning this energy, so they bounce better than others.

If there is an upward force on the egg when it hits the table, why doesn’t it bounce upward?

The enormous upward force on the egg when it hits the table does cause the egg to accelerate upward briefly. The egg loses all of its downward velocity during this upward acceleration. But the egg breaks before it has a chance to acquire any upward velocity and, having broken, it wastes all of its energy ripping itself apart into a mess. If the egg had survived the impact and stored its energy, it probably would have bounced, at least a little. But the upward force from the table diminished abruptly when the egg broke and the egg never began to head upward for a real bounce.

What effects do forces acting on an object which are not from the same pair have on one another? i.e. the force pulling the egg downward and the potential force of the table? Are they equal upon impact and there a pair?

Different forces acting on a single object are not official pairs; not the pairs associated with Newton’s third law of action-reaction. While it is possible for an object to experience two different forces that happen to be exactly equal in magnitude (amount) but opposite in direction, that doesn’t have to be the case. When an egg falls and hits a table, the egg’s downward weight and the table’s upward support force on the egg are equal in magnitude only for a fleeting instant during the collision. That’s because the table’s support force starts at zero while the egg is falling and then increases rapidly as the egg begins to push against the table’s surface. For just an instant the table pushes upward on the egg with a force equal in magnitude to the egg’s weight. But the upward support force continues to increase in strength and eventually pushes a hole in the egg’s bottom.