Category: spring scales

I enjoy watching the pole-vaulters at the Olympics, especially Daly Thompson. Could you explain the physics of the pole vault for me? — ZG, Bullcreek, West Australia

The pole vault is all about energy and energy storage. Lifting a person upward takes energy because there is an energy associated with altitude—gravitational potential energy. Lifting a person 5 or 6 meters upward takes a considerable amount of energy and that energy has to come from somewhere. In the case of a pole-vaulter, most of the lifting energy comes from the pole. But the pole also had to get the energy from somewhere and that somewhere is the vaulter himself. Here is the story as it unfolds:

When the pole-vaulter stands ready to begin his jump, he is motionless on the ground and he has no kinetic energy (energy of motion), minimal gravitational potential energy (energy of height), and no elastic energy in his pole. All he has is chemical potential energy in his body, energy that he got by eating food. Now he begins to run down the path toward the jump. As he does so, he converts chemical potential energy into kinetic energy. By the time he plants his pole at the jump, his kinetic energy is quite large.

But once he plants the pole, the pole begins to bend. As it does, he slows down and his kinetic energy is partially transferred to the pole, where it becomes elastic potential energy. The pole then begins to lift the vaulter upward, returning its stored energy to him as gravitational potential energy. By the time the vaulter clears the bar, 5 or 6 meters above the ground, almost all of the energy in the situation is in the form of gravitational potential energy. The vaulter has only just enough kinetic energy to carry him past the bar before he falls. On his way down, his gravitational potential energy becomes kinetic energy and he hits the pit at high speed. The pit’s padding extracts his kinetic energy from him gently and converts that energy into thermal energy. This thermal energy then floats off into the air as heat.

One interesting point about jumping technique involves body shape. The vaulter bends his body as he passes over the bar so that his average height (his center of gravity) never actually gets above the bar. Since his gravitational potential energy depends on his average height, rather than the height of his highest part, this technique allows him to use less overall energy to clear the bar.

Is it true that gravity is stronger at the north pole than at the equator. Does that mean that a person would be able to jump higher at the equator?

Yes. Because of its rotation, the earth isn’t quite spherical and people near the poles of the earth experience stronger gravity than at the equator. At the equator, they would experience an apparent weight that was 1% less than at the poles and would be able to jump higher as a result. The Olympic committee should take note.

When you transfer momentum between two objects, why is it that the change in total momentum is 0?

Suppose you are standing motionless on extremely slippery ice. If you now take off your shoe and throw it northward as hard as you can, you will transfer momentum to it. Since you and your shoe were initially motionless, your combined momentum was 0. Neither of you nor the shoe was moving, so the product of mass times velocity was 0. But after you throw the shoe, both you and the shoe have momentum. Your momentum is equal to your mass times your velocity, so your momentum points in the direction you are going. The shoe also has momentum, equal to its mass times its velocity. But since it is heading in the opposite direction from you, it has the opposite momentum from you. Together, your combined momentum remains exactly 0—it didn’t change. In general, momentum is transferred from one object to another so that any change in momentum in one object is always compensated for by an opposite change in momentum in the other object.

How can you measure weight and/or mass through distance?

With a spring scale, the distortion of the spring is proportional to how much force it is exerting. If you measure that distortion, you can determine how hard it is pulling or pushing on whatever is attached to it. If it’s supporting the weight of an object, you can determine that object’s weight by measuring how far the spring distorts while supporting it.

If a spring scale measures weight, what does a mass scale use to figure out mass? Are weight and mass measured the same way?

A spring scale measures weight. It does this by reporting how much upward force it needs to exert on an object to keep that object from accelerating. Since this upward force exactly balances the object’s weight (assuming the object isn’t accelerating), the upward force reported by the scale is exactly equal to the object’s weight. If the scale reports that the object has a certain mass (in kilograms), then it is taking advantage of the fact that, near the earth’s surface, each kilogram of mass weighs 9.8 newtons. But it is still measuring weight and using the relationship between mass and weight to determine the object’s mass. If you were to move the “mass” scale to a new location, such as the moon’s surface, the scale would read incorrectly because the relationship between mass and weight would have changed.

If you hang a weight from a scale ten feet up and the weight descends 2 feet, is the loss in gravitational potential energy equal to the elastic potential energy gained?

Not quite. When you first let go of the weight, it falls freely because the spring isn’t stretched and doesn’t exert any upward force on the weight. The spring won’t support the weight fully until the weight has fallen 2 feet. By then, the weight has acquired a lot of kinetic energy and it overshoots the 2-foot level. The weight begins to bounce up and down around that 2 foot point and takes a while to settle down. The weight is vibrating up and down because it has too much energy at the 2-foot point. Eventually, it converts its extra energy into thermal energy and becomes motionless at the 2-foot point. At that point, it has turned exactly 1/2 of the missing gravitational potential energy into elastic potential energy and the other 1/2 into thermal energy. This 50/50 conversion is a remarkable result related to the exact proportionality between the spring’s distortion and the force it exerts.

If you lifted an object with a hanging scale on earth and it read 15 N, would it read the same on Jupiter? What about the gravitational force pulling the object down? Wouldn’t that alter the reading on the scale? Would you have to calibrate another scale to measure mass on Jupiter?

No, the scale would not read the same on Jupiter, and there would be nothing wrong with the scale! On Jupiter, the object would simply weigh more than on earth. Its mass wouldn’t have changed and it would still contain the same number of atoms, but Jupiter would pull on it harder. As a result, the scale would have to pull upward on it harder and the scale would read a larger number of newtons. You wouldn’t want to recalibrate the scale because it would be doing its job: it would correctly report that the object weighed about 40 N.