Dot Physics

Sport Science, Pulling and Friction

Forgive me for all the posts on ESPN’s Sport Science (example: Pulling and Power). I can’t help myself.

In the short episode recently, Sport Science compared a football player pulling a sled with huge tires on it to a truck pulling stuff. I think their goal was to compare the power per kg from for the player and the truck to show how awesome humans are (and let me just say that humans ARE awesome). The problem was that they really didn’t give the truck a fair chance.

For the first test, they had Marshawn Lynch pull 585 pounds of stuff. The real question should be: how hard does he have to pull? They could have measured this with the tension in the cable used to pull the tires. Instead they hooked up sensors all over his body to make an animated skeleton move the same way he does.

The key to this pulling problem is friction. Let me draw a force diagram for the sled and for the player. Here is the sled moving at a constant speed:

i-39f057f3f2baafd4cebcf7382643d1f7-2010-02-09_untitled_1.jpg

I put the tension for angled up a bit because that is the way it was. However, I going to pretend it is parallel to the ground because it just doesn’t matter much. Here is the diagram for the football player.

i-74ae80d5159bd3db4c53c75881497189-2010-02-09_untitled_2.jpg

I drew those forces at the point that they are acting on the player – maybe it looks too confusing that way. But here are the important points.

Tsled on the player and Tdude on the sled are the same force (Newton’s third law). So. If you look at the player plus the sled AND they are moving at a constant speed, then the friction force on the player must be equal and opposite to the friction force on the sled. Here is a fairly normal model for sliding friction:

i-b65181a2fe47be30422048ce4dce53e2-2010-02-09_la_te_xi_t_1_1.jpg

This is just the magnitude of the friction force. It depends on how hard the two surfaces are pushing together (N) and some coefficient (mu). This coefficient depends on the two materials sliding (so for the player, it is the rubber of his shoes and the astro turf – for the sled it is something like steel and astro turf). The important point is that the contact forces (normal forces) on the two objects (the player and the sled) are different. Also the coefficients are different. If I assume that the tension is horizontal, then for the player the vertical forces must add up to zero. The same is true for the sled. This means that:

i-b47162af6ae46c5fe3e931780011fb3b-2010-02-09_la_te_xi_t_1_2.jpg

I hope you can figure out my notation – the “p” stands for the player and the “s” for the sled. If I use this with the model for friction AND assuming the two frictional forces are equal, then:

i-9db145050e3f1b8205d0ac147c3856ef-2010-02-09_la_te_xi_t_1_3.jpg

Maybe it will help if I assume that the coefficients of friction are the same (which would be like the football player wearing steel shoes). If that were true, he would not be able to pull the sled. The maximum frictional force that could be exerted on him would be smaller than the frictional force on the sled. The sled-player object would not be able to start moving.

Let me summarize this: when the player pulls, you have to consider the different coefficients of friction between the floor and the two objects. Now, one more thing – I said that it really doesn’t matter that the rope is not horizontal. Actually, it does a little. If the player is pulling up on the rope then his normal force (and thus his frictional force) would increase. The opposite would happen with the sled. So even if he has steel shoes, he could pull something more massive than him – this can be your homework assignment.

Truck pulling something

In the Sport Science show, they were trying to compare the power that Marshawn had to a truck. To make things fair, they wanted something similar to a person pulling tires – this was 2.6 times Marshawn’s body weight. They used a 6,700 pound truck. 2.6 times its weight is around 17,000 pounds. There, it’s fair – right? Ah ha! It is not fair. First, the truck was on asphalt and it was pulling some concrete barriers on asphalt. Check out this picture of the truck trying to pull this load:

i-e49b88d949dac1185247c2da9ea510ee-2010-02-09_0206101741_00jpg.jpg

Notice that only one wheel is slipping? That can’t be good (or fair). First, the truck is slipping. If it had better traction and couldn’t pull the load, that would be one thing. Also, even if both rear wheels were pulling, the normal force on these back tires is much smaller than the total (especially since most of the weight is in the front). This means that the resulting frictional force is smaller.

How should you calculate the power per kg for the truck?

Really, this is what they want – right? They state the truck has a 325 horsepower engine. Essentially, this is the power of the truck. One horsepower is equivalent to 746 watts. This means the output of the truck would be 2.4 x 105 Watts. Its power per kg would be about 80 Watts per kg.

Comments

  1. #1 Jean-Denis
    February 10, 2010

    Also, it might be worth mentioning that the static friction coefficient is larger than the dynamic friction coefficient, which means that it’s “harder” to get the sled moving than to keep it moving.

    Of course, that affects the truck and the human being to the same extent.

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