Dogsled Question

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Hoping that someone on this forum can help me with this problem that has been stealing my sleep at night:rolleyes:

I'm trying to determine the effect of various coefficients of friction on the 'slideability' of a typical dogsled. So imagne we have three identical dogsleds - as you probably know a dogsled can be either a basket- or a toboggan style but essentially it sledes over the ice or snow on two runners. For the purpose of my question I'll establish the following parameters:
1 - Each (of the two) runners of each sled are in contact with the surface over a length of say 6feet long and 2" wide.
2 - The runners are spaced 18" apart (centre to centre) looking at them from the front of the sled
3 - The sled is travelling at a constant speed of exactly 10 mph
4 - The combined weight of the sled & the driver on it is say exactly 200lbs
5 - The surface on which it is running is for all intents and purposes icey Alaskan trails so more ice than snow - let;'s assume it's a reasonably flat grade and the surface relatively smooth
6 - Lets assume it's a constant 0 degree F and no headwind outside some drag caused by the forward motion
7 - The sled is being pulled by 10 sleddogs of identical speed, build, stamina and ability - not that it matters too much but let's say they each weigh 45lbs
8 - Do not be too concerned about the dog team composition, for the sake of this analysis you may assume the sled to be pulled by one constant pulling force (although it will have 40 furry feet:rofl:)


my comparitive parameters:
- Sled number one has runners made of mild steel, nothing fancy, just old fashioned black
steel
- Sled number two has runners made of wood, pick one but probably birch is a likely product being considered (can also be ash, hickory, or even red oak)
- Sled number three has its runners sheathed with UHMW plastic (coefficeint of friction of this material is aparently similar to Teflon)

SOOOOOOOOOOOOO,

my questions:
1 - Incorporating the coefficient of friction for the above 3 materials used on the runners what difference in force would be required to propel the sled(s) at the same constant 10mph - the answer would look something like "wood runners require 25% more force to maintain same speed.." or similar
2 - Alternatively If the same force is applied by the dog team what would the reduced relative difference in speed be for the 3 options
3 - Last question : What increase in force is typically required if the sled driver's weight is increased. Let's say the combined sled/driver weight is increased from 200lbs to 250lbs


thanks in advance guys!!!!!
 

Answers and Replies

kuruman
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1. Propelling the sled at constant speed means zero acceleration. This happens when the pulling force ##F_p## matches the force of kinetic friction ##f_k=\mu_k W## where ##W## is the weight of the sled and its contents. So the equation to look at is $$F_p=\mu_k W$$Note that the pulling force has a linear dependence on the coefficient of kinetic friction ##\mu_k##. If ##\mu_k## for steel on Alaskan snow is 25% less that birch wood on Alaskan snow then the pulling force ##F_p## needed to pull the sled at constant speed will be 25% smaller.

2. The value of the constant speed though is limited by the power that the dogs can develop. Say the maximum power the dogs can develop is ##P_{max}##. The power required to maintain the speed constant is the product of the pulling force and the speed. So assuming that the doggies are doing their very best$$P_{max}=F_p v~\rightarrow~v=\frac{P_{max}}{F_p}=\frac{P_{max}}{\mu_k W}.$$3. The required pulling force to move the sled at constant speed is proportional to the weight that this force is pulling. If the weight is increased by 25% from 200 lbs to 250 lbs, the pulling force must also increase by 25%. If the dogs are already running at maximum power with speed 10 mph, then the extra weight will reduce their speed to 8 mph according to the equation in 2 above.

I hope you will be able to get some sleep now after more than 10 years of sleeplessness. :oldsmile:
 
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