nil1996 Messages 301 Reaction score 7 Thread starter Jun 25, 2013 #1 What is the relation between friction and gravity?Is friction independent of gravity?
genxium Messages 137 Reaction score 2 Jun 25, 2013 #2 Would you please make the statement more clear? If the question is involving high school physics problems, I would say that [itex]\text{friction is proportional to pressure not gravity, given a certain coefficient of friction} \; \mu.[/itex]
Would you please make the statement more clear? If the question is involving high school physics problems, I would say that [itex]\text{friction is proportional to pressure not gravity, given a certain coefficient of friction} \; \mu.[/itex]
CWatters Science Advisor Homework Helper Gold Member Messages 10,546 Reaction score 2,324 Jun 25, 2013 #3 Friction is pretty complicated. Several assumptions are usually made to try and simplify it. For example.. Friction is independent of velocity Friction is independent of surface area Friction is proportional to the normal reaction force (eg weight) The normal force is the mass * acceleration due to gravity so in short yes, the max friction force depends on gravity. EDIT: Astronauts using a treadmill in space have to be pulled down onto it using springs/elastic. Last edited: Jun 25, 2013
Friction is pretty complicated. Several assumptions are usually made to try and simplify it. For example.. Friction is independent of velocity Friction is independent of surface area Friction is proportional to the normal reaction force (eg weight) The normal force is the mass * acceleration due to gravity so in short yes, the max friction force depends on gravity. EDIT: Astronauts using a treadmill in space have to be pulled down onto it using springs/elastic.
CWatters Science Advisor Homework Helper Gold Member Messages 10,546 Reaction score 2,324 Jun 25, 2013 #4 More down here... http://hyperphysics.phy-astr.gsu.edu/hbase/frict3.html
nil1996 Messages 301 Reaction score 7 Jun 25, 2013 #5 thanks thanks for all your replies.You are helping me a lot.