110 likes | 215 Views
FINDING THE COEFFICIENT OF FRICTION. Also the well known “piece of cake” It really is easy. Trust me. Of course you will need a CALCULATOR!!!!!!. WE NEED A PROBLEM.
E N D
FINDING THE COEFFICIENT OF FRICTION • Also the well known “piece of cake” • It really is easy. • Trust me. Of course you will need a CALCULATOR!!!!!!
WE NEED A PROBLEM • An engine provides 5000N of force on a 1600 kg car moving at uniform speed. What is the coefficient of friction between the road and the tires? • Time to work an equation • Put it on your sheet.
HERE IT IS • First uniform tells us there is no acceleration. • So we have: • F= 5000 N • m = 1600kg • g = 9.8 • Coefficient = ?
So we put them together • Ff =mg x cof • Using algebra isolate cof • Now we have cof = Ff/m x g • In numbers cof = 5000 / (1600)(9.8) • Do the math in order!!!!! • Answer will be .3188….. • Round up .32
NOW WE GET COMPLICATED • Static Friction-(Mus) We know what this is. • Kinetic Friction(Muk) We know this. • So how do we work a problem? • SIMPLE!!!!
Like so: • We have a box that has a mass of 50kg. • If it takes 156.8 N of force to move it what is the static coefficient ? • First convert the mass ( 50 x 9.8) (490N) • Then we divide force needed (156.8) by force mass (490) 156.8/490 = .326….. Round to • 0.33
EASY NO? NOW FOR KINETIC FRICTION • Kinetic means moving. • You push a 100 kg rock down the road. • ( You are STRONG!!!) • If the coefficient of kinetic friction (Muk) is .25 between the road and rock how much force is needed to keep the rock moving? • First we find Fn for the rock • (100kg) ( 9.8 m/s^2) = 980N
NOW we plug in our formula (Muk) (Fn) = ? (.25) (980N) = 245N So it takes 245 N of force to keep the rock moving. Easy!!!!!!!!!!!!
Now you do it. • A car with a mass of 1500 kg needs a force of 2000 N to get it moving. What is the Mus? • .1360…………. (.14) • Andrea wants to move a 70 kg John. If the Mus is .20 how much force does she need? • 137.2 N • Simple?????
Last Ones • 60 kg Joe is moving down the hall, but looks like he wants to stop. If the Muk is .50 how much force must Coach Ahrens apply to keep him moving? • (60N) (9.8) x .50 = 294 N • 110 kg Colton is headed for the end zone. If Tristain has to apply 460 N of force to help him score what is the Muk? • 460/ (110 x 9.8) = .4267….. Round .43
OH NO • Review tomorrow • TEST THURSDAY!!!!!!!!!!!!!!!!