Download Calculating Kinetic Friction

Practice with Kinetic Friction
Renate Fiora
• Let’s try solving a problem involving kinetic
friction.
• Remember the equation for the force of
friction:
Ff = mFN
• Don’t forget to include a drawing and freebody diagram.
Bonnie and Clyde are sliding a 300. kg bank safe across the floor to their
getaway car. The safe slides with a constant speed if Clyde pushes from
behind with 385 N of force while Bonnie pulls forward on a rope with
350 N of force. What is the safe’s coefficient of kinetic friction on the
bank floor?
Try it on your own, then advance to the next slide
to see the solution.
Bonnie and Clyde are sliding a 300. kg bank safe across the floor to their
getaway car. The safe slides with a constant speed if Clyde pushes from
behind with 385 N of force while Bonnie pulls forward on a rope with
350 N of force. What is the safe’s coefficient of kinetic friction on the
bank floor?
FN
a=0
FB
FC
Ff
FW
+y
FN
Fnet = 0
FC
Ff
FB
FW
+x
Because there’s no net force, we can say:
SFy = FN – FW = 0  FN = FW = mg
SFx = FB + FC – Ff = 0  FB + FC = Ff
So,
m = Ff /FN = (FB + FC)/mg
= (350N + 385N)/(300kg)(9.8 m/s2)
m = 0.25
A man slides a 45 g crate at a constant velocity across a horizontal floor
by pulling on a rope attached to the crate. The angle  between the rope
and horizontal is 33 and the coefficient of kinetic friction between crate
and floor is 0.63. Determine the tension in the rope.
Try it on your own, then advance to the next slide
to see the solution.
A man slides a 45 kg crate at a constant velocity across a horizontal floor
by pulling on a rope attached to the crate. The angle  between the rope
and horizontal is 33 and the coefficient of kinetic friction between crate
and floor is 0.63. Determine the tension in the rope.
a=0
FN
FT
Ff
FW
+y
FN
FT
Fnet = 0

Ff
FW
+x
There’s no net force, so we can say:
SFy = FN + FTy – FW = 0

FN + FTy = FW = mg
FN = mg - FT sin 
SFx = FTx – Ff = 0 Now, plug in the values:
FT = 0.63(45)(9.8)/(cos
FT cos  = Ff
33 + 0.63 sin 33 )
So, Ff = mFN
FT = 240 N
FT cos  = mmg - mFT sin 
FT (cos  + m sin ) = mmg
FT = mmg/(cos  + m sin )