Transcript Aim: More friction

Aim: More friction
Do Now:
Why do objects at rest have more friction than
objects in motion?
A piece of steel whose mass is 100 kg is placed
on a steel beam as a force is applied to it
causing it to move with a constant velocity.
Draw a free-body diagram labeling all the forces
FN
FF
F
Fg
What is the weight?
Fg = mg
FN = 980 N
Fg = (100 kg)(9.8 m/s2)
Fg = 980 N
FF
F
What is the normal force?
FN = Fg
FN = 980 N
What is the net force?
0 N (velocity is constant)
Fg = 980 N
A new force of 700 N is applied. Does
steel move?
Solve for the maximum static force of
friction
FF = µsFN
FF = (0.74)(980N)
FF = 725.2 N
The steel does not move.
To get it to move, you must have a force
greater than 725.2 N
A car whose mass is 1,000 kg is accelerating at a
rate of 5 m/s2 with a force of 10,684 N on a surface
containing friction.
Draw and label a free-body diagram
FN
FF
F = 10,684 N
Fg
What is the weight?
FN = 9,800 N
Fg = mg
Fg = (1,000 kg)(9.8 m/s2)
Fg = 9,800 N
FF
F = 10,684 N
What is the normal force?
FN = Fg
FN = 9,800 N
What is the net force?
FNet = ma
FNet = (1,000 kg)(5 m/s2)
FNet = 5,000 N
Fg = 9,800 N
What is the force of friction?
FN = 9,800 N
FF = 5,684 N
F = 10,684 N
Fg = 9,800 N
FNet = F – FF
5,000 N = 10,684 N – FF
FF = 5,684 N
What is the coefficient of friction?
FF = µkFN
5,684 N = µk(9,800 N)
µk = 0.58
What two objects are in contact with one another?
Rubber on wet concrete
An 80-kilogram skier slides on waxed skis along a
horizontal surface of snow at constant velocity
while pushing with his poles. What is the
horizontal component of the force pushing him
forward?
F = Ff (constant velocity)
F = μkFN
F = μkFg (FN = Fg)
F = μkmg
F = (0.05)(80 kg)(9.8 m/s2)
F = 39.2 N
A car’s performance is tested on various
horizontal road surfaces. The brakes are
applied, causing the rubber tires of the car
to slide along the road without rolling. The
tires encounter the greatest force of
friction to stop the car on
(1)dry concrete
(2)wet concrete
(3)dry asphalt
(4)wet asphalt