Forces of Friction
!
Forces of Friction
The term friction refers to the resistive
forces that arise to oppose the motion
of a body past another with which it is
in contact.
Example: You need to move a box of books into your
dormitory room. To do so, you attach a rope to the box
and pull on it with a force of 90.0 N at an angle of 30.0°.
Forces of Friction
The box of books has a mass of 20.0 kg, and the
coefficient of friction between the bottom of the box and
the hallway surface is 0.5. Find the acceleration of the
box.
n
Ff=!n
!
!
!
Where
Sliding friction (kinetic friction) is the
frictional resistance a body in motion
experiences.
! Static friction is the frictional
resistance a stationary body must
overcome in order to be set in motion.
!
! = coefficient of friction
n = normal force
T=90 N
Example: You need to move a box of books into your dormitory room. To do
so, you attach a rope to the box and pull on it with a force of 90.0 N at an
angle of 30.0°. The box of books has a mass of 20.0 kg, and the coefficient
of friction between the bottom of the box and the hallway surface is 0.500.
Find the acceleration of the box.
Example: A hockey puck is given an initial speed of
20.0 m/s on a frozen pond. The puck remains on the
ice and slides 120. m before coming to rest. Determine
the coeficient of friction between the puck and the ice.
n
n = 151 N
n
Fkf = !n = (0.5)(151 N)
!Fx = max = 90cos30°-75.5
T=90 N
Fkf
d=120 m
Fkf
30°
(20)(ax) = 90cos30°-75.5
ax = +0.122 m/s2
30°
W=mg
!Fy = 0 = n - 90sin30°-(20)(9.8)
Fkf = 75.5 N (to the left)
Fkf
30°
The magnitude of ! depends on the nature
of the surfaces in contact
NOTE: friction does NOT depend on the
area of contact!
W=mg
W=mg
1
Example: A hockey puck is given an initial speed of
20.0 m/s on a frozen pond. The puck remains on the
ice and slides 120. m before coming to rest. Determine
the coeficient of friction between the puck and the ice.
Example: A hockey puck is given an initial speed of
20.0 m/s on a frozen pond. The puck remains on the
ice and slides 120. m before coming to rest. Determine
the coeficient of friction between the puck and the ice.
n
vi = 20 m/s
Fkf
vf = 0 m/s
0 = (20)2 + 2(a)(120)
d = 120 m
a = -1.67 m/s2
Fkf
4.00 kg object:
!Fx = ma = T - F kf
T
4a = T - F kf
T
m1g
Fkf
7 kg
m2g
7 kg
n = 39.2 N
m2g
T = (7)(9.8) – (7)(a)
T = 68.6 – 7a
Eq 2
Fkf = !n
Fkf = (0.300)(39.2N)
Fkf = 11.8 N
T = 4a + Fkf
T = 4a + 11.8 Eq 1
Fkf
4 kg
T = 4a + 11.8 Eq 1
T
T = 68.6 – 7a
T
T = mg - ma
4a = T - F kf
Example: A 4.00-kg object is connected to a 7.00-kg object by a light
string that passes over a frictionless pulley. The coefficient of sliding
friction between the 4.00-kg object and the surface is 0.300. Find the
acceleration of the two objects and the tension in the string.
N
!Fy = ma = mg - T
T
n = 39.2 N
T
m1g
7.00 kg object:
T
4 kg
4.00 kg object:
T
Example: A 4.00-kg object is connected to a 7.00-kg object by a light
string that passes over a frictionless pulley. The coefficient of sliding
friction between the 4.00-kg object and the surface is 0.300. Find the
acceleration of the two objects and the tension in the string.
N
N
n = (4.00 kg)(9.8m/s2)
7 kg
! = 0.170
Example: A 4.00-kg object is connected to a 7.00-kg object by a light
string that passes over a frictionless pulley. The coefficient of sliding
friction between the 4.00-kg object and the surface is 0.300. Find the
acceleration of the two objects and the tension in the string.
!Fy = 0 = n – mg
m2g
m1g
(m)(-1.67) = -(9.8)(!)(m)
d=120 m
N
4 kg
!Fx = max = -F kf
W=mg
Example: A 4.00-kg object is connected to a 7.00-kg object by a light
string that passes over a frictionless pulley. The coefficient of sliding
friction between the 4.00-kg object and the surface is 0.300. Find the
acceleration of the two objects and the tension in the string.
Fkf
Thus, Fkf = !n =!mg
Fkf = !mg = 9.8!m
d=120 m
4 kg
n = W = mg
vf2 = vi2 + 2ad
W=mg
Fkf
!Fy = 0 = n - W
n
m1g
7 kg
m2g
Eq 2
4a + 11.8 = 68.6 – 7a
11a = 56.8
a = 5.16 m/s2
T = 32.5 N
(found by plugging a into Eq 1 or Eq 2)
2