Transcript HTPIB04F Friction

Friction
Friction - the force needed to drag one object across another.
(at a constant velocity)
Depends on:
•How hard the
surfaces are held
together
•What type of
surface it is (i.e.
rough, smooth)
Not supposed to depend on:
•Surface area (pressure)
•Speed (low speeds)
FFr = μFN
Force of Friction
in N
Coefficient of Friction.
0<μ<1
Normal Force - Force exerted
a surface to maintain its
integrity
Usually the weight (level surfaces)
Kinetic Friction - Force needed to keep it going at a constant
velocity.
FFr = μkFN
Always in opposition to velocity
Static Friction - Force needed to start motion.
FFr < μsFN
Keeps the object from moving if it can.
Only relevant when object is stationary.
Always in opposition to applied force.
Calculated value is a maximum
A 5.0 kg block of wood has a μs of 0.41, and a μk of .24 between
itself and the level floor.
a. Calculate the limit of static friction, and the kinetic friction.
5.0 kg
20.1105 ≈ 20. N, 11.772 ≈ 12 N
A 5.0 kg block of wood has a μs of 0.41, and a μk of .24 between
itself and the level floor. (20.1105 N, 11.772 N)
b. The block is at rest, and you exert a force of 15 N to the right to
try to make it move. Draw and label all the forces acting on the
block
5.0 kg
A 5.0 kg block of wood has a μs of 0.41, and a μk of .24 between
itself and the level floor. (20.1105 N, 11.772 N)
c. If the block is initially at rest, and we exert a force of 35 N to the
right, calculate the block’s acceleration.
5.0 kg
4.6456 ≈ 4.6 m/s/s
A 5.0 kg block of wood has a μs of 0.41, and a μk of .24 between itself and
the level floor. (20.1105 N, 11.772 N)
d. Kyle and Sally both exert a force on the block, and it accelerates from
rest to a 5.2 m/s over a distance of 10. m to the right. If Kyle exerts a
force of 25 N to the right, what force in what direction does Sally exert?
5.0 kg
-6.468 ≈ -6.5 N
Whiteboards:
Friction
1|2|3|4|5|6
TOC
What is the force needed to drag a 12 kg chunk of rubber at a
constant velocity across dry concrete?
F = ma, FFr = μkFN
FN = weight = mg = (12 kg)(9.8 N/kg) = 117.6 N
FFr = μkFN = (.8)(117.6 N) = 94.08 N = 90 N
90 N
What is the force needed to start a 150 kg cart sliding across wet
concrete from rest if the wheels are locked up?
F = ma, FFr < μsFN
FN = weight = mg = (150 kg)(9.8 N/kg) = 1470 N
FFr = μkFN = (.7)(1470) = 1029 N = 1000 N
1000 N
μs = .62, μk = .48
What is the acceleration if there is a force of 72 N in the
direction an 8.5 kg block is already sliding?
v
FFr
72 N
8.5 kg
FFr = μkFN, FN = mg, FFr = μkmg
FFr = (.48)(8.5 kg)(9.8 N/kg) = 39.984 N
F = ma
<72 N - 39.984 N> = (8.5 kg)a, a = 3.77 = 3.8 ms-2
3.8 m/s/s
μs = .62, μk = .48
A 22 kg block is sliding on a level surface initially at 12 m/s.
What time to stop?
FFr
v=12m/s
22 kg
FFr = μkFN, FN = mg, FFr = μkmg
FFr = (.48)(22 kg)(9.8 N/kg) = 103.488 N
F = ma
< -103.488 N> = (22 kg)a, a = -4.704 ms-2
v = u + at, v = 0, u = 12, a = -4.704 ms-2, t = 2.55 s = 2.6 s
2.6 s
μs = .62, μk = .48
A 6.5 kg box accelerates and moves to the right at 3.2
m/s/s, what force must be applied?
FFr
v
6.5 kg
FFr = μkFN, m = 6.5 kg
FFr = (.48)(6.5 kg)(9.8 N/kg) = 30.576 N
F = ma
< F - 30.576 N > = (6.5 kg)(3.2 ms-2), F = 51.376 = 51 N
51 N
a = 3.2 ms-2
F=?
μs = .62, μk = .48
A 22 kg block is sliding on a level surface initially at 12 m/s
stops in 2.1 seconds. What external force is acting on it besides
friction??
v=12m/s
FFr
F=?
22 kg
v = u + at, v = 0, u = 12 m/s, t = 2.1 s, a = -5.7143 ms-2
FFr = μkFN, FN = mg, FFr = μkmg
FFr = (.48)(22 kg)(9.8 N/kg) = 103.488 N
F = ma
< -103.488 N + F> = (22 kg)(-5.7143 ms-2), F = -22 N (left)
-22 N (to the left)
μs = .62, μk = .48
A force of 35 N in the direction of motion accelerates a
block at 1.2 m/s/s in the same direction What is the
mass of the block?
FFr
v
m
FFr = μkFN, FN = mg, FFr = μkmg
FFr = (.48)m(9.8 ms-2) = m(4.704 ms-2)
F = ma
< 35 N - m(4.704 ms-2) > = m(1.2 ms-2)
35 N = m(4.704 ms-2) + m(1.2 ms-2) = m(4.704 ms-2 + 1.2 ms-2)
m = (35 N)/(5.904 ms-2) = 5.928 kg = 5.9 kg
5.9 kg
a = 1.2 ms-2
35 N