Friction
Friction - the force needed to drag one object across
another. (at a constant velocity)
(R)
Not supposed to depend on:
•Surface area (pressure)
•Speed (low speeds)
Depends on:
•How hard the
surfaces are held
together
•What type of
surface it is (I.e.
rough, smooth)
Ff = μR
Force of
Friction in N
Coefficient of
Friction.
0<μ<1
(Specific to a
surface) - in your
book (Table 4-2)
Reaction Force - Force
exerted by a surface to
maintain its integrity (FN)
(R)
Usually the weight (level surfaces)
IB calls kinetic
“dynamic”
These are approximate
What’s up with the two columns?
How are they related (one bigger = demo with force probe)
Dynamic Friction - Force needed to keep it going at a
constant velocity. (AKA Kinetic friction)
F F = μ dR
Always in opposition to velocity (direction it is
sliding)
(Demo, example calculation)
Static Friction - Force needed to start motion.
F F < μs R
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
(Demo, example calculation, examples of less than maximum)
A 5.00 kg block rests on a level table where there is a static
coefficient of friction of 0.470, and a dynamic of 0.170.
a) What are the dynamic and maximum static forces of friction?
b) if it is at rest and you exert a force of 12.0 N sideways on it
what happens?
c) if it is at rest and you exert a force of 35.0 N to the right on it,
what is the acceleration of the block?
d) If it is sliding to the right and you exert a force of 7.50 N to the
left, what is the acceleration of the block?
e) If it is sliding to the right, but decelerating at 0.950 m/s/s, what
force is acting on the block?
f) It is sliding to the right at 7.20 m/s and it slides to a halt in a
distance of 12.0 m. What other force besides friction is exerted
on the block as it slides to a halt?
a) 23.1 N, 8.34N; b) Nada; c) 5.33 m/s/s; d) -3.17 m/s/s, e) +3.59 N, f) -2.46 N
Whiteboards:
Friction
a-k
A 6.00 kg block rests on a level counter where the
static coefficient of friction is 0.620, and the dynamic
is 0.310.
a) Calculate the dynamic and maximum static friction
between the block and the counter. (Store these in your
calculator for later use)
6.00 kg
36.4932 ≈ 36.5 N, 18.2466 ≈ 18.2 N
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
b) if the block is initially at rest, and you exert a force of
30.0 N to the right, does the block move? Draw a force
diagram with four arrows – weight, R, applied force, and
static friction force. Use longer arrows for bigger forces,
indicate congruencies.
R = 58.9 N
Static Friction (-30.0 N)
6.00 kg
Applied (30.0 N)
Weight = -58.9 N
yup yup nothin to see here
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
c) Suppose the block is sliding to the right, and there is a
force of 25.0 N to the right, what is the acceleration of the
block?
6.00 kg
+1.13 m/s/s
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
d) If the block is sliding to the right, and there is a force of
12.0 N to the left, what is the acceleration of the block?
6.00 kg
-5.04 m/s/s
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
e) If the block is sliding to the left and there is a force of
45.0 N to the right, what is the acceleration?
6.00 kg
+10.5 m/s/s
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
f) What applied force would make the block slide to the
right and accelerate to the right at 5.60 m/s/s?
6.00 kg
+51.8 N
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
g) What force is being applied to the block if it is sliding to
the right but decelerating at 2.40 m/s/s?
6.00 kg
+3.85 N
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
h) What force is being applied to the block if it is sliding to
the left, but decelerating at 5.20 m/s/s?
6.00 kg
+13.0 N
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
i) Suppose the block is sliding freely to the right at 8.90
m/s. In what time will the block be brought to a halt?
6.00 kg
2.93 s
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
j) If the block begins to slide to the right, and there is a
force of 42.0 N acting to the right, what is the
displacement of the block when it reaches a speed of 15.0
m/s?
6.00 kg
+28.4 m
W
m = 6.00 kg, μs = 0.620, μd = 0.310 (36.4932, 18.2466)
k) If the block is sliding to the right at 9.00 m/s, and is
brought to a halt in a distance of 18.0 m, what force
besides friction acted horizontally as it was stopping?
6.00 kg
+4.75 N
W
What is the force needed to drag a 12
kg chunk of rubber at a constant
velocity across dry concrete?
F = ma, FFr = kFN
FN = weight = mg = (12 kg)(9.8 N/kg) = 117.6 N
FFr = kFN = (.8)(117.6 N) = 94.08 N = 90 N
90 N
W
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
W
μs = .62, μk = .48
A 22 kg block is sliding on a level surface
initially at 12 m/s. What time to stop?
v=12m/s
FFr
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
W
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?
v
FFr
a = 3.2 ms-2
F=?
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
W
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)
W
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?
v
FFr
m
a = 1.2 ms-2
35 N
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
W