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Unformatted text preview: FRICTION
Direction of frictional forces ... (not always obvious) ... CHAPTER 5
APPLICATIONS OF NEWTON’S LAWS
o static friction
o kinetic friction
• Circular motion
o centripetal acceleration
o centripetal force
• roller coaster
• Drag forces
* You study using the notes provided. r
of the ground on the man r
of the ground on the crate Here’s the easy way to remember ... The frictional force is opposite to the direction
of the motion without friction.
• crate moves to right. ∴ f gc • without friction foot would slip to the
left. ∴ f gm r
frictional force r
f normal force
applied force r
w ( = mg) r
frictional force weight
w ( = mg) r
fk Forces acting ON the crate Forces acting ON the crate  Block moving:
 Block initially at rest:
A “normal force” N is always present when an object When the object is moving, the frictional force is now
given by: r
fk = µk N , rests on a surface. You can think of it as a reaction
force of the surface to the force of the object on the
surface. Note ∑ y Fy = N − mg = 0. ∴ N = mg .
The frictional force that has to be overcome is:
f s ⇒ µs N ,
where µ s is called the coefficient of static friction.
If F > f s the block will move. Therefore, a minimum
applied force ( = µ s N ) is required to start moving an
object over a surface. If F ≤ f s the net force on the
block is zero (since f = F ). normal force
applied force where µ k is the coefficient of kinetic friction. Thus,
the net force on the object is then
F − fk r
towards the right ( = ma ). Note, if a force F is applied,
the net force is zero, i.e., F = f , until F > f s ( = µ s N ).
fs = µs N f r
fk = µk N f =F F
at rest in motion Here is some experimental data:
Force sensor r
N Values of µ s and µ k depend on the two surfaces in normal force
applied force r
frictional force contact: B
2.0 Time (s)
0 2 No motion
(static ) At A:
i.e., 4 6 Motion at constant velocity
(kinetic) mg =5 ( 2 × 9.81) 0.9 0.4 0.04 0.04 1.0 0.8 0.30 0.25 0.10 0.05 8 F = µ sN = µ s mg
µs = F 0.4 Wax/snow C 0.5 Rubber/wet concrete 4.0 0.6 Brass/steel A 0.7 Rubber/dry concrete 6.0 µk Steel/steel weight
w ( = mg) µs Teflon/steel 2kg Materials Glass/glass r
F Note: µ s > µ k.
Note also that the NORMAL FORCE N is always perpendicular to the surface even if the surface is
= 0.255 From B to C: F = µ kN = µ k mg
µk = F
( 2 × 9.81) inclined: r
N • mg r
N = mg r
N • θ
N = mgcos θ Question 1: The coefficient of friction between the tires
DISCUSSION PROBLEM [5.1]:
The “friction” equation of a car and the road on a particular day is 0.70. What is
the steepest slope of the road on which the car is parked
if the car is not to slide down the hill? Note, the brakes r
f = µN fully on so the wheels are locked. is not a vector equation. Why not?
N θ y
w = mg Consider when the car is just about to move:
∴ ∑ Fx = 0 and ∑ Fy = 0. θ x
w = mg Along x: mg sin θ − f s = 0 ... ... (i) After the car starts to move we have an additional force
in the x-direction ( = ma x ) and we replace f s by f k: Along y: N − mgcos θ = 0 ... ... (ii) Then: But f s = µ s N. ∴ mg sin θ − µ s N = 0 . Substituting for N from (ii) we get:
mg sin θ − µ s mg cos θ = 0 . ∴ tan θ = µ s .
So, at this angle θ the car just begins to slide
If µ s = 0.70, then θ = tan −1 (0.70) = 35o.
Note: it is independent of the mass of the car and has
nothing to do with the “strength” of the brakes! Along x: mg sin θ − f k = ma x ... ... (iii) with f k = µ k N
Along y: N − mgcos θ = 0 (unchanged) So (iii) becomes: ma x = mg sin θ − µ k mg cos θ.
∴ a x = g(sin θ − µ k cos θ ) .
In this case (with µ k = 0.50 ): ∴ a x = 9.81 × (sin 35o − 0.50 cos 35o ) = 1.61 m/s2. We need to analyze two scenarios;  when m1 is about
to move up the slope, and  when m1 is about to move
down the slope. WHY??
 Assume the 4kg is just about to move up the slope:
N T x T
x fs Question 2: In the figure above, m1 = 4 kg and the
coefficient of static friction between the block and the m1g m2 g o
∑ 1 Fx = T + ( −f s ) + ( − m1g sin 30 ) = 0: f s = µ s N inclined surface is 0.40. (a) Find the range of possible o
∑1 Fy = N + ( − m1g cos 30 ) = 0 ... ... (1) values of m2 for which the blocks are stationary, i.e., in ∑2 Fy = T + ( − m2 g ) = 0 ... ... ... ... (2) equilibrium. (b) What is the frictional force on m1 if m2 = 1kg? But m1 = 4 kg and µ s = 0.40. So, from (1):
N = 34.0N.
∴ f s = 0.40 × 34.0 = 13.6 N, and T = f s + m1g sin 30o = 13.6 + 19.6 = 33.2 N ∴ m2 = T = 33.2
= 3.39 kg.
g  Assume the 4kg is just about to move down the slope:
N T x T fs
m2 g Now:
∑1 Fx = T + f s + ( − m1g sin 30 ) = 0 : f s = µ s N.
∑1 Fy = N + ( − m1g cos 30 ) = 0 unchanged. ∑2 Fy = T + ( − m2 g ) = 0 unchanged. But m1 = 4 kg and µ s = 0.40. So, we find:
N = 34.0N and f s = 0.40 × 34.0 = 13.6 N.
Also T = −f s + m1g sin 30o = −13.6 + 19.6 = 6.0 N.
∴ m2 = T = 6.0
g Thus, m1 remains stationary if 0.61kg < m2 < 3.39 kg. (b) If m2 = 1kg: T = 1 × 9.81 = 9.81N.
But T + f s + ( − m1g sin 30o ) = 0 .
∴ f s = m1g sin 30o − T = 19.62 − 9.81 = 9.81N, i.e., f s = T. m vo positive
direction l xo x
fk mg Question 3: A block on a horizontal surface is given an
initial velocity v o. It comes to rest after sliding a
distance l. Find an expression for the coefficient of
kinetic friction that only involves the given quantities. Find the acceleration. v2
We have 0 = v o2 + 2a ( x − x o ) ⇒ a = − o .
Also, since the only force in the x-direction is the kinetic
frictional force, that is the net force, so, by Newton’s
−f k = ma = − m o = − µ kN = − µ k mg
∴ µk = o .
2lg Wheels on surfaces: DISCUSSION PROBLEM [5.2]: Two possibilities: A B Two people (A and B) are tugging at each other. By
Newton’s 3rd law the force that A applies on B is equal
to the force that B applies on A . How come one of them
can pull the other off their feet? What is the most • rolling wheel - no slipping - STATIC FRICTION
• locked wheel - skidding - KINETIC FRICTION important factor? (Hint: consider only the forces acting
on each person.) Since µ k < µ s then f k < f s and so there is less frictional
force in a skid ⇒ you travel further!
Look at the web-site ⇒ anti-lock (abs) brakes CIRCULAR MOTION
P1 v( t)
r ( t) r • r
P2 v( t + ∆t) ∆θ r ( t + ∆t) O
r =r r
→ a r , and Limit ∆t→0
Now, Limit ∆t→0
v ( t) A
v ( t + ∆t ) B
C where a r is the instantaneous acceleration (towards the
center), and v is the speed, since ∆ r is the distance. ∴ar = v
r a r is called the centripetal (i.e., center-directed)
acceleration. The corresponding (radial) force on an
Shown here are the position and velocity vector diagrams object of mass m is given by Newton’s 2nd Law, of an object in uniform circular motion, i.e., motion at
constant speed (v) and constant radius (r). Since
r ( t )⊥v ( t ) and r ( t + ∆t )⊥v ( t + ∆t ), then ∠P1OP2 = ∠BAC = ∆θ, OP1 = OP2 and AB = AC .
So, ∆ABC and ∆P1OP2 are similar triangles. Now
∆ r ∆v
, i.e., r = .
∆v v ∆ r
Multiply both sides by
acceleration ( a r ∆v ) i.e., F = ma r = m v2
r where F is called the centripetal force. The direction of
this force is parallel to a r so it is always directed towards
the center of the circular motion. In the absence of the
centripetal force, the object would continue in a straight
line (1st Law). Examples of a centripetal force are the
action of the tension in a string spinning a stone and the
gravitational force of the Earth on the Moon. However, there is a difference between a horizontal and
a vertical orbit.  Vertical circular orbit  Horizontal circular orbit r
mg Top θ y T T r
v mg R x
mg Horizontal circular orbit
conical pendulum The weight force is always perpendicular to the plane of Example of a vertical
looping-the-loop Bottom r
mg the orbit and so cannot produce the centripetal force. It
is the x − component of the tension,
Tx = T sin θ = Fr , that produces the centripetal force. The weight force and normal force of the seat combine
to produce the centripetal force on the pilot. y 30o T 30o T 0.35m m x
v mg Question 5: A stone, of mass 0.75kg, is attached to a In the y − direction: T cos 30o + ( − mg ) = 0 ... (i) string and whirled in a horizontal circle of radius 35cm In the x − direction: T sin 30o = ma x ... ... (ii) (like a conical pendulum). If the string makes an angle
of 30o with the vertical, find the tension in the string, the
speed of the stone and the time for one revolution, i.e., centripetal force From earlier, the centripetal acceleration a x = the orbital period. From (i): T = mg
cos 30o From (ii): a x = = 8.50 N. T sin 30o
= 5.66 m/s2.
m ∴ v = ra x = 1.41m/s. Orbital time = 2πr v = 1.56s . v2
r Analysis of the forces acting on a pilot during a loop.
1. At the top (inside the loop) Since At the top the two forces
Nt there are two possibilities to consider: acting on the pilot are (1)
the normal force ( N t ) due r
Top to the seat, and (2) the
gravitational force due to the
Earth. In executing the mv 2
− mg ,
R • If mv 2
≥ mg , Nt > 0, i.e., the seat applies a force
R on the pilot, but it is less than his true weight (mg). By maneuver, the net downward Newton’s 3rd Law this is the same as the force exerted force produces the centripetal R by the pilot on the seat. Therefore, the pilot experiences force, i.e.,
Ftop = Nt + mg = mv 2
R So, the magnitude of the force of the seat on the pilot,
i.e., the nomal force, is:
Nt = an apparent weight (Nt) that is less than mg. • If mg > mv 2
, Nt < 0, the pilot will “fall” from the
R seat unless restrained by a seat-belt! (The seat can only
− mg .
R provide a positive force.) Analysis of the forces acting on the riders on a roller 2. At the bottom (inside the loop): coaster ride doing a loop: At the bottom, the free 3. At the top (inside the loop) body diagram tells us the
centripetal force is:
R mv 2
= Nb − mg .
R Bottom v So, the magnitude of the r
N mg force exerted by the seat
on the pilot is
mg Nb = mg + mv 2
r which is, clearly, greater than mg, the true weight of the
pilot. But, by the 3rd Law, Nb is also the magnitude of
the force the pilot on the seat, i.e., the apparent weight
of the pilot. Thus, his apparent weight is greater than his
true weight. R The analysis is the same as case #1, i.e., the centripetal
force experienced by the rider is
= N + mg ,
R and so the force applied by the seat on the rider is
R 4. At the top (outside the loop)
mg v R 2 So, if mg > mv
, N < 0, and the riders will “fall” from
R their seat (as a seat can only provide a positive force!) The free body diagram gives:
= mg − N,
R Consequently, there is a minimum speed ( v min ) for a
roller coaster to carry out this maneuver safely, i.e., v min = Rg . so the force applied by the seat on the rider is With a speed less than this value, the riders will fall N = mg − from their seats. (Note, it does not depend on the rider’s
mass.) mv 2
R Providing v < Rg , then N > 0 , so the rider remains in
their seat. However, if v > Rg , then N < 0 the rider
“flies” from their seat (unless restrained)! Thus, there is
a maximum speed for this maneuver to be safely
executed. z N fs N
15o 15o v 70m • r θ fs mg
(a ) mg Vertical
( b) (c) Figures (a) and (b) show the real life scenario; (c) is the Question 5: A curve in the road, with a radius of 70.0m, free body diagram of (a). Note that the frictional force f s
is a static frictional force as there is no relative motion is banked at an angle of 15o. If the coefficient of friction between the tire and the road, i.e., no skidding or wheel between the road and tires is 0.70, what is the maximum spin. Take radial and vertical components in (c). Note speed a car can make the corner without sliding? that the net radial force is the centripetal force. In (a),
2 Radial direction: Nsin15o + f s cos15o = mv r .
z-direction: N cos15o − mg − f s sin15o = 0 .
Re-arranging these equations we get:
2 N sin15o = mv r − f s cos15o ... ... (1)
N cos15o = mg + f s sin15o
But f s = µ s N = µ s mg cos15o. ... ... (2) DRAG FORCES
Substituting for f s and dividing eq. (1) by eq. (2) we get:
o mv 2 FD v o
r − µ s mg cos15 cos15 .
mg + µ s mg cos15o sin15o tan15 = ( ( ) fluid
resistance ) Different from “ordinary” friction because it depends on Note that m cancels, so the result is independent of the speed and type of “flow”: mass of the car! Inserting the numerical values we get
v2 ( ) 2
70 − 0.70 × 9.81 × (0.966)
9.81 + (0.70 × 9.81 × 0.966) × 0.259
2 i.e., v 70 = 3.089 + 6.407 = 9.496 . ∴ v = 25.8 m/s. • at low speed (non-turbulent or laminar flow)
v FD = kv FD • at high speed (turbulent flow) Show that:
(1) if θ = 0 and µ s = 0.70, then v = 21.9 m/s,
i.e., no banking but static friction.
(2) if θ = 15o and µ s = 0 , then v = 13.6 m/s,
i.e., banked corner but zero static friction. v FD FD = bv 2 The constants depend on shape, size, etc. For example:
Large b (large area) v
Small b (small area) Sky diving FD ( = bv 2 ) no drag
velocity (v) with drag w ( = mg) Velocity (m/s) terminal velocity Terminal velocity
~ 70 m/s
~ 155 mph
dt time =g FD ( = bv 2 ) Time (s)
~ 15-20 s acceleration (a)
g The downward acceleration of a sky diver is gradually no drag
with drag reduced because of the upward resistance force due to time drag (FD), which increases with velocity. The net force w ( = mg) mg − bv 2 = ma y . ∴a y = on the sky-diver is: mg − bv 2
m Fnet = w − FD ( = mg − bv 2 ) = ma , hence g ≥ a ≥ 0. When FD = w there is no net force so
a = 0 , and the terminal velocity is: vt = mg
b NOTE: this is not “free fall” ... WHY NOT ?? displacement (y) no drag From rest: 1
( y − yo ) = a y t 2
= g − v 2 t 2. time
m So, why does a ball fall faster than a feather?
... the full answer!
From earlier: v t ≈ mg
b • Ball has larger m and smaller b than the feather ∴ v t (ball) > v t (feather).
v t (ball) ALL objects
= g (no drag)
ball v t (feather) feather
tb The feather has a smaller terminal velocity than the ball
and so it reaches its terminal velocity in less time ( t f )
than the ball reaches its terminal velocity ( t b). Hence,
the average velocity of the ball is much greater than the
feather, so, if they are dropped from the same height the
ball hits the ground first! ...
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