08 newton's law of motion
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Newton’s Laws of Motion
Topic 4 (cont.)
Lecture Outline
• Newton’s Law Involving Friction
Newton’s Laws Involving Friction• Friction is always present when two solid
surfaces slide along each other.– Even the smoothest looking surface is quite
rough on a microscopic scale• Even for a rolling object, there is a friction called
rolling friction• In this topic we are focusing on sliding friction,
kinetic friction(kinetic – ‘moving’ in Greek)
• Kinetic friction act on the opposite direction of the object’s velocity
• Magnitude of the kinetic friction depend on– Normal force between the two surface– Nature of the two sliding surface – coefficient of
static friction
• Approximation of the frictional force:Ffr = μkFN .
• Here, FN is the normal force, and μk is the coefficient of kinetic friction, which is different for each pair of surfaces.
• Static friction applies when two surfaces are at rest with respect to each other (such as a book sitting on a table).
Ffr ≤ μsFN
μs is the coefficient of static friction
• The static frictional force is as big as it needs to be to prevent slipping, up to a maximum value.
• Usually it is easier to keep an object sliding than it is to get it started.
Note that, in general, μs > μk.
Example 5-1: Friction: static and kinetic.Our 10.0-kg mystery box rests on a horizontal floor. The coefficient of static friction is 0.40 and the coefficient of kinetic friction is 0.30. Determine the force of friction acting on the box if a horizontal external applied force is exerted on it of magnitude:
(a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.
Conceptual Example 5-2: A box against a wall.You can hold a box against a rough wall and prevent it from slipping down by pressing hard horizontally. How does the application of a horizontal force keep an object from moving vertically?
Example 5-3: Pulling against friction.
A 10.0-kg box is pulled along a horizontal surface by a force of 40.0 N applied at a 30.0° angle above horizontal. The coefficient of kinetic friction is 0.30. Calculate the acceleration.
Conceptual Example 5-4: To push or to pull a sled?Your little sister wants a ride on her sled. If you are on flat ground, will you exert less force if you push her or pull her? Assume the same angle θ in each case.
Example 5-5: Two boxes and a pulley.
Two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration, a, of the system, which will have the same magnitude for both boxes assuming the cord doesn’t stretch. As box B moves down, box A moves to the right.
Example 5-6: The skier.
This skier is descending a 30° slope, at constant speed. What can you say about the coefficient of kinetic friction?
Example 5-7: A ramp, a pulley, and two boxes.Box A, of mass 10.0 kg, rests on a surface
inclined at 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown.
(a)If the coefficient of static friction is 0.40, determine what range of values for mass B will keep the system at rest.
(b)If the coefficient of kinetic friction is 0.30, and mB = 10.0 kg, determine the acceleration of the system.
ConcepTest 5.1ConcepTest 5.1 FrictionFriction
1) the force from the rushing air
pushed it off
2) the force of friction pushed it off
3) no net force acted on the box
4) truck went into reverse by accident
5) none of the above
A box sits in a pickup truck
on a frictionless truck bed.
When the truck accelerates
forward, the box slides off
the back of the truck
because:
Generally, the reason that the box in the truck bed would move
with the truck is due to frictionfriction between the box and the bed.
If there is no friction, there is no force to push the box along, If there is no friction, there is no force to push the box along,
and it remains at rest.and it remains at rest. The truck accelerated away, essentially
leaving the box behind!!
ConcepTest 5.1ConcepTest 5.1 FrictionFriction
1) the force from the rushing air
pushed it off
2) the force of friction pushed it off
3) no net force acted on the box
4) truck went into reverse by accident
5) none of the above
A box sits in a pickup truck
on a frictionless truck bed.
When the truck accelerates
forward, the box slides off
the back of the truck
because:
Antilock brakes keep the
car wheels from locking
and skidding during a
sudden stop. Why does
this help slow the car
down?
1) k > s so sliding friction is better
2) k > s so static friction is better
3) s > k so sliding friction is better
4) s > k so static friction is better
5) none of the above
ConcepTest 5.2ConcepTest 5.2 Antilock BrakesAntilock Brakes
Antilock brakes keep the
car wheels from locking
and skidding during a
sudden stop. Why does
this help slow the car
down?
1) k > s so sliding friction is better
2) k > s so static friction is better
3) s > k so sliding friction is better
4) s > k so static friction is better
5) none of the above
Static friction is greater than sliding frictionStatic friction is greater than sliding friction, so
by keeping the wheels from skidding, the static
friction force will help slow the car down more
efficiently than the sliding friction that occurs
during a skid.
ConcepTest 5.2ConcepTest 5.2 Antilock BrakesAntilock Brakes
ConcepTest 5.3ConcepTest 5.3 Going SleddingGoing Sledding
1
2
1) pushing her from behind
2) pulling her from the front
3) both are equivalent
4) it is impossible to move the sled
5) tell her to get out and walk
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
easiest way to
accomplish this?
ConcepTest 5.3ConcepTest 5.3 Going SleddingGoing Sledding
1
2
In Case 1, the force F is pushing downpushing down
(in addition to mg), so the normal normal
force is largerforce is larger. In Case 2, the force F
is pulling uppulling up, against gravity, so the
normal force is lessenednormal force is lessened. Recall that
the frictional force is proportional to
the normal force.
1) pushing her from behind
2) pulling her from the front
3) both are equivalent
4) it is impossible to move the sled
5) tell her to get out and walk
Your little sister wants
you to give her a ride
on her sled. On level
ground, what is the
easiest way to
accomplish this?
ConcepTest 5.4ConcepTest 5.4 Will it Budge?Will it Budge?
1) moves to the left
2) moves to the right
3) moves up
4) moves down
5) the box does not move
A box of weight 100 N is at
rest on a floor where s = 0.4.
A rope is attached to the box
and pulled horizontally with
tension T = 30 N. Which way
does the box move?
Tm
Static friction
(s= 0.4)
The static friction force has a
maximummaximum of ssN = 40 NN = 40 N. The
tension in the rope is only 30 N30 N.
So the pulling force is not big
enough to overcome friction.
ConcepTest 5.4ConcepTest 5.4 Will it Budge?Will it Budge?
1) moves to the left
2) moves to the right
3) moves up
4) moves down
5) the box does not move
A box of weight 100 N is at
rest on a floor where s = 0.4.
A rope is attached to the box
and pulled horizontally with
tension T = 30 N. Which way
does the box move?
Tm
Static friction
(s= 0.4)
Follow-up:Follow-up: What happens if the tension is What happens if the tension is 35 N35 N? What about ? What about 45 N45 N??
1) component of the gravity force parallel to the plane increased
2) coeff. of static friction decreased
3) normal force exerted by the board decreased
4) both #1 and #3
5) all of #1, #2 and #3
A box sits on a flat board.
You lift one end of the
board, making an angle
with the floor. As you
increase the angle, the
box will eventually begin
to slide down. Why?
Net Force
Normal
Weight
ConcepTest 5.5aConcepTest 5.5a Sliding Down ISliding Down I
1) component of the gravity force parallel to the plane increased
2) coeff. of static friction decreased
3) normal force exerted by the board decreased
4) both #1 and #3
5) all of #1, #2 and #3
A box sits on a flat board.
You lift one end of the
board, making an angle
with the floor. As you
increase the angle, the
box will eventually begin
to slide down. Why?
Net Force
Normal
Weight
As the angle increases, the component component
of weight parallel to the plane increasesof weight parallel to the plane increases
and the component perpendicular to the component perpendicular to the
plane decreasesplane decreases (and so does the Normal
force). Since friction depends on Normal
force, we see that the friction force gets friction force gets
smallersmaller and the force pulling the box force pulling the box
down the plane gets biggerdown the plane gets bigger.
ConcepTest 5.5aConcepTest 5.5a Sliding Down ISliding Down I
m
1) not move at all
2) slide a bit, slow down, then stop
3) accelerate down the incline
4) slide down at constant speed
5) slide up at constant speed
A mass m is placed on an
inclined plane ( > 0) and
slides down the plane with
constant speed. If a similar
block (same ) of mass 2m
were placed on the same
incline, it would:
ConcepTest 5.5bConcepTest 5.5b Sliding Down IISliding Down II
The component of gravity acting down
the plane is doubledouble for 2m. However,
the normal force (and hence the friction
force) is also doubledouble (the same factor!).
This means the two forces still cancel
to give a net force of zero.
A mass m is placed on an
inclined plane ( > 0) and
slides down the plane with
constant speed. If a similar
block (same ) of mass 2m
were placed on the same
incline, it would:
W
Nf
Wx
Wy
1) not move at all
2) slide a bit, slow down, then stop
3) accelerate down the incline
4) slide down at constant speed
5) slide up at constant speed
ConcepTest 5.5bConcepTest 5.5b Sliding Down IISliding Down II