FRICTION
and Newton’s second law
The “Normal” Force, N
When an object is pressed against asurface, the surface pushes back.
(That’s Newton’s 3rd Law)
This “push back” from the surface iscalled the Normal Force, N
The word “normal” in math terminologymeans “perpendicular”
The surface pushes back in a direction thatis perpendicular to the surface.
If the box is not accelerating upor down, then the net force iszero and the Normal forcemust balance the Weight
Weight = mg
Normal force, N
Weight = mg
Normal force, N
FA = 20 N f = 4 N m = 2 kg
a = ?
Fnet = ma
20 N – 4 N = ma
a = Fnet / m
a = 8 m/s2
FA = 20 N
f = 4 N
You pull on a box with an applied force of 20 N. The frictional force opposingthe motion is 4 N. If the mass of the box is 2 kg, what is its acceleration?
1.Draw the free body diagram.
2.Write what you know and don’t know.
3.Write the equation, Fnet = ma
4.Calculate the value of the net Force and then theacceleration.
Friction, f
•A force that always opposes motion
•Depends on two things: the roughness ofthe surfaces and how hard they arepressed together.
f = N
, mu- the “coefficient of friction” tells howrough the surfaces are.
N, the Normal force tells how hard thesurfaces are pressed together
Example: How large is the frictionalforce between 2 surfaces if thecoefficient of friction is 0.2 and theNormal force is 80 N?
f = N
f = 0.2 x 80 N
f = 16 N
There are two kinds of friction:
“static friction” (not moving) mustbe overcome to initiate motion.
“kinetic friction” must beovercome while an object ismoving
Static friction > Kinetic friction
IF you pull just hard enough to make the objectmove, you overcome static friction:
Your applied force = static friction force
Your applied force = staticN
Once an object is moving, if you pull in such a waythat the object is moving with constant velocity,then
Your applied force = kinetic friction force
Your applied force = kineticN
And… what is the normal force, N?
N = mg
AppliedForce
Friction = N
A 75 kg firefighter is slidingdown the pole at the fire station.If he is moving at a constantvelocity, what is the frictionforce?
Fnet = ma
friction – weight = ma = 0
Therefore,
friction = weight
friction = mg = 75 kg x 10 m/s2
friction = 750 N
weight
friction
Weight = mg
Normal force, N
FA = 20 N f = 4 N m = 2 kg
a = ?
Fnet = ma
20 N – 4 N = ma
a = Fnet / m
a = 8 m/s2
FA = 20 N
f = 4 N
You pull on a box with an applied force of 20 N. The frictional force opposingthe motion is 4 N. If the mass of the box is 2 kg, what is its acceleration?
1.Draw the free body diagram.
2.Write what you know and don’t know.
3.Write the equation, Fnet = ma
4.Calculate the value of the net Force and then theacceleration.
Weight = mg
Normal force, N
FA = 30 Nm = 2 kg
= 0.4a = ?
Fnet = maf = N
N = mg = 2 kg x 10 m/s2 = 20 N
f = N = 0.4(20 N) = 8 N
Horizontal : Fnet = FA - f
Fnet = 30 N – 8 N = 22 N
a = Fnet / m
a = 11 m/s2
FA
f = N
= 30 N
You pull on a box with an applied force of 30 N. The coefficient of friction is0.4. If the mass of the box is 2 kg, what is its acceleration?
1.Draw the free body diagram.
2.Write what you know and don’t know.
3.Write the equations, Fnet = ma and f = N
4.Calculate the Normal force and the friction force.
5.Calculate the value of the net Force and then theacceleration.
Pre-APonly…
Weight = mg
Normal force, N
FA = 25Nm = 2 kg = 43 = 0.4
a = ?
Fx = maf = N
Fx = - N + FA cos = ma
(-0.4(36.65) + 25cos 43 ) / 2= a
a = 1.81 m/s2
FA
f
N = mg + Fsin
Weight = mg
Normal force, N
FA = 15 Nm = 2 kg = 43
= 0.4a = ?
Fx = maf = N
Fx = - N + FA cos = ma
(-0.4(9.77) + 15cos 43) / 2 = a
a = 3.53 m/s2
FA
f
N = mg – Fsin
Weight = mg
Normal force, N
If the box is moving atconstant velocity, there isno acceleration,
Therefore the net force mustbe zero…
so the horizontal forces mustcancel each other.
N = Fcos
If you push hard enough tojust get the box moving, theacceleration is zero in thatcase also, but the friction isstatic, not kinetic.
FA
f
Friction Labfrictional force = (Normal forcef = N
Normal force, N = Wt1
frictional force, f = Wt2
Weight = mg
N
Wt1
Frictional force, f
Pre-AP Only…
More examples…
Xavier, who was stopped at a light, acceleratedforward at 6 m/s2 when the light changed. He haddice hanging from his rear view mirror. What angledid they make with the vertical during thisacceleration?
1- draw free body diagram
2- write Newton’s 2nd law for BOTH directions
Fx = max
Tsin = max
Tcos = mg
tan = a/g
= 31.47 degrees
Fy = may
mg
T
Tcos – mg = may = 0
An “Atwood’s Machine”
Two masses of 5 kg and 2 kg are suspended from amassless, frictionless pulley, When released from rest,what is their acceleration? What is the Tension in thestring?
1- Draw a free body diagram
2- Write Newton’s Second Law for EXTERNAL forcesacting on the whole system.
Fext = mtotala
The Tension in the string is an “internal” force. Theonly “external” forces are the gravitational forces ofweight, which oppose each other. So…
Fext = m1g – m2g = mtotala
Solving for the acceleration yields
a = 4.2 m/s2
Now, to find the Tension, we must “zoom in” onmass 2:
F= T – m2g = m2a
T = m2a + m2g
T = 28 N
a
+
Objects on Inclines- sliding down, no friction
mg
The free-body diagram ALWAYS comes first:
Draw the weight vector, mg
Draw the Normal force vector.
Are there any other forces???
Since the motion is parallel to the plane, ROTATE the axis from horizontal andvertical to “parallel” and “perpendicular”. Then draw the components of theweight vector, both perpendicular and parallel to the incline. The components helpform a right triangle. Label the angle.
Does the Normal force balance with the force of weight, mg?
NO! What force must balance the Normal force?
N
Does all of the weight, mg, pull thebox down the incline?
Write Newton’s Second Law for theforces on box parallel to the planewith “down” being negative.
F = ma
-mgsin = ma
N = mgcos
No! Only mgsin
Inclines: pushed upward, nofriction
mg
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to theincline. The components help form a right triangle. Label the angle.
N
Write Newton’s Second Law for thebox.
F = ma
FA - mgsin = ma
FA
Inclines: pushed downward, nofriction
mg
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to theincline. The components help form a right triangle. Label the angle.
N
Write Newton’s Second Law for thebox.
F = ma
- FA - mgsin = ma
FA
Inclines: pushed upward, nofriction
mg
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector AND the applied force, bothperpendicular and parallel to the incline. The components help form a righttriangle. Label the angle.
N
Write Newton’s Second Law for allforces acting parallel to the incline.
F = ma
FA cos - mgsin = ma
FA
Inclines: With friction, case 1: atrest or sliding down
mg
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to theincline. The components help form a right triangle. Label the angle.
N
Write Newton’s Second Law for thebox.
F = ma
f - mgsin = ma
mgcos - mgsin = ma
gcos - gsin = a
f
What is friction?
f = N
What is N?
N = mgcos
f = mgcos
Inclines: With friction, case 2:pushed downward
mg
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to theincline. The components help form a right triangle. Label the angle.
N
Write Newton’s Second Law for thebox.
F = ma
f - mgsin – FA = ma
mgcos - mgsin – FA = ma
f
What is friction?
f = N
f = mgcos
FA
Inclines: With friction, case 3:pushed upward
mg
Draw the weight vector, mg
Draw the Normal force vector.
Draw the components of the weight vector, both perpendicular and parallel to theincline. The components help form a right triangle. Label the angle.
N
Write Newton’s Second Law for thebox.
F = ma
FA - mgsin – f = ma
FA - mgsin - mgcos = ma
f
What is friction?
f = N
f = mgcos
FA
Friction along an incline
An object placed along an incline willeventually slide down if the incline iselevated high enough. The angle at whichit slides depends on how rough the inclinesurface is. To find the angle where itslides:
Since it doesn’t move:
mgsin = mgcos
Therefore:
Tan max= max, the coefficient of static friction
Or
The anglemax= tan -1 max
SpringForces
Spring Force
If a mass is suspended from a spring,
two forces act on the mass:
its weight and the spring force.
The spring force for many springs
is given by
Fs = kx,
Where x is the distance the spring is stretched (orcompressed) from its normal length and “k” is calledthe spring constant, which tells the stiffness of thespring. The stiffer the spring, the larger the springconstant k.
This relationship is known as “Hooke’s Law”.
mg
Fs
Spring Force
If the mass is at rest, then the twoforces are balanced, so that
kx = mg
These balanced forces provide aquick way to determine the springconstant of a spring, or anythingspringy, like a rubber band:
Hang a known mass from the springand measure how much it stretches,then solve for “k” !k = mg/x
mg
Fs= kx