PHYS121 Lecture Notes - Lecture 7: Friction, Net Force

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sayheyfei

11 Oct 2016

School

University of Waterloo

Department

Physics

Course

PHYS121

Professor

Richard Epp

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Document Summary

Is the normal force, always perpendicular to the surface. Is the weight force . Static friction prevents the block from increasing in speed from rest, hence the acceleration is 0. Maximum static friction occurs at a critical angle. Maximum static friction is proportional to the normal force. Kinetic friction occurs to prevent acceleration when the object is already in motion. At a critical angle, the kinetic friction is big enough such that the net force is zero, and the acceleration is zero: this is proportional to the normal force. The maximum and minimum horizontal force you could apply without the block moving from rest. The block is on the verge of moving up the incline, hence the friction is downward. Can go to infinite when , since when the incline is ninety degrees, the horizontal applied force is the normal force: hence, the proportional static friction force should equal to the weight.

Related Questions

I am so lost onwere to even begin. I just need help trying to find the Ñ² (angle)for part part 1 of the experiment if anyone can just explain how todo that and maybe give me a step-by-step example so I can finishthe rest. Thank you! Experiment 6 ~ Friction

Purpose:

In this lab, you will make some basic measurements of friction.You will measure the coefficients of staticfriction between several combinations of surfaces using aninclined plane, and you will measure the coefficient ofkinetic friction between two of the combinationsof surfaces you used in the static friction part of thisexperiment.

Part 1: StaticFriction

Theory:

The coefficient of static frictionm_s can be measured experimentally foran object placed on an inclined plane (a.k.a. ramp, a.k.a. hill).The coefficient of static friction is related to the criticalangle q_c for the ramp, at whichthe object just begins to slide. Using what we havecovered in class, you can derive this relationshipyourself! At this critical angle, static friction preventing theobject from sliding down the hill is just exactly equal to thecomponent of the objectâs weight along the hill. If the componentof the weight along the hill were just a little greater, it wouldovercome friction, and the object would start to slide down. Thefree body diagram for this situation will look like the figuresketched below. Use this diagram to find an equation that relatesm_s toq_c.

F_N = NormalForce = mg cos(Î¸)

F_F =Î¼_s x F_N = Î¼_s mg cos(Î¸)

Î£F_x = mA_x = F_Gx - F_F = ?

F_Gx = mgsin(Î¸)

Hint #1: the angle q of the incline is the same as theangle between the normal force N and the weightmg. Why?

Hint #2: The weight mg can be broken into twocomponents. One component is along the incline(W_H). The other component is âpressinginto the hillâ, and is equal in magnitude to the normal forceN. (This is an example of a situation where the normalforce is not equal to the weight!) Remember that you need to usethe normal force to calculate friction, sincef=mN. Now calculate what the angle ofthe incline should be, so that the component of the weight down thehill is just exactly balanced by friction.

Equipment:

For this lab you will need to share the following ramps andblocks with other lab groups:

Ramps Blocks

Aluminum Aluminum

ParticleBoard Brass

StainlessSteel Plexiglass

Plastic Copper

Rubber

Steel

Wood

Experiment:

Test at least 5 different combinations of materials. Record thematerial of the block and the ramp for each combination.

1. Measure and record the length of theinclined plane you will be using. (This is thehypotenuse of the right triangle!)

2. Place the block on the plane and begin tiltingthe plane just until the point when the block begins to slide.

3. Measure the height of the raisedend of the inclined plane at this point. (This is theopposite side of the right triangle!) You cancalculate the angle from the measurements of the two sides.

4. Repeat the height measurement at least 3 timesfor each combination of materials. Record all the measurements inthe table below. Then,

(a) calculate the angle from the measurements of the hypotenuseand opposite side you just obtained

(b) use the equation you derived above in order to calculate thecoefficient of kinetic friction from the angle.

Block: woodenblock Block: cooper

Ramp:particleboard Ramp: particle board

Length of ramp:0.61(m) Length of ramp: 0.61(m)

Height(m)	Ñ²	Ms	Height(m)	Ñ²	Ms
0.24			0.26
0.25			0.31
0.245			0.345

avg m_s: ______ s : _______ avg m_s: ______ s : _______

Block:aluminum Block:particle board block

Ramp: solidewood Ramp: solide wood

Length of ramp:0.605(m) Length of ramp: 0.605(m)

Height(m)	Ñ²	s	Height(m)	Ñ²	s
0.15			0.23
0.16			0.165
0.16			0.195

avg m_s: ______ s : _______ avg m_s: ______ s : _______

5. Remember that in class we said that friction does not dependon surface area. You can easily test whether or not this iscorrect. Choose a block/ramp combination not already tested, andmake sure the block is not a cube or cylinder. Place the block sothat the side with the larger surface area is in contact with theboard. Now measure the angle at which the block just begins toslide (by measuring the hypotenuse and opposite sides of thecorresponding right triangle). As before, calculate the coefficientof friction from the angle you measured, using your equation.Repeat this process with the side of the block with the smallersurface area in contact with the board. Enter this data in thesmall tables below.

Block (larger end) :_________ Block (smaller end):__________

Ramp:_______________ Ramp: ________________

Length of ramp:______(m) Length of ramp: ______(m)

Height(m)	s	Height(m)	s

avg m_s: ______ s : _______ avg m_s: ______ s : _______

Part 2: KineticFriction

Theory:

You can calculate the coefficient of kinetic friction,m_k, using a variation of the methodyou used for the coefficient of static friction. For thecoefficient of kinetic friction, you can use the same free bodydiagram as the one drawn on the first page. But now, thecombination of W_H and the force offriction will need to add up such that the block will slide at aconstant speed. Think of Newtonâs first and second laws when youset up this equation. Derive a relationship between the criticalangle and the coefficient of kinetic friction.

Experiment:

You can measure m_k using aprocedure similar to the one you used to measurem_s. This time, just pick twocombinations of materials. The combinations you pick for this partof the experiment have to be combinations you alreadyused for the static friction part of the experiment, because thepoint is to compare m_sandm_k!

1. Measure and record the length ofthe inclined plane you will be using.

2. Raise the plane high enough for the blockto begin moving. Begin lowering the plane just until the pointwhen the block begins to slow down. (Alternatively, raise theboard while tapping the block until block moves at a constantrate.)

3. Measure the height of the raisedend of the inclined plane at this point.

4. Repeat the height measurement at least 3times for each combination of materials. Record all themeasurements in the table below. Then,

(a) calculate the angle from the measurements of the hypotenuseand opposite side you just obtained

(b) use the equation you derived above in order to calculate thecoefficient of kinetic friction from the angle.

Block (from part 1) :_________ Block (from part 1):__________

Ramp:_______________ Ramp: ________________

Length of ramp:______(m) Length of ramp: ______(m)

Height(m)	k	Height(m)	k

avg m_k: ______ s : _______ avg m_k: ______ s : _______

Analysis:

Calculate the mean value and standard deviation of thecoefficient of static friction that you measured, for each set ofmaterials.

ramp/board pair		Standard deviation




(larger end)
(smaller end)

Calculate the mean value and standard deviation of thecoefficient of kinetic friction that you measured, for each set ofmaterials.

ramp/board

pair

Standard

deviation

Questions:

1. How do the values of m_s compareto the values of m_k? (Of course, youcan only compare them for the same pairs of materials.)

2. Is the relationship betweenm_sand m_kwhat you expected?

3. Of the two parts of the experiment, measurement ofm_sand measurement ofm_k, which had more sources of error?What were some of the sources of error?

4. Could m_k orm_s ever be greater than 1? Explain whyor why not.

5. Is the coefficient of friction the same as when the block wasstanding on its larger (or smaller) end?

6. Think about your results. Do they make sense when youconsider your everyday experiences?

1. A block is placed on a plane whose angle of inclination is 30˚. The coefficients of static and kinetic friction for the block on the inclined plane are both 0.2. The block
(A) remains stationary on the inclined plane.
(B) accelerates down the inclined plane.
(C) travels down the inclined plane at a constant velocity.
(D) travels up the inclined plane at a constant velocity.
(E) accelerates up the inclined plane.

2. An object with a mass of 5.5 kg is allowed to slide from rest down an inclined plane. The plane makes an angle of 30˚ with the horizontal and is 72 m long. The coefficient of friction between the plane and the object is 0.35. The speed of the object at the bottom of the plane is?

3. A 5-kg object slides down a frictionless surface inclined at an angle of 30˚ from the horizontal. The total distance moved by the object along the plane is 10 meters. The work done on the object by the normal force of the surface is?

4. A mass m = 2.5 kg is sliding along a frictionless table with initial speed v as shown in the figure. It strikes a coiled spring that has a force constant k = 500 N/m and
compresses it a distance x=5.0 cm. The initial speed v of the block was..?

5. The graph shows how the position of a particle depends on time. Which choice is closest to the instantaneous speed of the particle at t = 3s?
(A) 0.40 m/s
(B) 0.67 m/s
(C) 0.75 m/s
(D) 1.50 m/s
(E) 2.22 m/s

6. Calculate the work done by a force, F = 2x i, along the
path (0, 0) -> P₁(2, 0) -> P2(2, 2).

turquoisearmadillo973

A boy in a wheelchair (total mass 50.5 kg) has a speed of 1.40 m/s at the crest of a slope 2.40 m high and 12.4 m long. At the bottom of the slope, his speed is 6.50 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

2 Answers

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electron1
Lv 7

4 years ago

There are two forces that cause the wheelchair to accelerate. These two forces are the component of the wheelchair’s weight that is parallel to the hill and the force the boy exerts. The net force on the wheelchair is equal to the sum of these two forces minus the friction force. Let’s use the following equation to determine the acceleration.

vf2 = vi^2 + 2 * a * d

6.5^2 = 1.4^2 + 2 * a * 12.4

24.8 * a = 40.29

a = 40.29 ÷ 24.8

This is approximately 1.62 m/s^2.

Net force = 50.5 * 40.29 ÷ 24.8 = 2034.645 ÷ 24.8

This is approximately 82 N.

Force parallel = 50.5 * 9.8 * 2.40 ÷ 12.4 = 1187.76 ÷ 12.4

This is approximately 95.8 N. This force and the force that the boy exerts are the forces that cause the wheelchair to accelerate.

Net force = F + (1187.76 ÷ 12.4) – 41

Set the two net forces equal to each other.

F + (1187.76 ÷ 12.4) – 41 = 2034.645 ÷ 24.8

F = 41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)

F is approximately 27.3 N

Work = [41 + (2034.645 ÷ 24.8) – (1187.76 ÷ 12.4)] * 12.4 = 337.9625 J
JupiterSky
Lv 6

4 years ago

His change in kinetic energy is due to gravitational- and his own work,

while the friction does negative work.

∆K = ∆U +∆W + ∆Wf

½m(Vf²−Vi²) = mgh + ∆W + Ff*d

∆W = ½m(Vf²−Vi²) − mgh − Ff*d

∆W = ½*50.5(6.5²−1.4²) − (50.5*9.8*2.4) − (−41*12.4)

∆W = 508 J

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---

asked in Science & MathematicsPhysics · 8 years ago

What is the spring constant? please help me!!!!?

A block of mass 487 g is pushed against the
spring (located on the left-hand side of the
track) and compresses the spring a distance
4.7 cm from its equilibrium position (as shown
in the ﬁgure below). The block starts from
rest, is accelerated by the compressed spring,
and slides across a frictionless horizontal track
(as shown in the ﬁgure below). It leaves the
track horizontally, ﬂies through the air, and
subsequently strikes the ground.
acceleration = 9.81
 
A) What is the spring constant?
B)What is the speed v of the block when it leaves
the track?
C) What is the total speed of the block when it
hits the ground?
...Show more

Update:

Height Above ground = 1.9 m
Lands = 4.59m away

What is the spring constant? please help me!!!!?

3 Answers

Relevance

Jánošík
Lv 7

8 years ago

Favorite Answer

I'm sure there is some crucial information in "the figure below" !!!

Edit:

OK... now we got something to work with ... ;-)

Time needed to fall a vertical distance of 1.9 m is:

t = √( 2h / g )

t = √[ 2×(1.9 m) / (9.81 m/s²) ]

t = 0.6224 s

In that time, the block must travel a horizontal distance of 4.59 m, so the horizontal velocity is:

r = v×t

(4.59 m) = v×(0.6224 s)

v = 7.37 m/s < - - - - - - - - - - - - - - - - - answer B

That is also the speed of the block as it leaves the spring.

Its kinetic energy at that time is:

KE = m×v²/2

KE = (0.487 kg)×(7.3749 m/s)²/2

KE = 13.244 J

That is also the elastic potential energy stored in the spring at maximum compression. So the spring constant is:

KE = PEe = k×d²/2

(13.244 J) = k×(0.047 m)²/2

k = 12 N/m < - - - - - - - - - - - - - - - - - - answer A

The gravitational potential energy of the block on the table is:

PEg = m×g×h

PEg = (0.487 kg)×(9.81 m/s²)×(1.9)

PEg = 9.077 J

The total energy of the block on the table is:

TE = PEg + PEe

TE = (9.077 J) + (13.244 J)

TE = 22.32 J

When hitting the ground, all that energy is converted to kinetic energy, so

TE = KE = m×v²/2

(22.32 J) = (0.487 kg)×v²/2

v = 9.57 m/s < - - - - - - - - - - - - - - - - - - answer C

...Show more
Technobuff
Lv 7

8 years ago

There must be a height above ground and a distance from the end of the track on the "as shown in the figure below".??
STALKER
Lv 4

8 years ago

use f=kl

where k is the spring constant and l is extension from the natural length.

k=f/l

since you have mass in grams. use w=mg

w=0.487 x 9.8 = 4.77N

Now use the equation,

k=4.77/4.7

k=1.012

Note: my answer could be wrong since I do not know the springs original length from the diagram shown. But, the method of working out spring constant is same.

...Show more

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ひいらぎ

Lv 5

asked in Science & MathematicsPhysics · 1 decade ago

Physics Momentum, Rolling?

Pat builds a track for his model car out of wood. The track is 10.00 cm wide and 3.00 m long, and is solid. The runway is cut such that it forms a parabola by the equation y=(x - 3)^2/9. Locate the horizontal position of the center of gravity of this track.
 
A washing machine load in the "spin" cycle, (much like a centrifuge used to separate fluids), can be modeled as a cylinder which has zero density toward the center, and starting from some inner radius where the clothes (or fluids) begin, increases in density as you move radially outward. The washing machine has density of clothes (in SI units) of p=14r+5, a height of 0.51 m, and only has clothes present from inner radius 0.2m out to 0.7m from the center.
-What is the moment of Inertia of the wash load?
-What is the power of motor that drives the clothes from rest up to 4 rev/sec, in 14 seconds if the basin has moment of Inertia of 4 kg*.m2
...Show more

Physics Momentum, Rolling?

1 Answer

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Anonymous

1 decade ago

Favorite Answer

I can't visualize the first problem.

The second: Moment of inertia is I = m r^2. In a shell of thickness dr there is a mass of p * 2 * pi * r * h * dr. The moment of inertia of the shell is 2 pi h p r^3 dr. Now integrate this from r = 0.2m to r = 0.7m.

Kinetic energy is 1/2 m v^2. In rotations the moment of inertia is like the mass and the angular velocity is like the velocity, so rotational energy is 1/2 I omega^2. Omega is in rad/s. The total inertia of the system is the sum of the inertia of the clothes and the basin. Divide the energy by the time needed to get the power.

...Show more

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Next Question

asked in Science & MathematicsPhysics · 1 decade ago

Physics:You and your friends push a 75-kg...Help!?

You and your friends push a 75-kg greased pig up an aluminum slide at the county fair, starting from the low end of the slide. The coefficient of kinetic friction between the pig and the slide is 0.070.
 
(a) All of you pushing together (parallel to the incline) manage to accelerate the pig from rest at the constant rate of 4.9 m/s2 over a distance of 1.8 m, at which point you release the pig. The pig continues up the slide, reaching a maximum vertical height above its release point of 50 cm. What is the angle of inclination of the slide? __degree
 
(b) At the maximum height the pig turns around and begins to slip down the slide, how fast is it moving when it arrives at the low end of the slide? __m/s
...Show more

Physics:You and your friends push a 75-kg...Help!?

1 Answer

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Anonymous

1 decade ago

Favorite Answer

♣the distance the pig was pushed up is

s=0.5*a*t^2, where s=1.8m, a=4.9m/s^2, t is time of push-up;

speed of the pig gained after s=1.8m is v=a*t; thus v^2=2as;

♠ kinetic energy of the pig at the moment it’s released after push-up is

E=0.5*m*v^2, which is enough for the pig to move by itself L distance more along the ramp, where L=h/sin(b), h=50cm=0.5m, b is angle in question;

♦pig’s potential energy at vertex (when stopped) is E1=mgh;

the energy E2=E-E1 is gone on friction, that is E2=f*L is work of friction, where

f=μ*w1, w1=mg*cos(b) is component of pig’s weight normal to the aluminum ramp; or;

f= μ*mg*cos(b);

♦Thus E2=E-E1; or; f*L = 0.5*m*v^2 –mgh; or;

(μ*mg*cos(b)) * (h/sin(b)) =0.5*m* 2as –mgh; or;

μ*gh*cos(b)/sin(b) =as –gh, hence cot(b) =(as –gh)/(μ*gh), hence

(a); b=atan(μ*gh/(as-gh))= atan(0.07*9.8*0.5/(4.9*1.8 -9.8*0.5)) =5°;

♥ pig’s potential energy above the low end of the ramp is

W= mg*(s*sin(b) +h); pig’s ramp position is p=s +L = s + h/sin(b);

Pig’s final kinetic energy is W1=W-W2, where W2=f*p is work of friction;

and W1=0.5*m*v^2; thus

0.5*m*v^2 = mg*(s*sin(b) +h) – μ*mg*cos(b)*(s + h/sin(b));

v^2 =2g*(s*sin(b) +h)*(1 –μ*cot(b)) =2.575;

(b); v =1.605 m/s;

...Show more

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PHYS121 Lecture Notes - Lecture 7: Friction, Net Force

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