Chapter 5: Friction, Drag, and Elasticity

Question 1

Define deformation

Answer 1

Change in shape due to the application of force.

Question 2

Define drag force in words.

Answer 2

The drag force is found to be proportional to the square of the speed of the object; mathematically

Question 3

What is the equation for drag force?

Answer 3

F_D∝v2 or F_D=1/2CρAv^2

F_D = Drag force (_ = subscript)
v = velocity
C = drag coefficient
A = area of the object facing the fluid
p = density of the fluid
∝ = directly proportional

Question 4

Define friction

Answer 4

A force that opposes relative motion or attempts at motion between systems in contact

Question 5

Define Hooke’s Law in works

Answer 5

Proportional relationship between the force on a material and the deformation it causes

Question 6

What is Hooke’s law in a mathematical expression?

Answer 6

F = kΔL

F = force
k= spring constant
ΔL = amount of deformation (change in length)

Question 7

Define kinetic friction

Answer 7

A force that opposes the motion of two systems that are in contact and moving relative to one another

Question 8

What is the equation for the magnitude of kinetic friction?

Answer 8

fk=μkN

fk = force of kinetic friction
uk = coefficient of kinetic friction
N = normal force

Question 9

What is the equation for the “magnitude of static friction”

Answer 9

fs≤μsN

fs = force of static friction
us =the coefficient of static firction
N = magnitude of normal force

Question 10

Define shear deformation

Answer 10

Deformation perpendicular to the original length of an object

Question 11

Define static friction

Answer 11

A force that opposes the motion of two systems that are in constant and are not moving relative to one another

Question 12

Define Stoke’s Law (equation)

Answer 12

Fs=6πrηv

r = radius of the object
η = is the viscosity of the fluid
v = velocity

Question 13

Define strain

Answer 13

Ratio of change in length to original length

Question 14

Define stress

Answer 14

Ratio of force to area

Question 15

Define tensile strength

Answer 15

The breaking stress that will cause permanent deformation of fraction of a metal

Question 16

Between the 3 drag force equation. Which is used for small objects, and which is used for big objects?

F_D∝v2
F_D=1/2CρAv^2
Fs = 6πηrv

Answer 16

Any object not in fluid: F_D∝v2

Big objects in fluid (air can be fluid): F_D=1/2CρAv^2

Small objects (such as bacteria in water): Fs = 6πηrv

Question 17

T/F: Stokes’ Law is a drag force equation

Answer 17

True

Question 18

How can the relationship between the deformation (ΔL) and the applied force (F) be written as other than Hooke’s Law?

Answer 18

ΔL=1/Y * F/A * L0

ΔL = deformation ( change in length)
Y = Young’s modulus depends on the substance
A = cross-sectional area
F = force
L0 = original length

Question 19

What does the “F/A” in ΔL=1/Y * F/A * L0 represent?

Answer 19

F/A is defined as “stress” and its unit is N/m^2

Question 20

What does ΔL/ L0 mean in ΔL=1/Y * F/A * L0

Answer 20

ΔL/L0 is defined as strain (has no unit)

Question 21

What is one way to calculate stress?

Answer 21

Stress = Y * strain

Y = young’s modulus, depends on the substance
Strain = ΔL/L0 (change in length / original length)

Question 22

What is the expression for shear deformation?

Answer 22

Δx=1/S * F/A *L0

S = shear modulus
F = force applied perpendicular to L0
A = cross-sectional area

Question 23

What is the relationship of change in volume to other physical quantities equation?

Answer 23

ΔV=1/B * F/A * V0

B = bulk modulus
V0 = original volume
F/A = force per unit area

Question 24

Albert pushes a 12kg box across a level floor. He pushes with a force of 43 N and observes that the box accelerates toward his push at a rate of 3 m/s^2.

What is the net force acting on the box?

Answer 24

36 N

Explanation:
I. Newton’s 2nd law of motion (discussed in Ch.4) says Fnet = ma

II. Net force = mass x acceleration -> F = 12kg x 3 m/s^2

III. F = 36 N (kgm/s^2 = N)

Question 25

Albert pushes a 12kg box across a level floor. He pushes with a force of 43 N and observes that the box accelerates toward his push at a rate of 3 m/s^2. The force acting on the box from Albert’s push is 36 N.

Assuming that Albert’s push and the friction with the floor are the only horizontal forces acting, what is the magnitude of the frictional force pushing backward on the box?

Answer 25

7 N.

Explanation:
I. Albert psuhes with 43 N, but only 36 N is acted on the box. The difference between these two forces is the friction.

II: Fnet = F1 + F2 (F1 = initial force, F2 = friction force)

III: Plug in the numbers: 43 N = 36 N + F2

IV: Do the algebra to get F2 by itself: F2 = 43 N - 36N

V: Solve, F2 = 7 N

Question 26

If you push a box across a horizontal floor toward the east, in which direction will friction act?

Answer 26

West.

Explanation: Friction will always act in the opposite direction of the main force

Question 27

If the frictional force is 23.0 N and you push a box with a positive force of 27.0 N, what will the net force be?

Answer 27

4.0 N

Explanation:
I. Assess what you know; you know the friction force and the push force, and you need the net force

II: Find the equation: Net force = (push) - (friction)

III: Plug in the numbers: Fnet = 27 N - 23 N

IV: Solve: Fnet = 4 N

Question 28

If a box has a mass of 2.0 kg, and a net force of 4 N, what will its acceleration be

Answer 28

2.0 m/s^2

Explanation:

I. Assess what you have. You know the mass, you know the net force, you need acceleration

II: Find the equation: Fnet = ma

III: Get acceleration by itself: a = Fnet / m

IV: Plug in the numbers: a = 4 N / 2kg

V: Solve: a = 2.0 m/s^2

-Units?
-N = kgm/s^2, since it was divided by kg, the kg cancel each other out and you’re left with m/s^2

Question 29

You need to push a heavy box across a rough floor, and you want to minimize the average force applied to the box during the time the box is being pushed.

Which method of pushing results in the minimum average force being applied to the box?

A. Keep pushing the box forward at a steady speed

B. Push the box so that it accelerates forward at a constant rate

C. Push the box forward short distance, rest, then repeat until finished

Answer 29

A. Keep pushing the box forward at a steady speed

Explanation: Option B would involve constantly putting in energy to maintain a force higher than the kinetic friction. Option C involves needing to overcome the static friction after every break. Only option A has the smallest average force applied to it by keeping the minimal energy required to stay above the kinetic friction and keep the box moving.

Question 30

A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602, and the coefficient of kinetic friction is 0.327.

How much force is needed to begin moving the book?

Answer 30

12N

Explanation:

I. Assess what you have: Mass, coefficient of static friction, coefficient of kinetic friction

II. Assess what you need: Force

III: Find a fitting equation: 𝐹s=𝜇s𝑁
a. N = mg

IV: Plug in the numbers: 𝐹s=(0.602)(2.03kg)(9.81m/s^2)

V: Solve: Fs = 11.9N or 12N if you round

Question 31

A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602, and the coefficient of kinetic friction is 0.327.

How much force is needed to keep the book moving at a constant speed once it begins moving?

Answer 31

6.51 N

Explanation:

I. Assess what you have: mass, static coefficient, kinetic coefficient

II. Assess what you need: Kinetic force (force needed for movement)

III: Find the right equation: 𝐹k=𝜇k𝑁
a. N = mg

IV: Plug in the numbers: Fk = (0.327)(2.03kg)(9.81m/s^2)

V. Solve: Fk = 6.51 N

Question 32

What are the units for mass?

Answer 32

Kg

Question 33

How do you convert g to kg?

Answer 33

Divide it by 1,000

Ex: A 2g cookie = 0.002kg (2 / 1000)

Question 34

What are the units of force?

Answer 34

Newtons (kgm/s^2)

Question 35

A block whose weight is 45.0 N rests on a horizontal table. A horizontal force of 36.0 N is applied to the block. The coefficients of static and kinetic friction are 0.650 and 0.420, respectively. Will the block move under the influence of the force, and, if so, what will be the block’s acceleration? Explain your reasoning

Answer 35

Answer: Yes, 3.73 m/s^2 to the right

Explanation:

Known variables
W (weight) = 45.0N
F = +36.0 N

Express N2 as y-component and as x-component equations (only need y for now)
y-components: ∑Fy = may -> N - 45N = m(0 m/s^2) -> N = 45N

    Ay = 0 because Voy = 0 m/s AND Vy = 0 m/s)

Use Normal force to determine the maximum possible friction force, using Us
□ (Ffric)max = UsN
□ (Ffric)max = (0.650)(45N) = 29.25 Newtons

Since (Ffric)max is less than the applied force, F = 36 N, the block WILL HAVE A NET FORCE THAT IS NONZERO. THE BLOCK MOVES!

However, as soon as the block begins to move, the kinetic friction coefficient UK determines the frictional force:
□ (Ffric)k = UkN
□ (Ffric)k = (0.420)(45N)
□ (Ffric)k = 18.9 N

Now consider x-component equation for N2
□ F - (Ffric)k = max

Since the weight of the block is 45N, m = w/g -> 45N/9.81m/s^2 -> 4.59 kg

36N - 19.9N = (4.59kg)ax

Ax = (36N-18.9N)/4.59 kg = 3.73m/s^2 (to the right)

Question 36

A cup of coffee is sitting on a table in an airplane that is flying at a constant altitude and a constant velocity. The coefficient of static friction between the cup and the table is 0.30. Suddenly, the plane accelerates, it’s altitude remaining constant. What is the maximum acceleration that the plane can have without the cup sliding backward on the table?

Answer 36

Answer: Ax = 2.94 m/s^2

Explanation:
The maximum acceleration with no slip, Fs =Fsmax

N2:
∑Fx = max
UsFN = max
Usmg = max
Usg = ax

So ax = (0.30)(9.81m/s^2)

Ax = 2.94m/s^2

Question 37

A 92 kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is Uk = 0.61

What is the magnitude of the frictional force?

If the player comes to rest after 1.2 s, what is his initial speed?

Answer 37

Answer: 550 Newtons; Vox = 7.2m/s

Explanation for Q1:
There is no acceleration in the vertical direction, so from N2:

∑Fy = may, we get FN = mg = (92kg)(9.81m/s^2) = 902 N

Kinetic friction: Fk = UkFN = (0.61)(902N)

Fk = 550 N

Explanation for Q2:
Note that the frictional force is in the direction opposite to the velocity of the sliding player

N2: ax = (∑Fx/m) = (-550N / 92kg) = -5.98 m/s^2

Now do the constant-acceleration problem:
Ax = 5.98 m/s^2
Vx = 0 m/s (he slides, slowing down, and stops on the base)
Vox = find
T = 1.2 seconds

Vx Vox + at
Vox = Vx - at
Vox = 0 m/s - (-5.98 m/s^2)(1.2s)
Vox = 7.2m/s

Question 38

A bicyclist is coasting down an 8.0o hill at a constant velocity of +8.9 m/s. The mass of the bicycle and rider is 85 kg. A force due to air resistance is directed opposite to the motion and has a magnitude fair that is proportional to the cyclist’s speed v, so that fair = cv, where c is a constant. Using the fact that the cyclist is coasting at a constant velocity, determine the numerical value (including units) for the constant c.

Answer 38

Answer: c = 13 kg/sec

Explanation:
Note how fair get longer for larger speeds
Constant veloctu v = 8.9 m/s
Bike & Rider = 85 kg

The result comes from considering N2 in the x direction
∑Fx = max
But ax = 0, since it’s constant velocity
0 = ∑Fx = mgsinꝋ - f
Cv = mgsinꝋ

Solve for c and plug in the given values

C = (mgsinꝋ)/v = [(85kg)(9.81m/s^2)sin(8.0o)]/8.9 m/s

C = kg/sec

Question 39

A truck is traveling at a speed of 25.9 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient of static friction between the crate and the truck bed is 0.650. Determine the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck.

Answer 39

Answer: 49 meters
Explanation:
Us = 0.650
The force that must slow the create down comes from the coefficient of static friction, giving a maximum possible slowing force of: Fsmax = USFN = Us (mcrateg)
The acceleration of the crate (and so of the truck too) will be from N2:
Ax = (-Fsmax)/ mcrate = (-Usmcrateg) / mcrate = Usg
Ax = -(0.650)(9.81m/s^2) = -6.37m/s^2
Do the constant acceleration problem
X = find
A = -6.37 m/s^2
V = 0 m/s (to stop truck)
Vf = +25 m/s

V^2 = Vo^2 + 2ax
X = (1/2a)(v^2-Vo^2)
X = [(0m/s)^2-(25m/s)^2]/ 2(-6.37m/s^2)
X = 49 meters