Chapter 5: Friction, Drag, and Elasticity Flashcards
Define deformation
Change in shape due to the application of force.
Define drag force in words.
The drag force is found to be proportional to the square of the speed of the object; mathematically
What is the equation for drag force?
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
Define friction
A force that opposes relative motion or attempts at motion between systems in contact
Define Hooke’s Law in works
Proportional relationship between the force on a material and the deformation it causes
What is Hooke’s law in a mathematical expression?
F = kΔL
F = force
k= spring constant
ΔL = amount of deformation (change in length)
Define kinetic friction
A force that opposes the motion of two systems that are in contact and moving relative to one another
What is the equation for the magnitude of kinetic friction?
fk=μkN
fk = force of kinetic friction
uk = coefficient of kinetic friction
N = normal force
What is the equation for the “magnitude of static friction”
fs≤μsN
fs = force of static friction
us =the coefficient of static firction
N = magnitude of normal force
Define shear deformation
Deformation perpendicular to the original length of an object
Define static friction
A force that opposes the motion of two systems that are in constant and are not moving relative to one another
Define Stoke’s Law (equation)
Fs=6πrηv
r = radius of the object
η = is the viscosity of the fluid
v = velocity
Define strain
Ratio of change in length to original length
Define stress
Ratio of force to area
Define tensile strength
The breaking stress that will cause permanent deformation of fraction of a metal
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
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
T/F: Stokes’ Law is a drag force equation
True
How can the relationship between the deformation (ΔL) and the applied force (F) be written as other than Hooke’s Law?
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
What does the “F/A” in ΔL=1/Y * F/A * L0 represent?
F/A is defined as “stress” and its unit is N/m^2
What does ΔL/ L0 mean in ΔL=1/Y * F/A * L0
ΔL/L0 is defined as strain (has no unit)
What is one way to calculate stress?
Stress = Y * strain
Y = young’s modulus, depends on the substance
Strain = ΔL/L0 (change in length / original length)
What is the expression for shear deformation?
Δx=1/S * F/A *L0
S = shear modulus
F = force applied perpendicular to L0
A = cross-sectional area
What is the relationship of change in volume to other physical quantities equation?
ΔV=1/B * F/A * V0
B = bulk modulus
V0 = original volume
F/A = force per unit area
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?
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)
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?
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
If you push a box across a horizontal floor toward the east, in which direction will friction act?
West.
Explanation: Friction will always act in the opposite direction of the main force
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?
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
If a box has a mass of 2.0 kg, and a net force of 4 N, what will its acceleration be
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
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
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.
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?
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
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?
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
What are the units for mass?
Kg
How do you convert g to kg?
Divide it by 1,000
Ex: A 2g cookie = 0.002kg (2 / 1000)
What are the units of force?
Newtons (kgm/s^2)
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: 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)
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: 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
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: 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
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: 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
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: 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
