Friction

Question 1

As a block of ice sits on a frictionless ramp, it experiences a force of gravity. Define the components of this force acting on the block. How is it different if the block was sitting on a flat surface?

Answer 1

A block sitting on a flat surface experiences only a downwards, vertical force from gravity. However, with the block sitting on a ramp, the block experiences force of gravity in both the vertical and horizontal direction. (note: vertical to the to the block sitting on the ramp and horizontal to the block on the ramp). These vectors can be isolated and defined by utilizing the angle at which the ramp sits at to the ground. This should be the same angle as the block sitting on the surface of the ramp, and therefore this is the angle you can use to do trig calculations

Question 2

A 10kg block sits at a 30 degree angle on a ramp. What is the horizontal force of gravity to the block/ramp surface?

Answer 2

S = O/H -> O = S(H) 
O = sin(30) *98N 
O = ½*98N = 49N

Question 3

If the vertical component of the force of gravity on a block sitting on a ramp at 30 degrees is 10 N, how much mass does the block have?

Answer 3

Fy = 10N 
Fh ?
CAH - because the want to find the hypotenuse
H = C/A 
H = 10N/cos(30)
H =  (10N) (√(3)/2) = 11.63N 
11.63N = m*g 
m = 11.63N / 9.8m/s^2 = 1.19 kg

Question 4

Cos (30)

Answer 4

√(3)/2

Question 5

Sin (30)

Answer 5

½

Question 6

After calculating the force of gravity and its components on a block sitting on a ramp, you conclude that the block should accelerate down the ramp. When you conduct this experience, this is not what happens. What is your error?

Answer 6

You did not calculate the force of friction! This would hinder the block from moving down a ramp.

Question 7

You are asked to find all the forces acting on a block stationary on a ramp. You come up with a force of gravity components and force of friction. Are you done?

Answer 7

No, you forgot to calculate the normal force. This force is perpendicular to the plane at which the object rests on.

Question 8

Assuming no friction, what would be the net force acting on a box on an incline angle at 30 degrees, with a weight of 10N?

A. 3
B. 5
C. 8
D. 15

Answer 8

Net force (Fnet) = F gravity, F normal, F acceleration
F = m*a = 10kg

S1: Break Fg into x and y components. Know that the x and y components of the block are going to cancel out, therefore the x component is the main focus here
S2: Trig -> SOH -> sin(30)*10 = 5. Therefore the net force is 5 Newtons parallel to the the surface of the plane

Question 9

As a block sits on a ramp, it experiences a force of gravity pushing against it. What force acts against the force of gravity’s component horizontal to the block? Why?

Answer 9

Force of friction acts against the parallel component. This is the component innate to the ramp and can prevent the object from sliding down.

Question 10

The normal force of an object sitting on an inclined plane acts against what force?

Answer 10

The force of gravity component perpendicular to the object sitting on the ramp.

Question 11

With an object accelerating constantly down a ramp, what specific force is the force of gravity’s component overcoming? What new force is experienced as a result of the acceleration?

Answer 11

The force of gravity’s horizontal component is overcoming the friction. With kinetic energy (due to acceleration), the object is experiencing kinetic friction (stationary friction is when the object is not moving).

Question 12

When can we assume μk is equal to tan θ with an object on an inclined plane?

Answer 12

This can be assumed when the object is accelerating at a constant acceleration and velocity down the ramp.

Question 13

What is one assumption you can always assume about friction and object motion?

Answer 13

Force of friction will always act opposite to the motion!

Question 14

How much force do you have to apply to a stationary object resting on an incline ramp experiencing a 98 N from the force of gravity in order for the object to start moving?

Answer 14

In order to calculate this, you have to multiply the coefficient of the static force by the normal force in order to find the budging force

μm = Budging force/normal force

Question 15

What is the coefficient of static friction? What is it dependent on?

Answer 15

This is calculation of the budging force / normal force. It is experimentally found and is based on material composition

Question 16

You have Angelie push a boulder up a hill. In the process of pushing the rock until it budges and starts moving, what is happening to the force of friction against the push?

Answer 16

The force of friction increases with increasing force you add onto the object. This process occurs until the object has enough energy to move. When it moves, the force overcomes the static force of friction.

Question 17

You are calculating the forces acting on a block sliding down a ramp. What are the forces at play?

Answer 17

Force of gravity x and y components, force normal, and kinetic friction (this will be going against the direction of the motion to prevent acceleration)

Question 18

An object is moving 5 m/s down a ramp. Is it following Newton’s Law of Inertia?

Answer 18

Yes, but can depend with additional information. Moving 5 m/s down a ramp implies the velocity is constant and therefore may not have an acceleration and therefore is following Newton’s law. This is possible because accounting for kinetic friction prevents the object from accelerating. Newton’s law states it has to accelerate.

Question 19

When does kinetic friction play a role in a scenario?

Answer 19

Kinetic friction acts to prevent the object from accelerating. This allows the object to still follow Newton’s first law and therefore the object can move at a constant velocity.

Question 20

What is the coefficient of kinetic friction if the force of gravity on the object is 98 N on a 30 degree ramp and the object is sliding at a constant velocity down the ramp?

Answer 20

||F_friction|| / ||normal force||
49N / 49N √(3)
= 1/√(3) = 0.58

Question 21

Compare and contrast μ_k and μ_s.

Answer 21

Both are coefficients of friction, kinetic and static respectively. Both play a role in maintaining Newton’s first law, however when the force on the object is greater than the coefficients, Newton’s first law will break. μ_k = ||F_friction|| / ||normal force|| and is active when the object is in motion. μ_s = || Budge Force|| / || normal force||

Question 22

What is one rule of thumb in regards to the relationship of μ_k and μ_s.

Answer 22

μ_k is less than μ_s. Never see a situation where μ_s is lower than μ_k. But can see scenarios where μ_k is lower than μ_s or equal to one another. Once something is moving, for some reason, friction is less potent than when the object is stationary

Question 23

Why is it harder to imagine why differences exist between the coefficient of kinetic friction and static friction on a macroscopic level?

Answer 23

Macroscopically doesn’t entail how the two objects are intimately interacting, therefore theorizing why the difference exists makes no sense in this school of thought.

Question 24

Though this is an active topic of research, what is one popular theory about why coefficients of kinetic and static friction differs?

Answer 24

From the microscopic level, it is theorized that when an object is at rest on a surface, the atoms of both settle in between the empty spaces of each other. Therefore in order to get the object to accelerate from stationary requires a lot more energy to break not only the interactions between the surface and object, but it also needs to break off atoms in between the spaces. Whereas, the energy required to accelerate a moving object is only necessary to overcome the interactions between the moving object and surface

Question 25

What is exactly Kinetic Friction and what creates it?

Answer 25

Looking at the interaction of two objects microscopically , you will see electrons of the object interacting with the protons of the atoms of the surface. These interactions, create kinetic friction commonly known as friction aka electrostatic force

Question 26

Compare and Contrast the mathematical definitions of kinetic and static forces.

Answer 26

Both use their respective coefficient of friction and the normal force.
F_kinetic = μ_k * F_n
F_static ≤ μ_s * F_n (remember that in the process of accelerating a stationary object, as you increase the force of push on an object, the force of friction increases as well until the force you apply to it overcomes the maximum static friction)

Question 27

True or False: Friction, both static and kinetic are dependent on the surface interacting with the object.

Answer 27

False. Though both are created by the interaction of the two, both are not mathematically dependent on the surface.

Question 28

Graphically depict the relationship of static force and kinetic force to one another.

Answer 28

On a friction force vs applied force graph, static force is a linear line, whereas on its own separate line, the kinetic friction is a horizontal line of same friction force after the static force has ended. Note: any applied force above the maximum force static friction directly goes to kinetic friction.

Question 29

If F_n = 100N, μ_s = 0.7 and μ_k = 0.3. Calculate the max static force and max kinetic force

Answer 29

Fstatic = 70N
Fkinetic = 30N

Question 30

For a moving object on a flat surface, how would you calculate the net force of kinetic force and applied force?

Answer 30

When an object is moving, to find net force on the object, you subtract the applied for from kinetic friction

Question 31

Analyze the problem [Ex: Block mass 5kg, μ_s = 0.6 and μ_k = 0.55. Push the mass to the left with force of 100N. Assume no air resistance. Calculate acceleration after budging] For this example, explain why we can’t just solve for the problem through F = m*a

Answer 31

F = m*a can only be used alone when there is no friction!!!!

Question 32

Describe the relationship of μ_s and static force. Can static force be used to solve for acceleration?

Answer 32

Coefficient of static friction (μ_s) = ||F_budging|| / || F_n||.

F_b = to static force (KHAN terminology) and once this is subtracted from the applied force on the object, can then utilize the net force found to find the acceleration through F = m*a.

Question 33

Find the momentary acceleration as the object budges from its place it had a mass of 5 kg with μ_s = 0.6 and an applied force of 100N.

Answer 33

F normal = 49N
F_budging = μ_sF_n = 0.649N = 29.4N
Applying 100N to the object, will lead to 100 - 29.4N = 70.6N momentarily when budge it.
Acceleration of the mass at the momentarily budging point = 70.6N / 5 kg = 14.12 m/s^2

Question 34

Find the acceleration of the object with a constant applied force of 100N to a 5kg with μ_k = 0.55.

Answer 34

F normal = 49N
F_f = μ_kF_n = 0.5549N = 26.95N
Once start moving, the net force after 100N - 26.95N = 73.05 N
F = m*a => F/m = a -> 73.05/5kg => 14.61 m/s^2

Question 35

In a scenario in which an applied force is constant vs non constant, how would this affect the acceleration of an object after budging?

Answer 35

With constant applied force, you can use the constant force to calculate the acceleration. In the scenario where the force was applied once, you have to account for the dissipated force to move the stationary object and then calculate acceleration of the object with the leftover net force for the acceleration

Question 36

Explain why Newton’s third law plays a role in kinetic friction but not static friction.

Answer 36

While both have normal forces in their equations, kinetic friction is conceptually related to Newton’s third law while the Static friction does not. The harder the two objects are pressed together will lead to how much friction is produced from the two objects. Static friction is not determined by how hard the two objects are pressed together. BUT again!! F_n is apart of both mathematically!!!

Question 37

Which of the following does kinetic and static friction not share?
A. Change in direction of changing motion
B. Dependent on normal force
C. Not dependent on Surface area
D. Increases with increase applied force

Answer 37

D. Increases with increasing applied force. Only Static force experiences this phenomenon. Both experience A - C together.

Question 38

Assuming no friction, what would be the net force acting on a box on an incline angled at 30 degrees, with a weight of 10N?
(A) 3
(B) 5
(C) 8
(D) 15

Answer 38

(B) 5

Angle = 30 degrees

• No friction.
• Normal force cancels out the perpendicular force of gravity.

Net force = Parallel force of gravity = 10N x sin30 = 5N

Question 39

If a box on an inclined plane is held stationary by friction, and the parallel component to the force of gravity has a force of 10 N, what is the force of friction?

Answer 39

Net force = 0

Force of friction = parallel force of gravity = 10 N

Question 40

A block is sitting on a flat surface, with a weight of 100 N. If it takes a force of 73 N to just barely start moving the block, what is the coefficient of static friction (μs)?

(A) .54
(B) .62
(C) .73
(D) .81

Answer 40

(C) .73

Fs ≤ μs·Fn
73 = μs x 100
μs = 73 N/100 N
μs = 0.73