Friction

Question 1

Draw the graph of external Force F and friction f? Explain the various segments?

Answer 1

1. Segment ab - Block is at rest, F is not sufficient to move block. Static friction opposes the applied F
2. At b, Static friction reaches its max value, also called at f_l
3. Segment cd : Block is moving Kinetic friction takes over. Kinetic friction is less than Max static friction. Kinetic friction is always constant and does not vary

Question 2

Which form of friction is higher ? What is the maximumim value of friction?

Answer 2

1. The Maximum value of static friction is always higher than the Kinectic friction
   f_s-max > f_k
2. The maximum value of friction is
   f_s-max = 𝛍_sN

Question 3

What is the direction of Friction?

Answer 3

Friction is always in the **opposite direction **to the motion of a body or the relation motion between two bodies.

Question 4

What is the formula for the two frictional forces in the below system?

Answer 4

N₂ = m₂g

N₁ = (m₁+m₂)g

Assuming slipping at all places
f₁ = 𝛍₁N₁ = 𝛍₁(m₁+m₂)g
f₂ = 𝛍₂N₂ = 𝛍₂m₂g

Question 5

What takes more effort to over to overcome friction - pushing or pulling?

Answer 5

Pushing takes more over effort to overcome friction

Question 6

Why does pushing take more effort to overcome friction.

Answer 6

When you are pushing a block, the pushing force exerts a component of force (FSinϴ) vertically down. This adds to the down force and hence the Normal acting on the body by the surface increases by this amount.
N_push = mg + FSinϴ

Where while pulling at the same angle, the pulling force exerts a component of force (FSinϴ) vertically updward direction. This reduces downwards and hence the Normal acting on the body by the surface decreases by this amount.
N_pull = mg - FSinϴ

Question 7

What is the minimum pulling force required to be applied for a block to start slipping?

Answer 7

To start slipping, horizontal component of applied force should be equal to the max frictional force

FCosϴ = f_s-max = 𝛍N
FCosϴ = 𝛍 (mg - FSinϴ)
F = 𝛍mg / (Cosϴ + 𝛍Sinϴ)

To minimise F, (Cosϴ + 𝛍Sinϴ) needs to maxmized. Which means differentiating (Cosϴ + 𝛍Sinϴ) and equating to zero. This gives us
tanϴ = 𝛍
F_min = 𝛍mg / (√(𝛍² + 1))

Question 8

Define the contact force and provide its formula? What are its minimum and maximum values?

Answer 8

Contact force is the resultant of the Normal and the Friction force.
It is given by
F_c = √(N²+f²)
F_{c, min} = N, as f = 0
F_{c, max} = N √(1+𝛍²), as f = 𝛍N

Question 9

What is the angle of repose? Derive the angle of repose.

Answer 9

Angle of repose is the minimum angle at which the body slips due to its weight.
Mathematically, block will start to slip when
mgsin𝝷 = f_{s, max},
mgsin𝝷 = 𝛍N, N = mgcos𝝷
mgsin𝝷 = 𝛍mgcos𝝷
tan𝝷 = 𝛍 or 𝝷 = tan^-1𝛍

1. If 𝝷 < tan^-1𝛍, then no slipping
2. 1. If 𝝷 > tan^-1𝛍, then block starts to slip

Question 10

What is the acceleration of the block on a rough inclined plane?

Answer 10

Using NLM,
mgsin𝝷 - f = ma
mgsin𝝷 - 𝛍N= ma
mgsin𝝷 - 𝛍mgcos𝝷= ma
⇒ a= gsin𝝷 - 𝛍gcos𝝷

Question 11

Briefly explain the movement and forces at play in a block over block system?

Answer 11

1. Block B will never move before Block A
2. Minimum force F required to move A = 𝛍 (m₁+m₂)g
3. On further increasing force F, upto a F_{max</max> , then m2 and m1 moves together (doesn’t slip).}

Question 12

Consider a block over block system acted upon by a force F on the lower block? What is the acceleration and the maximum Force required to move both blocks without the top block slipping?

Answer 12

Since the top block isnt slipping, its acceleration is same as of the system.
Applying NLM on top block, we get
f_{2, max} = m₂a,
𝛍₂m₂g = m₂a,
⇒ a = 𝛍₂g….(i)

Applying NLM to lower block, we get
F_max - f₁ = (m₁+m₂)a, substituting for a from (i)
F_max - f₁ = (m₁+m₂)𝛍₂g

Question 13

What is the maximum hanging length of the chain after which it starts to slip?

Answer 13

Using NLM on the non-hanging portion of chain on the surface, we get
f_max = 𝛍N,
∵ N = mass of non-hanging length * g, we have
f_max = 𝛍(m/L * (L-x)) g
we also know that
f_max = m₁g where m₁ is the hanging mass.
Substituting we get
m₁g = 𝛍(m/L * (L-x)) g
∴ x = 𝛍L / (1+𝛍)