Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to working with friction. The questions are categorized in terms of the characterizing features of the subject matter/ question :

Equilibrium

Stack of blocks

Problem 1 : The coefficient of friction between sand particles of uniform size is “μ”. Find the maximum height of the sand pile that can be made in a circular area of radius “r”.

Solution : We consider a sand particle on the slant side of the sand pile. We observe here that the particle on the sand pile resembles the situation that of the motion of a particle on an incline of certain angle. The pile will have the maximum height for which the friction between the particles has maximum friction i.e. limiting friction. Beyond this angle, the sand particle will slide down and spill outside the circular base boundary of the sand pile.

Figure 1: The slant side of the sand pile can be treated as an inline.

Sand pile

We know that the angle of incline corresponding to the limiting friction is the angle of repose,

tan θ = μ tan θ = μ

From the geometry of right angle ABC, we have :

tan θ = μ = A B B C = h max r tan θ = μ = A B B C = h max r

⇒ h max = μ r ⇒ h max = μ r

Problem 2 : A homogeneous chain of length “L” lies on a horizontal surface of a table. The coefficient of friction between table and chain is “μ”. Find the maximum length of the chain, which can hang over the table in equilibrium.

Figure 2: The chain is in equilibrium.

Chain lying on a table

Solution : Here, we have to consider the translational equilibrium of the chain for which net force on the chain should be zero. Now, the chain is homogeneous. It means that the chain has uniform linear mass density. Let it be “λ”. Further, let a length “y” hangs down the table in equilibrium. The length of chain residing on the table, then, is "L – y".

Figure 3: A length "y" of the chain hangs from the table.

Chain lying on a table

The mass of the chain,” m 1 m 1 ”, on the table is

⇒ m 1 = L − y λ ⇒ m 1 = L − y λ

The mass of the chain,” m 2 m 2 ”, hanging is

m 2 = y λ m 2 = y λ

Free body diagram of the mass of chain, “ m 1 m 1 ”, on the table is shown in the figure above. This mass is pulled down by the weight of the hanging chain. For balanced force condition, the friction force on the chain on the table should be equal to the weight of the hanging chain,

μ m 1 g = m 2 g μ m 1 g = m 2 g

⇒ μ L − y λ g = y λ g ⇒ μ L − y λ g = y λ g

⇒ μ L − y = y ⇒ μ L − y = y

μ L − μ y = y μ L − μ y = y

y 1 + μ = μ L y 1 + μ = μ L

⇒ y = μ L 1 + μ ⇒ y = μ L 1 + μ

Problem 3 : In the arrangement of three blocks on a smooth horizontal surface, the friction between blocks “A” and “B” is negligible, whereas coefficient of friction between blocks “B” and “C” is “μ”. Find the minimum force “F” to prevent middle block from sliding down.

Figure 4: The blocks are pushed together towards right.

Three blocks system

Solution : The block “B” can be held in position by the friction force between “B” and “C”, which operates in vertically upward direction. The friction acts in the upward direction on block “B” as it has the tendency to fall down vertically. We need to evaluate the minimum external force, “F”, which corresponds to a situation when friction in upward direction at the interface is equal to the weight of the block “B”.

Figure 5: The blocks are in equilibrium.

Three blocks system

F S = μ N F S = μ N

The friction force in vertically upward direction at the interface, however, depends on the normal force applied by the block “C” on the block “B”. It is important to understand that “N” is the normal force at the interface between blocks “B” and “C”.

We can find horizontal normal force on block “C”, using second law of motion. Referring to the free body diagram shown in the right side of the figure above.

N = m C a C N = m C a C

Here, m C = 2 m m C = 2 m . The acceleration “ m C m C ” is equal to the acceleration of the combined body as three blocks move together. Let “a” be the acceleration of the combined body. Then,

N = m C a C = 2 m a N = m C a C = 2 m a

We can find the common acceleration of the blocks, by considering three blocks together. The common acceleration for the combined body is given by :

a = F 3 m + m + 2 m = F 6 m a = F 3 m + m + 2 m = F 6 m

Hence,

⇒ N = 2 m X F 6 m ⇒ N = 2 m X F 6 m

⇒ N = F 3 ⇒ N = F 3

For preventing the middle block “B” to fall down, the forces in vertical direction on block “B” should form a balanced force system.

F S μ N = m B g F S μ N = m B g

⇒ μ F 3 = m g ⇒ μ F 3 = m g

⇒ F = 3 m g μ ⇒ F = 3 m g μ

Acceleration along incline

Problem 4 : A block “A” is placed over block “B”, whose base matches with the incline as shown in the figure. The incline is fixed on the horizontal plane. The friction between blocks is “μ”, whereas friction between block “B” and incline is negligible. Find the angle of incline for which block “A” is about to move over block “B”.

Figure 6: Block "A" is placed on a horizontal surface.

Blocks on an incline

Solution : The block “A” is about to move on the block “B”. It means that the friction between the blocks is equal to limiting friction,

F S = μ N F S = μ N

Now, we need to know the value of “N” in order to determine limiting friction. The block “A” is about to move relative to block “B”. As such, block “A” is not having relative motion with respect “B”. The two blocks move together with common acceleration, “a”, given by :

Figure 7: Acceleration of block "A".

Blocks on an incline

a = g sin θ a = g sin θ

Since we require to find normal force "N", we shall consider forces in vertical direction. For this, we need to find acceleration in vertical direction. Hence, resolving acceleration in “x” and “y’ direction, we have :

a x = a cos θ = g sin θ cos θ a x = a cos θ = g sin θ cos θ

a y = a sin θ = g sin 2 θ a y = a sin θ = g sin 2 θ

Now, the forces on block “A” in vertical (“y”) direction are (i) its weight and (ii) normal force. Thus,

m g − N = m a y = m g sin 2 θ m g − N = m a y = m g sin 2 θ

⇒ N = m g 1 − sin 2 θ = m g cos 2 θ ⇒ N = m g 1 − sin 2 θ = m g cos 2 θ

Putting expression of “N” in the equation of limiting friction, we have :

⇒ F S = μ m g cos 2 θ ⇒ F S = μ m g cos 2 θ

The limiting friction provides the acceleration in horizontal direction (note that blocks have no relative motion between them, but have accelerations with respect to ground),

F S = μ m g cos 2 θ = m a x = m g sin θ cos θ F S = μ m g cos 2 θ = m a x = m g sin θ cos θ

⇒ tan θ = μ ⇒ tan θ = μ

θ = tan - 1 μ θ = tan - 1 μ

Motion with friction

Problem 5 : A car, starting from rest, is brought to rest during which it covers a linear distance of 125 m. If the friction between tyre and road is 0.5, then what is the minimum time taken by the car to cover the distance?

Solution : The acceleration or deceleration of the car during travel is determined by the friction. The car will cover the distance in minimum time, if it moves at the maximum acceleration and breaks motion with the maximum deceleration. The journey takes equal time (say “ t 0 t 0 ”) in two phases of motion for minimum time of journey.

The limiting friction on the car is given by,

F S = μ N = μ m g F S = μ N = μ m g

Acceleration/ deceleration of car corresponding to limiting friction is given by :

a = μ m g m = μ g a = μ m g m = μ g

The velocity – time plot corresponding to the situation is given here. From the plot, the maximum speed, " v 0 v 0 ", is given by the vertical height of the triangle. The magnitude of acceleration/ deceleration is equal to the slope of velocity – time plot.

Figure 8: The car accelerates and decelerates for equal time.

Velocity time plot

⇒ a = μ g = tan θ = v 0 t 0 ⇒ a = μ g = tan θ = v 0 t 0

⇒ v 0 = μ g t 0 ⇒ v 0 = μ g t 0

The total displacement during journey is equal to the area of the triangle on velocity – time plot,

⇒ 125 = 1 2 X 2 t 0 X μ g t 0 = t 0 X 0.5 X 10 t 0 = 5 t 0 2 ⇒ 125 = 1 2 X 2 t 0 X μ g t 0 = t 0 X 0.5 X 10 t 0 = 5 t 0 2

⇒ t 0 2 = 25 ⇒ t 0 2 = 25

⇒ t 0 = 5 s ⇒ t 0 = 5 s

Hence, time taken to cover the distance is :

⇒ 2 t 0 = 2 X 5 = 10 s ⇒ 2 t 0 = 2 X 5 = 10 s

Contact force on an incline

Problem 6 : A block is placed on a rough incline whose angle can be varied as shown in the figure. Draw a plot between contact force and angle of incline (θ) for 0°≤θ≤90° (only show the nature of the plot).

Figure 9: Angle of incline varies.

A block on an incline

Solution : The contact force comprises of normal force and friction. When θ = 90°, there is no normal force. Hence, there is no friction either. The block is under free fall. And,

F C = 0 F C = 0

We have studied that the block does not slide till the angle of incline is equal to angle of repose. The angle of repose is given by :

⇒ θ = tan - 1 μ ⇒ θ = tan - 1 μ

Till this angle, the block is stationary and the contact forces (normal and friction forces) and gravity form the balanced force system. It means that contact force is equal to the weight of the block in magnitude. Hence,

Figure 10: Before motion, the contact force is equal to the weight of the block in magnitude.

A block on an incline

F C = m g = N 2 + f s 2 F C = m g = N 2 + f s 2

When the angle of incline exceeds the value of angle of repose, then the component of weight parallel to the incline is greater than the friction force. As such, block begins to slide down with kinetic friction. The contact force beyond angle of repose, therefore, is given by :

⇒ F C = { m g cos θ 2 + μ m g cos θ 2 } ⇒ F C = { m g cos θ 2 + μ m g cos θ 2 }

⇒ F C = m g cos θ { 1 + μ 2 } ⇒ F C = m g cos θ { 1 + μ 2 }

The plot in this region is a portion of cosine curve for ⇒ θ > tan - 1 μ ⇒ θ > tan - 1 μ The figure below shows the plot for the complete range.

Figure 11: Plot of contact force and angle of incline.

Contact force