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 :

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

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

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