Solutions to Exercises on Friction HC Verma's Concepts of Physics Part 1 (17-21)

 Q#17

A car starts from rest on a half km long bridge. The coefficient of friction between the tyre and the road is 1.0. Show that one can not drive through the bridge in less than 10 s.

Answer:
Let mass of the car = m, weight = mg = Normal force

Maximum force of friction available that will help drive the car = µmg = mg     (Given µ = 1)
So, the maximum acceleration the car can have = Force/mass = mg/m = g =10 m/s²

Now initial velocity u = 0, distance s = 500 m, acceleration a =  10 m/s², time taken t= ?

From s = ut + ½at²

500 = 0 + ½(10)t²

t² = 100

t = 10 s

So, one can not drive through the bridge in less than 10 s.

Q#18
Figure (6-E6) shows two blocks in contact sliding down an inclined surface of inclination 30°. The friction coefficient between the block of mass 2.0 kg and the incline is µ₁, and that between the block of mass 4.0 kg  and the incline is µ₂. Calculate the acceleration of 2.0 kg block if  (a) µ₁= 0.20 and µ₂ = 0.30 , (b) µ₁ = 0.30 and µ₂ = 0.20. Take g = 10 m/s².

Answer:
Let us first calculate the accelerations of blocks assuming that they are not in contact. For 2 kg block,
Force down the incline = component of weight-friction force

= 2gsin30° – µ₁(2g)cos30°

acceleration a₁ = Force/mass = gsin30° – µ₁gcos30° = 10(½ – ½µ₁√3)

Similarly, For 4 kg block,

acceleration a₂ = gsin30° – µ₂gcos30° = 10(½ – ½µ₂√3)

(a) For µ₁ = 0.20 and µ₂ = 0.30

a₁ = 10(½ - ½ µ₁√3) = 5 – 10(0.20)(√3/2) = 3.27 m/s²

a₂ = 10(½ – ½µ₂√3) = 5 – 10(0.30)(√3/2) = 2.40 m/s²

Since a₁ > a₂, so sliding together upper and lower blocks will push each other, let the joint acceleration be 'a'. Now considering the 2 kg block an extra force 'F' exerted by the lower block will act on it. Now force on the block

= 2gsin30° – µ₁(2g)cos30° – F

and acceleration a = (2gsin30° –µ₁(2g)cos30° –F)/2                         (i)

Considering 4 kg block,

force on the block = 4gsin30°– µ₂(4g)cos30° + F

and acceleration a = (4gsin30° – µ₂(4g)cos30° + F)/4                       (ii)

Equating (i) and (ii)

[2gsin30° – µ₁(2g)cos30° – F]/2 = [4gsin30° – µ₂(4g)cos30° + F]/4

sin30° – µ₁cos30° – F/2g = sin30° – µ₂cos30° + F/4g

3F/4g  = µ₂ cos30° – µ₁cos30°

3F/4g = (0.30 – 0.20)(√3/2) = 0.05√3

F = (4g/3)(0.05√3) = 2√3/3

So, acceleration

a = (2gsin30° – µ₁(2g)cos30° –F)/2             from                  (i)

= g(½) – 0.20g(√3/2) – √3/3

a = 5 – √3 – √3/3 = 2.70 m/s²

Acceleration of of 2.0 kg block  a = 2.70 m/s².

(b) For µ₁ = 0.30 and µ₂ = 0.20
a₁ = 10(½ – ½µ₁√3) = 5 – 10(0.30)(√3/2) = 2.41 m/s²

a₂ = 10(½ – ½µ₂√3) = 5 – 10(0.20)(√3/2) = 3.27 m/s²

Since acceleration of lower block is more than the upper one, so both blocks would not remain in contact. Acceleration of of 2.0 kg block  a₁ = 2.41 m/s².

Q#19
Two masses M₁ and M₂ are connected by a light rod and the system is slipping down a rough incline of angle θ with the horizontal. The friction coefficient at both the contacts is µ. Find the acceleration of the system and the force by the rod on one of the blocks.

Answer:
Let the force by the rod on one of the blocks be F and acceleration of the system = a.
For upper block of mass M₁, Net Force down the plane

= M₁gsinθ – µM₁gcosθ – F

Acceleration
a = (M₁gsinθ – µM₁gcosθ – F)/M₁

= gsinθ – µgcosθ – F/M₁

For lower block M₂, F will act along the motion, So net force down the plane = Weight component – Friction force + F

= M₂gsinθ – µM₂gcosθ + F

Acceleration
a = (M₂gsinθ – µM₂gcosθ + F)/M₂

= gsinθ – µgcosθ + F/M₂

Since both blocks are connected, their accelerations will be the same.
Equating the accelerations,

Gsinθ – µgcosθ – F/M₁ = gsinθ – µgcosθ + F/M₂

F/M₁ + F/M₂ = 0   

F(1/M₁ +1/M₂) = 0

F = 0

So the force by the rod on one of the blocks is zero.
Now acceleration

a = gsinθ – µgcosθ + F/M₁

 = g(sinθ – µcosθ)

Q#20
A block of mass M is kept on a rough horizontal surface. The coefficient of static friction between the block and the surface is µ. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block? In which direction should this force act?

Answer:
Let the pull force F be applied to the block at an angle θ from the horizontal, See figure below:

We resolve force F in vertical and horizontal directions as Fsinθ and Fcosθ. Force applied by the block on the surface

= Weight – vertical component of F

= Mg – Fsinθ = Normal force by the surface on the block

Given Static friction coefficient = µ,

So, Maximum Friction force = µ(Mg – Fsinθ)   {In horizontal direction}

Hence to move the block horizontal component of F must be just greater than this friction force,
i.e.
Fcosθ = µ(Mg – Fsinθ)

Fcosθ + µFsinθ = µMg

F = µMg/(cosθ + µsinθ)

Since µ, M and g are constant in this case, so numerator of this expression is fixed. The force F will be minimum if denominator of the above expression is maximum,

i.e. (cosθ + µsinθ) should be maximum. To find the θ for which it is maximum we differentiate it with respect to θ and equate to zero.

Let y = (cosθ + µsinθ)

dy/dθ = –sinθ + µcosθ = 0

tanθ = µ

θ = tan^-1µ

Since dy/dθ is zero for both at maxima and minima, to know whether it is maxima and minima we again differentiate it,

d²y/dθ² = –(cosθ + µsinθ)

For a positive value of µ, 0°< θ <90°, and cosθ+µsinθ will be positive, Hence d²y/dθ² = –ve which gives that it is a maxima value.

So, the force should be applied at an angle θ = tan^-1µ from the horizontal to move the block.
And minimum force needed to slide the block is

= µMg/(cosθ + µsinθ)

= µMgsecθ/(1 + µtanθ)         {Divide numerator &denominator by cosθ}

= µMg√sec²θ/(1 + µ²)

= µMg√(1 + tan²θ)/(1+µ²)

= µMg√(1 + µ²)/(1 + µ²)

= µMg/√(1+µ²)

Q#21
The friction coefficient between the board and the floor shown in figure (6-E7) is µ. Find the maximum force that the man can exert on the rope so that the board does not slip on the floor.

Answer:
Let the maximum force be P. At the moment when P is applied by the man his effective weight on the board reduces to Mg-P. Total force applied on the floor is equal to weight of the man plus weight of the board, i.e.

= Mg – P + mg = Normal force by the plane on the board.

In this case maximum friction force = µ(Mg – P + mg)

This maximum friction force is equal to horizontal pull by the string which in turn is equal to P, equating these two,

µ(Mg – P + mg) = P

µP + P = µ(M + m)g

P = µ(M + m)g/(1 + µ)

Share :