Solutions to problems on Friction-"Ten New Questions"

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Solutions to problems on Friction-"Ten New Questions"-See_Last_Blog".

EXERCISE on FRICTION  --Ten New Questions 

1.     A bamboo ladder having mass 'm' kg is resting against a smooth wall. The coefficient of friction between the ladder and the floor is µ. What are the Normal forces at each end of the ladder if it is about to slip on the floor. (Assume the weight of ladder acts at the center of the ladder). 
Answer: Let us draw a Free Body Diagram of the Ladder as below:- 

Free Body Diagram of the Ladder

        Since the ladder is stationary, the sum of all vertical forces will be zero,
weight mg acts downwards while the normal reaction at bottom R acts upwards, so
R-mg=0
→ R=mg  Newton.
Force of friction at limiting state at bottom of the ladder=µR, H is the normal force on the upper end of the ladder. Sum of these two horizontal forces must also be zero, so
H-µR = 0
→ H=µR   Newton.
    So the normal force on the top end of the ladder = µR (Horizontal) and at the bottom end = mg (Vertical) as the shown direction in the diagram.   
  

2.   A bamboo ladder of mass 20 kg is resting on a floor (having coefficient of friction µ₁ between the surfaces) against a wall (having coefficient of friction µ₂ between the surfaces). The ladder's both ends are in a limiting state of friction. Find, 

(i) Normal forces on the ladder at both ends.
(ii) Total forces at both ends.
(Assume the weight of ladder acts at the center of the ladder).      

Answer: Let us draw a Free Body Diagram of the Ladder as below:- 

Free Body Diagram of Ladder

     Normal force at the lower end of the ladder be R, so friction force = µ₁R   
        Normal force at the upper end of the ladder be H, so friction force = µ₂H  
Sum of horizontal forces will be zero, so
H-µ₁R =0
→ H = µ₁R   ...............................(i)         
Sum of vertical forces will be zero, so
R+µ₂H -mg=0
(Since weight mg acts at the center of the ladder, so R and µ₂H will be equal. It can be proved by equating moments of all forces at upper and lower ends to zero.)
→2R=mg
→R=mg/2 = 20g/2 =10g   N
and From (i), H =  µ₁R =10µ₁g    N
     So the normal force on the top end of the ladder = 10µ₁g    N (Horizontal) and at the bottom end = 10g   N (Vertical) as the shown direction in the diagram.  

3.    A person of mass 'm' kg is running on a long flat plywood board placed on a plane horizontal surface at maximum speed without slipping on the board. The coefficient of friction between his feet and the board is µ₁  and between the board and the floor is µ₂. What should be the minimum mass of the board 'M' so that it does not slip on the floor. 

Answer: Normal force on the feet of the man = weight of the man = mg. Maximum friction force on the feet = µ₁mg (in the direction of motion). An equal and opposite force will be applied on the board. Normal force on the board = (m+M)g=Total weight. At limiting state of static friction, Frictional force = µ₂(m+M)g by the floor on the board will be equal to frictional force by the feet on the board. i.e.
µ₂(m+M)g = µ₁mg
m+M = µ₁mg/µ₂g    
M = µ₁m/µ₂ -m = m(µ₁/µ₂ -1) = m(µ₁ -µ₂)/µ₂  
    

4.     Two blocks connected by a string are placed on two opposite planes as shown in the figure below. What should be the value of the coefficient of static friction µ for both blocks (with respect to the surface) to be at a critical state of static friction.

Diagram for question number -4

Answer: Let such value of the coefficient of friction be µ and the tension in the string be 'T'. The weight of 5 kg block = 5g   N and 2 kg block = 2g  N. Normal forces on each block will be equal to the component of weight perpendicular to the surface. Consider the Free Body Diagram of 5 kg block.

FBD of 5 kg block

R= Normal force=5g.cos45° 
Frictional force= µ.R= µ5g.cos45° upward along the plane. 
Component of weight along the plane = 5g.cos45° 
At the limiting friction sum of all forces along the plane will be zero, ie
5g.cos45° -T- µ5g.cos45°= 0
→T= 5g.cos45°- µ5g.cos45°=(1-µ) 5g.cos45°

Now consider the Free Body Diagram of 2 kg block,

Free body diagram of 2 kg block

 Normal force on the block =2g.cos45° 
Frictional force=µ.2g.cos45°  
Component of weight along the surface =2g.cos45°
Since the block is in limiting state, sum of forces along the surface will be zero, ie,
T-2g.cos45°-µ.2g.cos45°=0
→T=(1+µ).2g.cos45°        
Equating the two values of 'T' we get,
(1+µ).2g.cos45°= (1-µ) 5g.cos45°   
→2(1+µ)=5(1-µ)
→2+2µ=5-5µ 
→7µ=3
→µ=3/7 =0.43   

         

5.     A car moving with a speed of 30 km/hr approaches a downslope of 30° from horizontal. The driver at once sees a big tree fell across the road 50 m ahead and decides to apply hard brakes but till then the speed of the car increases to 60 km/hr on the slope. Will the car stop before hitting the tree. If yes find the distance between the tree and the stopped car. Otherwise, calculate the speed of the car at which it hits the tree. Take the mass of the car 600 kg and coefficient of friction between the tyres and the road 0.20. 

Answer: Initial velocity of the car at the time of applying brakes u=60 km/hr=60000/3600 m/s=100/6 m/s=50/3 m/s
Normal force on the car =600g.cos30° = 5196 N
Frictional force =µ.5196 N=0.2x5196 N= 1039 N
Component of weight along slope= 600g.sin30°= 6000x½=3000N
Net force along the slope=3000 N-1039 N= 1961 N
Since the net force is down the slope, the car will still accelerate along the direction of net force and it will hit the tree. Let the velocity at the time of hitting the tree be 'v'. 
Acceleration of the car 'a'= Force/mass =1961/600=3.27 m/s²
Distance of tree 's'=50 m
Using the equation v²=u²+2as
v²=(50/3)²+2x3.27x50 =2500/9+327 =278+327=605  
v=24.6 m/s =88.5 km/hr
The car will hit the tree at 88.5 km/hr. 
      

6.  A dome in the shape of a hemisphere having a radius equal to 15 m is constructed over a building. A Mason of weight 60 kg rides to the top of the dome to repair it. How far away from the crown of the dome in terms of angular distance from vertical (taking the center of the hemisphere as origin) can he stand without slipping. The coefficient of static friction between his feet and the surface of the dome is 0.3. 

Answer: Given that mass of the mason = m = 60 kg 
Coefficient of static friction = µ = 0.3 
The weight of the mason = mg
Component of weight perpendicular to surface = mg.cosθ = R= Normal force.
(θ is the angle between the vertical line and position radius of the mason, as shown in the Free body diagram below)

Free body Diagram of Mason

Component of weight tangential to the point = mg.sinθ  
Frictional force=µR=µ.mg.cosθ
At the limiting state of static friction, Component of weight mg.sinθ will be equal to Frictional force µ.mg.cosθ  
Equating the two we get, 
mg.sinθ=µ.mg.cosθ, 
tanθ=µ
θ=tan^-1µ =tan^-10.3 = 16.7° 
    

7. Seeing a crowded bus a boy rides to its roof holding window bars but there is nothing to hold there when the bus speeds up. Find the maximum acceleration of the bus on a straight horizontal road so that the boy does not slide on the roof. Mass of the boy is 50 kg and the coefficient of static friction between the body of the boy and the roof surface is 0.4. 

Answer: When the bus speeds up with an acceleration, the body of the boy feels a  backward force due to inertia. This force  P=ma,
where 'm' is the mass of boy and 'a' is the acceleration of the bus.  
Weight of the boy = mg=Normal force R
Maximum Frictional force= µR  
Just before slipping, P=µR
ma=µ.mg
→a=µg=0.4x10 =4 m/s²  (Taking g=10 m/s²)
So the maximum acceleration of the bus without slipping the boy
=a=4 m/s²
      

8.  Solve the problem no 7 if the bus accelerates on a down sloping road with an angle of inclination 30° from the horizontal. 

Answer: When the bus speeds up on a downslope of 30° the normal force on the boy changes. Component of weight perpendicular to surface = mg.cos30°
So, the Normal force on the boy = mg.cos30°
Maximum friction force at limiting stage = µ.mg.cos30°
Inertial force P=ma
Component of weight along the slope =mg.sin30°
See the Free Body Diagram of the boy

Free Body Diagram of the boy on down sloping road

Just before the boy slips, the sum of force along the slope=zero
ie, 
P-µR-mg.sin30°=0
→P = µR+mg.sin30°
→ma = µ.mg.cos30°+mg.sin30°
→a=g(µcos30°+sin30°)
→a=10(0.4x0.866 + 0.5 )
→a=8.46 m/s²
         

9.   A person is trying to pull a block of wood having a mass of 20 kg up along a sloping road with the help of a rope that makes an angle of 30° from the road. The road itself is sloping at an angle of 30° from the horizontal. What should be minimum tension in the rope to overcome the resistance and just slide up the block? The coefficient of static friction between the block and the road is 0.25.  

Answer: Let the tension in the rope be T. Total force by which the block pushes the ground perpendicularly = mg.cos30°-T.sin30° =Normal force by the ground on the block=R.
Maximum frictional force on the block =µR
=µ(mg.cos30°-T.sin30°)
Component of the weight of block along the plane = mg.sin30°
See diagram below:-

FBD of the block

At the limiting state forces along the plane will sum up to zero, ie
T.cos30°-µR-mg.sin30°=0
→T.cos30°- µ(mg.cos30°-T.sin30°) -mg.sin30°=0
→T.cos30°+µT.sin30°= mg.sin30°+µmg.cos30° 
→T=mg(sin30°+µ.cos30°)/(cos30°+µ.sin30°) 
→T=mg(0.5+0.25x0.866)/(0.866+0.25x0.5)
→T=mg(0.72/0.990) =20x10x0.73
→T=146  N 
       

10.   In the problem number 9 what should be the minimum coefficient of friction between the road and the feet of the man so that he does not slip on the road. Take the weight of the man=50 kg   

Answer: Force by which the man pushes the ground 
=mg.cos30°+T.sin30°
So the Normal force is also =mg.cos30°+T.sin30°=R
Frictional force= µR   (µ is coefficient of friction)
Weight component of man along the slope=mg.sin30°
Tension component along the slope=T.cos30°
At the limiting state total forces along the slope sum up to zero.
T.cos30°+ mg.sin30°=µR
→146x0.866+mgx0.5=µ(mg.cos30°+T.sin30°)
→126.4+5m=µ(8.7m+146x0.5)
→µ= (5m+126.4)/ (8.7m+73)=376.4/508=0.74
(Putting the value of m)

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QUESTIONS FOR SHORT ANSWER

1.         You are driving motorcycle on a horizontal road. It is moving with a horizontal velocity. Is it possible to accelerate the motorcycle without putting higher petrol input rate into the engine ?

Answer:     Yes. Since acceleration is the rate of change of velocity and velocity can even be changed by changing its direction keeping the magnitude constant. So keeping the the petrol input rate constant if the motorcycle is turned along an arc of circle it will have an acceleration towards the center of arc.        

2.           Some washing machines have cloth driers. It contains a drum in which wet clothes are kept. As the drum rotates, the water particles get separated from the cloth. The general description of this action is that "the centrifugal force throws the water particles away from the drum". Comment on this statement from the viewpoint of an observer rotating with the drum and the observer who is washing the clothes.    

Answer:   From the view point of an observer rotating with drum his frame of reference is non-inertial as it has a centripetal acceleration. To apply Newton's Laws in this frame a pseudo force has to be applied on the particle which amounts to mass multiplied by the acceleration of the frame but the direction is just opposite to the acceleration of the frame. This pseudo force is called "Centrifugal force". So from observer's viewpoint rotating with drum this statement is true. 

          But from the observer's viewpoint who is washing the clothes this statement is not true as his frame is inertial and in inertial frame there is no pseudo force. From his frame to keep a water particle in circular motion a force towards the center is needed. In the rotating drum of washing machine there is not enough force towards the center on the water particle to keep it in circular motion. So from the Newton's First Law of Motion the inertial force expels the water particles away from the drum.            

3.         A small coin is put on the record rotating at 33⅓ rev/minute. The coin does not slip on the record. Where does it gets the required centripetal force from ? 

Answer:    The required centripetal force is the force of friction.      

4.          A bird while flying takes a left turn. Where does it get the the required centripetal force from ? 

Answer:     When a bird takes left or right turn it does not remain horizontal but tilts like a plane. Thus pressing the air with greater area of the underside. As per Newton's Third Law of Motion the air underneath also exerts equal and opposite force on the bird, like normal force on a banked circular road. The horizontal component of this force is the required centripetal force for the bird.      

5.            Is it necessary to express all angles in radian while using the equation ω=ω₀+αt ?    

Answer:    No, but all angles should be in the same unit.        

6.           After a good meal at a party you wash your hands and find that you have forgotten to bring your handkerchief. You shake your hands vigorously to remove the water as much as you can. Why is water removed in this process ?

Answer:  Shaking the hand gives acceleration to he hand and to remain the water particles attached to hand require a resultant force in the direction of acceleration. The force of the adhesion between the water particles and the hand is not sufficient to transfer the force applied by hand to the particle. So due to inertial force water is removed.         

7.          A smooth block loosely fits in a circular tube placed on a horizontal surface. The block moves in a uniform circular motion along the tube (Figure 7-Q1). Which wall (inner or outer) will exert a nonzero normal contact force on the block ?

Figure for Problem no 7

   

Answer: Since the uniform circular motion has a centripetal acceleration and there must be a resultant nonzero force in the direction of acceleration ie a centripetal force. In this case the outer wall will exert a nonzero normal force on the block in the direction of center of the tube.

8.          Consider the circular motion of earth around the sun. Which of the following statements is more appropriate ?

(a) Gravitational attraction of the sun on the earth is equal to the centripetal force. 

(b) Gravitational attraction of the sun on the earth is the centripetal force.

Answer:  (b)         

9.              A car driver going at some speed 'v' suddenly finds a wide wall at distance 'r'. Should he apply brakes or turn the car in a circle of radius 'r' to avoid hitting the wall ?  

Answer:   To suggest the right action, the values of 'v', 'r' and coefficient of friction between the road and the tyres of the car must be known. Since the wall is found suddenly it is better to turn the car in a circle of radius 'r' as well as apply the brakes simultaneously to avoid hitting the wall.            

10.            A heavy mass 'm' is hanging from a string in equilibrium without breaking it. When this same mass is set into oscillation, the string breaks. Explain.   

Answer: When the heavy mass is hanging from a string and is in equilibrium without breaking it, the weight of mass is balanced by tension in the string. But when the same mass is set into oscillation it moves on an arc of a circle having radius equal to length of the string ie it has a circular motion. Due to the circular motion the mass has a centripetal acceleration and to maintain this acceleration the string applies a centripetal force on the mass in addition to the tension in equilibrium position. thus the tension in the string increases due to the oscillation and it breaks.

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