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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:-
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.
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,
Answer: Let us draw a Free Body Diagram of the Ladder as below:-
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.
(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).
(ii) Total forces at both ends.
(Assume the weight of ladder acts at the center of the ladder).
 |
| 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.
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.
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,
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
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| Diagram for question number -4 |
 |
| FBD of 5 kg block |
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.
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)
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°
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
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
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
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:-
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
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)
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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