In the previous section, we saw the condition for rolling without slipping. We also saw kinetic energy of a rolling body. In this section, we will see the 'effect of friction' in rolling motion
For that, first we must understand how friction helps us to walk on the ground. We will write it in steps:
1. We know that, it is difficult (and dangerous) to walk on an icy floor
■ Why is that so?
2. Let us say, a person take the first forward step with his right foot
• The blue rectangle in fig.7.140(a) represents his right shoe
• He presses the right shoe onto the floor to get a grip
3. The floor must hold the shoe in position
• If proper grip is available, the man can pull himself forward
4. 'Pulling oneself forward' is a horizontal motion
• The man will be exerting a horizontal force $\mathbf\small{\vec{F}}$ through his right foot. This is shown in fig.a
• So the floor must hold the shoe in such a way that, the shoe does not move horizontally in the direction of $\mathbf\small{\vec{F}}$
5. But the icy floor will not be able to hold the shoe in such a manner
• This is because, there is no friction
6. So when he tries to pull himself forward, the shoe will slip backwards
• He will not be able to move forward
• In fact, if he does not hold on to some proper support with his hands, he will slip and fall, causing serious injuries
7. Things are different if he wears spiked shoes which are manufactured specially for icy conditions
• The spikes will penetrate into the ice. This is shown in fig.b
• So the ice will be able to keep his feet in position
8. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the man pulls himself forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the backward direction
(iv) The ice in turn, applies an equal forward horizontal force (in accordance with Newton's third law) on the spikes
(v) This forward force is the thrust that helps the man to move forward
9. Let us see how this forward thrust is related to friction:
(i) We saw that, the forward thrust is produced due to the interlocking between the spikes and the ice
(ii) On ordinary floors, spikes are not required because there are already numerous ridges and valleys
(iii) These ridges and valleys will do the interlocking
(iv) These ridges and valleys is the cause of friction. We saw this in an earlier chapter (Fig.5.76 in chapter 5.19). There we also saw that, frictional force will always be parallel to the surfaces in contact
(v) So the 'forward thrust that helps in walking', is in fact the frictional force. This frictional force is denoted as $\mathbf\small{\vec{f_s}}$ in fig.b
10. Next we want to determine whether this frictional force is static or kinetic
(i) Consider the following instant:
The instant at which the man pulls himself forward
(ii) At that instant, the frictional force acts
• But at that instant, actual motion has not begun. In other words, at that instant, there is no relative motion between the shoe and the ice
(iii) So the friction which helps the forward motion is static friction
11. Also recall that static friction is a self adjusting force
♦ It increases when the applied force increases
♦ It decreases when the applied force decreases
• The maximum possible friction that can be expected is $\mathbf\small{\vec{f}_{s,max}}$
• This maximum force depends on the nature of the ice and the spikes
• When the man moves forward, the full $\mathbf\small{\vec{f}_{s,max}}$ need not be necessarily mobilized
• A force lesser than $\mathbf\small{\vec{f}_{s,max}}$ may be sufficient
12 Another interesting aspect:
(i) We saw that, the spikes on the right shoe helps the man to pull himself forward
(ii) Once he has pulled himself forward, he puts the left foot forward and lifts the right foot
(iii) Now, it is the duty of the spikes on the left shoe to provide the necessary forward force
(iv) Thus we see that, the left and right spikes 'alternately and continuously' provide the necessary forward force. So the man is able to walk properly
• In fig.140 (c), the shoe is being pulled by an external force $\mathbf\small{\vec{F}}$
♦ This external pulling force can be provided by a snowmobile for example
• If the man can safely provide proper support to the upper part of his body (like by holding on to a rope attached to the snowmobile), he can slide over the snow
• In this case, the man does not have to put up any effort for the forward movement
♦ The effort is done by the snow mobile
14. But if he uses spiked shoes, things will be different
• The spikes will penetrate into the ice. This is shown in fig.d
• So the ice will try to keep his feet in position
15. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the shoe is pulled forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the forward direction
(iv) The ice in turn, applies an equal backward horizontal force (in accordance with Newton's third law) on the spikes
(v) This backward force tries to prevent the shoe from moving forward
• And we know that, this force on the spikes is the frictional force
16. Here, the force which tries to prevent motion is the static friction
• But once the motion begins, kinetic friction takes over
17. Now we will write a summary:
• When the man propels himself, the friction acts in the forward direction
• When an external force propels him, the friction acts in the backward direction
• While walking, the left and right feet are in contact with the ground alternately and continuously
• For tires, succeeding portions of the tire comes into contact with the ground continuously
• Each portion of the tire must receive proper grip from the ground
• Icy floor cannot provide this grip
• So the 'portion of the tire' in contact with ice will slip
• Succeeding portions of the tire will also slip
• This causes the tire to spin. It will not roll
We will consider two cases where vehicle tires spin:
A. The case of a car
B. The case of a cart
A. The case of a car
1. Consider fig.7.141 below:
• It is a schematic arrangement of the rear wheels of a car
♦ The two rear wheels are shown in yellow color
♦ The axle connecting the two wheels is shown in cyan color
♦ The engine is shown as a block in grey color
2. Remember that, this is only a schematic arrangement. The actual arrangements in a car will be different
• But all the information that is relevant for our present discussion are the following four:
(i) The engine rotates the axle
♦ The rotation of the axle is indicated by the white curved arrow
(ii) Since the wheels are connected to the axle, the wheels also will rotate
♦ The rotation of the wheel is indicated by the red curved arrow
(iii) If there is no slipping between the wheels and the ground, the wheels will begin to roll forward
(iv) Thus the car as a whole moves forward
3. So we see that, there is no 'push' or 'pull' on the wheels of a car
• The movement of the wheels is achieved by providing a rotational motion to the wheels
B. The case of a cart
1. Consider fig.7.142 below:
• It is a schematic arrangement of the rear wheels of a cart
♦ The two rear wheels are shown in yellow color
♦ The axle connecting the two wheels is shown in cyan color
♦ There is no engine. Instead, a cyan rod is connected to the axle
2. Remember that, this is only a schematic arrangement. The actual arrangements in a cart will be different
• But all the information that is relevant for our present discussion are the following four:
(i) A pulling force is applied on the rod connected to the axle. As a result, the axle moves forward
(ii) Since the wheels are connected to the axle, the wheels also move forward
(iii) If there is no slipping between the wheels and the ground, the wheels will begin to roll forward
(iv) Thus the cart as a whole moves forward
3. So we see that, there is indeed a 'push' or 'pull' on the wheels of a cart
• The push or pull is applied at the center of the wheels
■ We can write a summary:
• In the case of a car, the wheels are made to roll by applying a rotational motion to the wheels
• In the case of a cart, the wheels are made to roll by applying a force at the center of the wheels
First we will consider the wheel of a car. We will write the analysis in steps:
1. Fig.7.143 (a) below shows one of the rear wheels of a car
• It is a side view. So the engine is not visible
• In the fig., the engine is transmitting power to the wheel. The wheel is spinning
• But the car is resting on icy floor
• The icy floor is so smooth that, there is no friction
2. consider the instant at which any random point on the periphery of the wheel is in contact with the icy floor
3. The floor must hold that point in position
• If proper grip is available, the 'wheel as a whole' can pull itself forward
4. 'Pulling oneself forward' is a horizontal motion
• The wheel will be exerting a horizontal force $\mathbf\small{\vec{F}}$ through the point of contact. This is shown in fig.a
• So the floor must hold the point in such a way that, it does not move horizontally in the direction of $\mathbf\small{\vec{F}}$
5. But the icy floor will not be able to hold the point in such a manner
• This is because, there is no friction
6. So when the wheel tries to pull itself forward, the point of contact on the wheel will slip backwards
• The succeeding point will come into contact with the ice. It also will experience the same 'no grip condition' and will slip
• This process continues and so the wheel will just spin. It will not be able to roll
7. Things are different if the wheels are spiked
• The spikes will penetrate into the ice. This is shown in fig.b
• So the ice will be able to keep the point in position
8. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the wheel spin further, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the backward direction
(iv) The ice in turn, applies an equal forward horizontal force (in accordance with Newton's third law) on the spikes
(v) This forward force is the thrust that helps the wheel to roll forward
9. Let us see how this forward thrust is related to friction:
• We saw that, the forward thrust is produced due to the interlocking between the spikes and the ice
• On ordinary floors, spikes are not required because there are already numerous ridges and valleys
• These ridges and valleys is the cause of friction
• We saw this in an earlier chapter (Fig.5.76 in chapter 5.19)
• Also we saw that, frictional force will always be parallel to the surfaces in contact
• So the 'forward thrust that helps in rolling', is in fact the frictional force
• This frictional force is denoted as $\mathbf\small{\vec{f_s}}$ in fig.b
10. Next we want to determine whether this frictional force is static or kinetic
• Consider the following instant:
The instant at which the wheel pulls itself forward
• At that instant, the frictional force acts
• But at that instant, actual motion has not begun. In other words, at that instant, there is no relative motion between the point and the ice
• So the friction which helps the forward motion is static friction
11. Also recall that static friction is a self adjusting force
♦ It increases when the applied force increases
♦ It decreases when the applied force decreases
• The maximum possible friction that can be expected is $\mathbf\small{\vec{f}_{s,max}}$
• This maximum force depends on the nature of the ice and the spikes
• When the wheel moves forward, the full $\mathbf\small{\vec{f}_{s,max}}$ need not be necessarily mobilized
• A force lesser than $\mathbf\small{\vec{f}_{s,max}}$ may be sufficient
12 Another interesting aspect:
• We saw that, the spikes on the wheel helps the wheel to pull itself forward
• Once the wheel has pulled itself forward, the succeeding point comes into contact with the ice
• Now, it is the duty of the 'spikes at the new point of contact' to provide the necessary forward force
• This process continues and so the wheel is able to roll forward
• The spikes will penetrate into the ice. This is shown in fig.b
• So the ice will try to keep the point of contact in position
4. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the wheel is pulled forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the forward direction
(iv) The ice in turn, applies an equal backward horizontal force (in accordance with Newton's third law) on the spikes
(v) Note that, this backward horizontal force is applied by the ice on the spikes
• That means, the spikes experience this force
• Since the spikes are attached rigidly at the periphery of the wheel, the wheel will experience this force
• The wheel will experience this force as:
A force acting tangentially at the periphery
• So this is the force that we were waiting for. It does not pass through the center 'C'
• The wheel will begin to rotate
5. We know that, this force on the spikes is the frictional force
■ So we can write:
Friction helps the wheel of a cart to roll. If there is no friction, the wheel will only slide
♦ The rear wheels of a car
♦ The wheels of a cart
• It can be written in two steps:
1. For the rear wheels of a car, rotation is already present. Because, the engine provides rotation to the axle, and so the wheels which are connected to the axle naturally rotates. If there is no friction, these rotating tires will just spin. The car will not move
• For the rear wheels of a cart, rotation is not readily present. Even if the cart is pulled forward, the wheels will rotate only if friction is present. In the absence of friction, the rear wheels of a cart can only slide
2. Frictional force, if present, acts on the rear wheel of a car
♦ The direction of this frictional force is same as the direction of motion of the car
• Frictional force, if present, acts on the wheels of a cart
♦ The direction of this frictional force is opposite to the direction of motion of the cart
1. A car with an engine can self propagate
• A man is also self propagating when he is walking
• So the two sets of figs. are comparable:
♦ Set 1: Figs.7.143 (a) and (b)
♦ Set 2: Figs.7.140 (a) and (b)
2. A cart without an engine cannot self propagate. It needs an external pulling force
• A man trying to slide behind a snowmobile also needs the external pulling force
• So the two sets of figs. are comparable:
♦ Set 1: Figs.7.144 (a) and (b)
♦ Set 2: Figs.7.140 (c) and (d)
For that, first we must understand how friction helps us to walk on the ground. We will write it in steps:
1. We know that, it is difficult (and dangerous) to walk on an icy floor
■ Why is that so?
2. Let us say, a person take the first forward step with his right foot
• The blue rectangle in fig.7.140(a) represents his right shoe
• He presses the right shoe onto the floor to get a grip
 |
| Fig.7.140 |
• If proper grip is available, the man can pull himself forward
4. 'Pulling oneself forward' is a horizontal motion
• The man will be exerting a horizontal force $\mathbf\small{\vec{F}}$ through his right foot. This is shown in fig.a
• So the floor must hold the shoe in such a way that, the shoe does not move horizontally in the direction of $\mathbf\small{\vec{F}}$
5. But the icy floor will not be able to hold the shoe in such a manner
• This is because, there is no friction
6. So when he tries to pull himself forward, the shoe will slip backwards
• He will not be able to move forward
• In fact, if he does not hold on to some proper support with his hands, he will slip and fall, causing serious injuries
7. Things are different if he wears spiked shoes which are manufactured specially for icy conditions
• The spikes will penetrate into the ice. This is shown in fig.b
• So the ice will be able to keep his feet in position
8. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the man pulls himself forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the backward direction
(iv) The ice in turn, applies an equal forward horizontal force (in accordance with Newton's third law) on the spikes
(v) This forward force is the thrust that helps the man to move forward
9. Let us see how this forward thrust is related to friction:
(i) We saw that, the forward thrust is produced due to the interlocking between the spikes and the ice
(ii) On ordinary floors, spikes are not required because there are already numerous ridges and valleys
(iii) These ridges and valleys will do the interlocking
(iv) These ridges and valleys is the cause of friction. We saw this in an earlier chapter (Fig.5.76 in chapter 5.19). There we also saw that, frictional force will always be parallel to the surfaces in contact
(v) So the 'forward thrust that helps in walking', is in fact the frictional force. This frictional force is denoted as $\mathbf\small{\vec{f_s}}$ in fig.b
10. Next we want to determine whether this frictional force is static or kinetic
(i) Consider the following instant:
The instant at which the man pulls himself forward
(ii) At that instant, the frictional force acts
• But at that instant, actual motion has not begun. In other words, at that instant, there is no relative motion between the shoe and the ice
(iii) So the friction which helps the forward motion is static friction
11. Also recall that static friction is a self adjusting force
♦ It increases when the applied force increases
♦ It decreases when the applied force decreases
• The maximum possible friction that can be expected is $\mathbf\small{\vec{f}_{s,max}}$
• This maximum force depends on the nature of the ice and the spikes
• When the man moves forward, the full $\mathbf\small{\vec{f}_{s,max}}$ need not be necessarily mobilized
• A force lesser than $\mathbf\small{\vec{f}_{s,max}}$ may be sufficient
12 Another interesting aspect:
(i) We saw that, the spikes on the right shoe helps the man to pull himself forward
(ii) Once he has pulled himself forward, he puts the left foot forward and lifts the right foot
(iii) Now, it is the duty of the spikes on the left shoe to provide the necessary forward force
(iv) Thus we see that, the left and right spikes 'alternately and continuously' provide the necessary forward force. So the man is able to walk properly
13. Now, instead of walking motion, let us see another type of motion:
♦ This external pulling force can be provided by a snowmobile for example
• If the man can safely provide proper support to the upper part of his body (like by holding on to a rope attached to the snowmobile), he can slide over the snow
• In this case, the man does not have to put up any effort for the forward movement
♦ The effort is done by the snow mobile
14. But if he uses spiked shoes, things will be different
• The spikes will penetrate into the ice. This is shown in fig.d
• So the ice will try to keep his feet in position
15. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the shoe is pulled forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the forward direction
(iv) The ice in turn, applies an equal backward horizontal force (in accordance with Newton's third law) on the spikes
(v) This backward force tries to prevent the shoe from moving forward
• And we know that, this force on the spikes is the frictional force
16. Here, the force which tries to prevent motion is the static friction
• But once the motion begins, kinetic friction takes over
17. Now we will write a summary:
• When the man propels himself, the friction acts in the forward direction
• When an external force propels him, the friction acts in the backward direction
■ The above situations arise in the case of vehicle tires also
• For tires, succeeding portions of the tire comes into contact with the ground continuously
• Each portion of the tire must receive proper grip from the ground
• Icy floor cannot provide this grip
• So the 'portion of the tire' in contact with ice will slip
• Succeeding portions of the tire will also slip
• This causes the tire to spin. It will not roll
We will consider two cases where vehicle tires spin:
A. The case of a car
B. The case of a cart
A. The case of a car
1. Consider fig.7.141 below:
 |
| Fig.7.141 |
♦ The two rear wheels are shown in yellow color
♦ The axle connecting the two wheels is shown in cyan color
♦ The engine is shown as a block in grey color
2. Remember that, this is only a schematic arrangement. The actual arrangements in a car will be different
• But all the information that is relevant for our present discussion are the following four:
(i) The engine rotates the axle
♦ The rotation of the axle is indicated by the white curved arrow
(ii) Since the wheels are connected to the axle, the wheels also will rotate
♦ The rotation of the wheel is indicated by the red curved arrow
(iii) If there is no slipping between the wheels and the ground, the wheels will begin to roll forward
(iv) Thus the car as a whole moves forward
3. So we see that, there is no 'push' or 'pull' on the wheels of a car
• The movement of the wheels is achieved by providing a rotational motion to the wheels
B. The case of a cart
1. Consider fig.7.142 below:
 |
| Fig.7.142 |
♦ The two rear wheels are shown in yellow color
♦ The axle connecting the two wheels is shown in cyan color
♦ There is no engine. Instead, a cyan rod is connected to the axle
2. Remember that, this is only a schematic arrangement. The actual arrangements in a cart will be different
• But all the information that is relevant for our present discussion are the following four:
(i) A pulling force is applied on the rod connected to the axle. As a result, the axle moves forward
(ii) Since the wheels are connected to the axle, the wheels also move forward
(iii) If there is no slipping between the wheels and the ground, the wheels will begin to roll forward
(iv) Thus the cart as a whole moves forward
3. So we see that, there is indeed a 'push' or 'pull' on the wheels of a cart
• The push or pull is applied at the center of the wheels
■ We can write a summary:
• In the case of a car, the wheels are made to roll by applying a rotational motion to the wheels
• In the case of a cart, the wheels are made to roll by applying a force at the center of the wheels
■ We will analyse the role of friction in each case
1. Fig.7.143 (a) below shows one of the rear wheels of a car
• It is a side view. So the engine is not visible
• In the fig., the engine is transmitting power to the wheel. The wheel is spinning
• But the car is resting on icy floor
• The icy floor is so smooth that, there is no friction
 |
| Fig.7.143 |
3. The floor must hold that point in position
• If proper grip is available, the 'wheel as a whole' can pull itself forward
4. 'Pulling oneself forward' is a horizontal motion
• The wheel will be exerting a horizontal force $\mathbf\small{\vec{F}}$ through the point of contact. This is shown in fig.a
• So the floor must hold the point in such a way that, it does not move horizontally in the direction of $\mathbf\small{\vec{F}}$
5. But the icy floor will not be able to hold the point in such a manner
• This is because, there is no friction
6. So when the wheel tries to pull itself forward, the point of contact on the wheel will slip backwards
• The succeeding point will come into contact with the ice. It also will experience the same 'no grip condition' and will slip
• This process continues and so the wheel will just spin. It will not be able to roll
7. Things are different if the wheels are spiked
• The spikes will penetrate into the ice. This is shown in fig.b
• So the ice will be able to keep the point in position
8. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the wheel spin further, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the backward direction
(iv) The ice in turn, applies an equal forward horizontal force (in accordance with Newton's third law) on the spikes
(v) This forward force is the thrust that helps the wheel to roll forward
9. Let us see how this forward thrust is related to friction:
• We saw that, the forward thrust is produced due to the interlocking between the spikes and the ice
• On ordinary floors, spikes are not required because there are already numerous ridges and valleys
• These ridges and valleys is the cause of friction
• We saw this in an earlier chapter (Fig.5.76 in chapter 5.19)
• Also we saw that, frictional force will always be parallel to the surfaces in contact
• So the 'forward thrust that helps in rolling', is in fact the frictional force
• This frictional force is denoted as $\mathbf\small{\vec{f_s}}$ in fig.b
10. Next we want to determine whether this frictional force is static or kinetic
• Consider the following instant:
The instant at which the wheel pulls itself forward
• At that instant, the frictional force acts
• But at that instant, actual motion has not begun. In other words, at that instant, there is no relative motion between the point and the ice
• So the friction which helps the forward motion is static friction
11. Also recall that static friction is a self adjusting force
♦ It increases when the applied force increases
♦ It decreases when the applied force decreases
• The maximum possible friction that can be expected is $\mathbf\small{\vec{f}_{s,max}}$
• This maximum force depends on the nature of the ice and the spikes
• When the wheel moves forward, the full $\mathbf\small{\vec{f}_{s,max}}$ need not be necessarily mobilized
• A force lesser than $\mathbf\small{\vec{f}_{s,max}}$ may be sufficient
12 Another interesting aspect:
• We saw that, the spikes on the wheel helps the wheel to pull itself forward
• Once the wheel has pulled itself forward, the succeeding point comes into contact with the ice
• Now, it is the duty of the 'spikes at the new point of contact' to provide the necessary forward force
• This process continues and so the wheel is able to roll forward
Now we will consider the cart
1. Fig.7.144 (a) below shows one of the rear wheels of a cart
• It is a side view.
• In the fig., a horizontal force is being applied at the center of the wheel
• But the cart is resting on icy floor
• The icy floor is so smooth that, there is no friction
2. We see the weight vector $\mathbf\small{m\vec{g}}$ in the down ward direction
♦ This vector passes through the center 'C' of the wheel
• We see the normal reaction vector $\mathbf\small{\vec{N}}$
♦ This vector also passes through the center 'C' of the wheel
• We see no other forces
• Forces passing through the center of mass of a body cannot cause it to rotate. So the wheel will not rotate
• When the cart is pulled. the wheel will just slide
3. But if the spiked wheels are used, things will be different
• It is a side view.
• In the fig., a horizontal force is being applied at the center of the wheel
• But the cart is resting on icy floor
• The icy floor is so smooth that, there is no friction
 |
| Fig.7.144 |
♦ This vector passes through the center 'C' of the wheel
• We see the normal reaction vector $\mathbf\small{\vec{N}}$
♦ This vector also passes through the center 'C' of the wheel
• We see no other forces
• Forces passing through the center of mass of a body cannot cause it to rotate. So the wheel will not rotate
• When the cart is pulled. the wheel will just slide
3. But if the spiked wheels are used, things will be different
• So the ice will try to keep the point of contact in position
4. Let us see what really happens when spikes are used:
(i) The spikes first penetrate into the ice
(ii) When the wheel is pulled forward, the spikes 'presses horizontally' against the ice
(iii) So a horizontal force is applied by the spikes on the ice. This force is in the forward direction
(iv) The ice in turn, applies an equal backward horizontal force (in accordance with Newton's third law) on the spikes
(v) Note that, this backward horizontal force is applied by the ice on the spikes
• That means, the spikes experience this force
• Since the spikes are attached rigidly at the periphery of the wheel, the wheel will experience this force
• The wheel will experience this force as:
A force acting tangentially at the periphery
• So this is the force that we were waiting for. It does not pass through the center 'C'
• The wheel will begin to rotate
5. We know that, this force on the spikes is the frictional force
■ So we can write:
Friction helps the wheel of a cart to roll. If there is no friction, the wheel will only slide
Let us write a comparison between the two types of wheels:
♦ The wheels of a cart
• It can be written in two steps:
1. For the rear wheels of a car, rotation is already present. Because, the engine provides rotation to the axle, and so the wheels which are connected to the axle naturally rotates. If there is no friction, these rotating tires will just spin. The car will not move
• For the rear wheels of a cart, rotation is not readily present. Even if the cart is pulled forward, the wheels will rotate only if friction is present. In the absence of friction, the rear wheels of a cart can only slide
2. Frictional force, if present, acts on the rear wheel of a car
♦ The direction of this frictional force is same as the direction of motion of the car
• Frictional force, if present, acts on the wheels of a cart
♦ The direction of this frictional force is opposite to the direction of motion of the cart
Also we see the following similarities:
• A man is also self propagating when he is walking
• So the two sets of figs. are comparable:
♦ Set 1: Figs.7.143 (a) and (b)
♦ Set 2: Figs.7.140 (a) and (b)
2. A cart without an engine cannot self propagate. It needs an external pulling force
• A man trying to slide behind a snowmobile also needs the external pulling force
• So the two sets of figs. are comparable:
♦ Set 1: Figs.7.144 (a) and (b)
♦ Set 2: Figs.7.140 (c) and (d)
In the next section, we will see situations where the wheels will spin/slide even when friction is present
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