Kinetic Friction: Difference between revisions
| (18 intermediate revisions by the same user not shown) | |||
| Line 37: | Line 37: | ||
The magnitude of the acceleration of the box is therefore | The magnitude of the acceleration of the box is therefore | ||
<math>a = \frac{f}{m} = \frac{17.64}{6} = 2.94</math>m/s by Newton's second law | <math>a = \frac{f}{m} = \frac{17.64}{6} = 2.94</math>m/s<sup>2</sup> by Newton's second law | ||
We can use this acceleration along with | We can use this acceleration along with [[Kinematics]] to answer each question. Let us define the positive direction as the direction of the box's velocity. | ||
<math>v_f = v_i + at</math> | <math>v_f = v_i + at</math> | ||
| Line 45: | Line 45: | ||
<math>0 = 10 - 2.94t</math> | <math>0 = 10 - 2.94t</math> | ||
<math>t = 3. | <math>t = 3.4</math>s | ||
<math>v_f^2 = v_i^2 + 2ad</math> | <math>v_f^2 = v_i^2 + 2ad</math> | ||
| Line 51: | Line 51: | ||
<math>0 = 10^2 + 2(-2.94)d</math> | <math>0 = 10^2 + 2(-2.94)d</math> | ||
<math>d = 17. | <math>d = 17.0</math>m | ||
===2. (Simple)=== | ===2. (Simple)=== | ||
A worker pushes a 4kg box across the floor at a constant 5m/s. The coefficient of kinetic friction <math> | A worker pushes a 4kg box across the floor at a constant 5m/s. The coefficient of kinetic friction <math>\mu_k</math> = .2. How much power does the worker apply to the box to keep it sliding? | ||
Solution: | |||
The magnitude of the kThe magnitude of the kinetic friction force is given by | |||
<math>F_k = \mu_k * N = .2 * 4 * 9.8 = 7.84</math>N (The normal force between the box and the floor must be the weight of the box or it would be accelerating vertically) | |||
[[Power]] is the dot product of force and velocity, so the power applied by the worker to the box is given by | |||
<math>P = 7.84 * 5 = 39.2</math>J/s. | |||
===3. (Middling)=== | ===3. (Middling)=== | ||
A box of mass <math>m</math> sits in an elevator. The coefficient of kinetic friction between it and the floor of the elevator is <math>\mu_k</math>. The elevator accelerates upwards with an acceleration of <math>a_1</math>. How much force does a worker need to apply to accelerate the box forwards at with an acceleration of <math>a_2</math> (relative to the elevator)? | |||
[[File:kineticfrictionelevator.png]] | |||
Solution: | |||
First, let us find the normal force between the box and the elevator floor: | |||
<math>F_{net} = m * a</math> | |||
<math>N - mg = m * a_1</math> | |||
<math>N = mg + ma_1</math> | |||
Next, let us find the kinetic friction force between the box and the elevator floor: | |||
<math>F_k = \mu_k * N</math> | |||
<math>F_k = \mu_k * (mg + ma_1)</math> | |||
Finally, let us find the applied force necessary to accelerate the box: | |||
<math>F_{net} = m * a</math> | |||
<math>F_{ext} - \mu_k * (mg + ma_1) = m * a_2</math> | |||
<math>F_{ext} = ma_2 + \mu_k * (mg + ma_1)</math> | |||
===4. (Difficult)=== | ===4. (Difficult)=== | ||
A sheet of rigid plastic is laid down on an ice skating rink. Its mass is .5kg and it is 8m long. There is negligible friction between it and the ice. Next, a .17kg hockey puck is slid across it from one end to the other. Its initial velocity is 12m/s. A kinetic friction force acts between the sheet and the puck. The coefficient of kinetic friction between them is <math>\mu_k = .15</math>. How long does it take for the puck to slide off the other end of the sheet? | |||
[[File:Kineticfrictionplasticsheet.png]] | |||
Solution: | |||
For this problem, it is important to realize that the plastic sheet is not fixed in place; as the kinetic friction force slows down the puck, it also accelerates the sheet. To find out when the puck crosses the end of the sheet, let us find expressions for the position of the puck, <math>x_{puck}</math>, as a function of time, as well as the position of the end of the sheet, <math>x_{end of sheet}</math>, as a function of time. That way, we can set their positions equal to each other and solve for <math>t</math>. Let us define <math>t=0</math> as the time when the puck begins to slide across the sheet and <math>x=0</math> as the position where the puck began (the left end of the sheet as it is depicted in the diagram). Let us define the positive direction as the direction of the puck's travel (towards the right as it is depicted in the diagram). To find the expressions, we will need the accelerations of the puck and of the sheet: | |||
<math>a_{puck} = \frac{-f_k}{m_{puck}} = \frac{- \mu_k * N}{m_{puck}} = \frac{-.15 * .17 * 9.8}{.17} = -1.47</math>m/s<sup>2</sup>. | |||
<math>a_{sheet} = \frac{f_k}{m_{sheet}} = \frac{\mu_k * N}{m_{sheet}} = \frac{.15 * .17 * 9.8}{.5} = .50</math>m/s<sup>2</sup>. | |||
To find the expressions for the objects' positions, let us use [[Kinematics]]: | |||
<math>x_{puck}(t) = x_0 + v_0 t + \frac{1}{2}at^2 = 12t - .735t^2</math>m | |||
<math>x_{end of sheet}(t) = x_0 + v_0 t + \frac{1}{2}at^2 = 8 + .25t^2</math>m | |||
Now let us set the expressions for the objects' positions equal to each other to see when the puck crosses the end of the sheet: | |||
<math>12t - .735t^2 = 8 + .25t^2</math> | |||
<math>0 = .985t^2 - 12t + 8</math> | |||
Using the quadratic equation (<math>t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>), | |||
<math>t = .708, t = 11.5</math>. | |||
We are interested in the lowest value for t, which is the first time the puck and the end of the sheet have the same position. At this time, the puck passes the end of the sheet. The second solution represents the time at which the sheet overtakes the puck, which would eventually occur if both objects were to continue accelerating as described by the position equations we used. However, in reality, both objects stop accelerating after the puck leaves the sheet because after they lose contact they no longer exert friction forces on each other. | |||
<b>The puck reaches the end of the sheet after .708s</b>. | |||
This answer makes sense because it is slightly longer than it would take for the puck to leave the sheet if there were no friction force; if that were the case, one could simply divide 8m by 12m/s to get .66s. | |||
==Connectedness== | ==Connectedness== | ||
Latest revision as of 17:39, 8 July 2019
This page defines and describes kinetic friction. Some of the information presented on this page is also present on the Friction page.
The Main Idea
Kinetic friction is a type of Friction that occurs between two touching objects that are moving with respect to each other at their point of contact. When two objects touch each other and there is sliding between their surfaces of contact, they exert a kinetic friction force on each other. The kinetic friction force acting on each object acts in a direction opposite to its direction of motion relative to the other object. If the kinetic friction force is the only force acting on each object, they will eventually come to rest relative to each other, at which point the friction force between them becomes static. For example, consider a crate sliding across the floor. It will slow down and eventually come to rest due to the kinetic friction force between it and the floor, unless another force acts on it, such as the pushing of a person that wants to keep the crate moving. Kinetic friction force within a system lowers the system's kinetic energy, converting it to Thermal Energy. In other words, sliding between surfaces causes the surfaces to heat up. Usually in mechanical physics, this thermal energy is no longer considered part of the system and is considered lost because it usually dissipates into the environment without ever being converted back into mechanical energy. This means kinetic friction does negative work on the system. In addition to dissipating energy as thermal energy, kinetic friction may also dissipate a smaller amount of energy as sound, and can wear down surfaces if sufficiently intense.
A Mathematical Model
The magnitude of the kinetic friction force between objects is given by
[math]\displaystyle{ F_k = \mu_k * N }[/math]
where [math]\displaystyle{ \mu_k }[/math] is the coefficient of kinetic friction of the objects and [math]\displaystyle{ N }[/math] is the normal force between the objects. [math]\displaystyle{ \mu_k }[/math] is a property of the materials of the surfaces in contact and is usually less than 1. [math]\displaystyle{ \mu_k }[/math] has no units because it is a ratio of one force to another. For any given pair of surfaces, their [math]\displaystyle{ \mu_s }[/math] value is greater than their [math]\displaystyle{ \mu_k }[/math] value.
Note that this formula indicates that the magnitude of the kinetic friction force does not depend on the speed of the sliding between the objects.
A Computational Model
This is a VPython simulation of a box sliding along a surface. At first it experiences no friction, but when it passes the red line, it experiences a constant kinetic friction due to the floor until it comes to rest.
Click "View this program" on the top left corner to view the VPython code.
Examples
1. (Simple)
A 6kg box is kicked, sending it sliding across the floor. Its initial speed is 10m/s. The coefficient of kinetic friction [math]\displaystyle{ \mu_k }[/math] = .3. How long does it take the box to come to rest due to friction? How far does it slide in this time?
Solution:
The magnitude of the kinetic friction force is given by
[math]\displaystyle{ F_k = \mu_k * N = .3 * 6 * 9.8 = 17.64 }[/math]N (The normal force between the box and the floor must be the weight of the box or it would be accelerating vertically)
The magnitude of the acceleration of the box is therefore
[math]\displaystyle{ a = \frac{f}{m} = \frac{17.64}{6} = 2.94 }[/math]m/s2 by Newton's second law
We can use this acceleration along with Kinematics to answer each question. Let us define the positive direction as the direction of the box's velocity.
[math]\displaystyle{ v_f = v_i + at }[/math]
[math]\displaystyle{ 0 = 10 - 2.94t }[/math]
[math]\displaystyle{ t = 3.4 }[/math]s
[math]\displaystyle{ v_f^2 = v_i^2 + 2ad }[/math]
[math]\displaystyle{ 0 = 10^2 + 2(-2.94)d }[/math]
[math]\displaystyle{ d = 17.0 }[/math]m
2. (Simple)
A worker pushes a 4kg box across the floor at a constant 5m/s. The coefficient of kinetic friction [math]\displaystyle{ \mu_k }[/math] = .2. How much power does the worker apply to the box to keep it sliding?
Solution:
The magnitude of the kThe magnitude of the kinetic friction force is given by
[math]\displaystyle{ F_k = \mu_k * N = .2 * 4 * 9.8 = 7.84 }[/math]N (The normal force between the box and the floor must be the weight of the box or it would be accelerating vertically)
Power is the dot product of force and velocity, so the power applied by the worker to the box is given by
[math]\displaystyle{ P = 7.84 * 5 = 39.2 }[/math]J/s.
3. (Middling)
A box of mass [math]\displaystyle{ m }[/math] sits in an elevator. The coefficient of kinetic friction between it and the floor of the elevator is [math]\displaystyle{ \mu_k }[/math]. The elevator accelerates upwards with an acceleration of [math]\displaystyle{ a_1 }[/math]. How much force does a worker need to apply to accelerate the box forwards at with an acceleration of [math]\displaystyle{ a_2 }[/math] (relative to the elevator)?
Solution:
First, let us find the normal force between the box and the elevator floor:
[math]\displaystyle{ F_{net} = m * a }[/math]
[math]\displaystyle{ N - mg = m * a_1 }[/math]
[math]\displaystyle{ N = mg + ma_1 }[/math]
Next, let us find the kinetic friction force between the box and the elevator floor:
[math]\displaystyle{ F_k = \mu_k * N }[/math]
[math]\displaystyle{ F_k = \mu_k * (mg + ma_1) }[/math]
Finally, let us find the applied force necessary to accelerate the box:
[math]\displaystyle{ F_{net} = m * a }[/math]
[math]\displaystyle{ F_{ext} - \mu_k * (mg + ma_1) = m * a_2 }[/math]
[math]\displaystyle{ F_{ext} = ma_2 + \mu_k * (mg + ma_1) }[/math]
4. (Difficult)
A sheet of rigid plastic is laid down on an ice skating rink. Its mass is .5kg and it is 8m long. There is negligible friction between it and the ice. Next, a .17kg hockey puck is slid across it from one end to the other. Its initial velocity is 12m/s. A kinetic friction force acts between the sheet and the puck. The coefficient of kinetic friction between them is [math]\displaystyle{ \mu_k = .15 }[/math]. How long does it take for the puck to slide off the other end of the sheet?
Solution:
For this problem, it is important to realize that the plastic sheet is not fixed in place; as the kinetic friction force slows down the puck, it also accelerates the sheet. To find out when the puck crosses the end of the sheet, let us find expressions for the position of the puck, [math]\displaystyle{ x_{puck} }[/math], as a function of time, as well as the position of the end of the sheet, [math]\displaystyle{ x_{end of sheet} }[/math], as a function of time. That way, we can set their positions equal to each other and solve for [math]\displaystyle{ t }[/math]. Let us define [math]\displaystyle{ t=0 }[/math] as the time when the puck begins to slide across the sheet and [math]\displaystyle{ x=0 }[/math] as the position where the puck began (the left end of the sheet as it is depicted in the diagram). Let us define the positive direction as the direction of the puck's travel (towards the right as it is depicted in the diagram). To find the expressions, we will need the accelerations of the puck and of the sheet:
[math]\displaystyle{ a_{puck} = \frac{-f_k}{m_{puck}} = \frac{- \mu_k * N}{m_{puck}} = \frac{-.15 * .17 * 9.8}{.17} = -1.47 }[/math]m/s2.
[math]\displaystyle{ a_{sheet} = \frac{f_k}{m_{sheet}} = \frac{\mu_k * N}{m_{sheet}} = \frac{.15 * .17 * 9.8}{.5} = .50 }[/math]m/s2.
To find the expressions for the objects' positions, let us use Kinematics:
[math]\displaystyle{ x_{puck}(t) = x_0 + v_0 t + \frac{1}{2}at^2 = 12t - .735t^2 }[/math]m
[math]\displaystyle{ x_{end of sheet}(t) = x_0 + v_0 t + \frac{1}{2}at^2 = 8 + .25t^2 }[/math]m
Now let us set the expressions for the objects' positions equal to each other to see when the puck crosses the end of the sheet:
[math]\displaystyle{ 12t - .735t^2 = 8 + .25t^2 }[/math]
[math]\displaystyle{ 0 = .985t^2 - 12t + 8 }[/math]
Using the quadratic equation ([math]\displaystyle{ t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} }[/math]),
[math]\displaystyle{ t = .708, t = 11.5 }[/math].
We are interested in the lowest value for t, which is the first time the puck and the end of the sheet have the same position. At this time, the puck passes the end of the sheet. The second solution represents the time at which the sheet overtakes the puck, which would eventually occur if both objects were to continue accelerating as described by the position equations we used. However, in reality, both objects stop accelerating after the puck leaves the sheet because after they lose contact they no longer exert friction forces on each other.
The puck reaches the end of the sheet after .708s.
This answer makes sense because it is slightly longer than it would take for the puck to leave the sheet if there were no friction force; if that were the case, one could simply divide 8m by 12m/s to get .66s.
Connectedness
Kinetic friction is present in many places in everyday life. Sometimes, it is an inconvenience, such as when a car engine loses power to kinetic friction, hurting gas mileage. Sometimes, it is necessary for certain machines and activities, such as car brakes. In addition to affecting motion, kinetic friction raises temperatures, can wear down objects, and can emit sound, which may also need to be taken into account, depending on the application. Below are some examples of scenarios in which kinetic friction plays an essential role, including inventions designed to reduce it.
Wheels
Sliding objects along the ground can be difficult. Enter one of the oldest inventions of mankind: the wheel. Wheels significantly reduce the power dissipated by kinetic friction by changing where it takes place, and, in doing so, changing the distance over which it is exerted. Consider a situation in which a person wants to push two boxes across the floor: one with wheels and one without. On the box without wheels, kinetic friction takes place between the floor and the bottom of the box. The friction force does negative work on the box equal to the product of its magnitude and the distance over which it was exerted, which is the distance traveled by the box. On the box with wheels, the friction at the point of contact between the wheels and the floor is static because the wheels roll rather than slide. This static friction force is not exerted over any distance, because the point of contact between the ground and each wheel changes rather than moves. It therefore does no work. Kinetic friction is still present; it now occurs between the wheels' axles and the brackets that hold them to the underside of the box. The axle rotates within its brackets, so sliding occurs. (Alternatively, if the axle is fixed and doesn't rotate, the kinetic friction occurs where each wheel slides around the axle that holds it.) The normal force between the brackets and the axle is the same as the normal force between the box without wheels and the floor; that is, the weight of the box. However, because the radius of the axle is significantly smaller than the radius of the wheel, this kinetic friction force is exerted over a much smaller distance. In other words, which each rotation an axle makes, the outer edge of the wheel travels much farther than the outer edge of the axle, so the distance over which the force is exerted is much smaller than the distance traveled by the box. This means that the negative work done on the box with wheels is smaller in magnitude than the negative work done on the sliding box. Another way of thinking about it is that the radius of the axle is so small that friction barely exerts any torque on it.
Ball bearings
Ball bearings reduce friction between a rotating axle and the bracket or frame that holds it. A ball bearing consists of a rotating inner cylinder inside a hollow outer cylinder that holds it. The inner and outer cylinders do not touch but are separated by a layer of spheres that turn so that there is only static friction. In fact, in a theoretical ideal ball bearing, there is no kinetic friction and therefore no loss of energy. In real ball bearings, some slipping does occur between the balls and the cylinders, and the balls may come into contact with each other or the casing that holds them in place. Ball bearings are often used around the axles of wheels to make pushing carts easier, or in conjunction with motors in robots to reduce the wasting of energy.
Rubbing hands together for warmth
Rubbing our hands together for warmth is possible because of friction. The kinetic friction between our hands causes the kinetic energy of our hands to be converted into thermal energy. Our muscles continually supply a new source of kinetic energy. Similarly, sticks can be rubbed together to create enough heat for fire.
Sandpaper
The abrasive nature of friction is often undesirable, but for sandpaper, it is essential. Sandpaper uses kinetic friction to wear off the uneven protrusions on a surface to make it smoother. Belt sanders can even be used to remove large volumes of material, for example to round a sharp corner.
See also
External links
References
Matter and Interactions 4th edition. Full Citation: Chabay, Ruth W., and Bruce A. Sherwood. Matter and Interactions. Hoboken, NJ: Wiley, 2011. Print.