Chapter 2 Review Measurements and Calculations Review Answer Key

Learning Objectives

By the end of this section, you will be able to:

• Describe the full general characteristics of friction
• List the various types of friction
• Calculate the magnitude of static and kinetic friction, and use these in problems involving Newton's laws of motion

When a torso is in motion, information technology has resistance considering the body interacts with its surroundings. This resistance is a strength of friction. Friction opposes relative motion betwixt systems in contact just as well allows us to move, a concept that becomes obvious if you try to walk on water ice. Friction is a common yet circuitous force, and its behavior still not completely understood. All the same, it is possible to empathize the circumstances in which it behaves.

Static and Kinetic Friction

The basic definition of friction is relatively simple to state.

Friction

Friction is a strength that opposes relative movement between systems in contact.

There are several forms of friction. One of the simpler characteristics of sliding friction is that it is parallel to the contact surfaces between systems and is e'er in a direction that opposes motion or attempted motion of the systems relative to each other. If two systems are in contact and moving relative to one some other, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice. When objects are stationary, static friction can act between them; the static friction is normally greater than the kinetic friction between two objects.

Static and Kinetic Friction

If two systems are in contact and stationary relative to one some other, then the friction between them is called static friction. If two systems are in contact and moving relative to i another, then the friction between them is called kinetic friction.

Imagine, for instance, trying to slide a heavy crate across a concrete flooring—yous might push very difficult on the crate and non move it at all. This ways that the static friction responds to what yous do—it increases to be equal to and in the opposite direction of your push. If you lot finally button hard plenty, the crate seems to slip suddenly and starts to movement. Now static friction gives manner to kinetic friction. Once in move, information technology is easier to continue information technology in motion than it was to get it started, indicating that the kinetic frictional force is less than the static frictional forcefulness. If you add mass to the crate, say by placing a box on top of it, you demand to button even harder to go it started and also to proceed it moving. Furthermore, if you oiled the concrete you would find it easier to get the crate started and keep it going (equally yous might expect).

Figure half-dozen.ten is a crude pictorial representation of how friction occurs at the interface between ii objects. Close-upward inspection of these surfaces shows them to be rough. Thus, when you button to get an object moving (in this case, a crate), you must enhance the object until it can skip along with only the tips of the surface striking, breaking off the points, or both. A considerable force can be resisted by friction with no credible motion. The harder the surfaces are pushed together (such as if another box is placed on the crate), the more force is needed to move them. Part of the friction is due to adhesive forces between the surface molecules of the ii objects, which explains the dependence of friction on the nature of the substances. For example, rubber-soled shoes slip less than those with leather soles. Adhesion varies with substances in contact and is a complicated aspect of surface physics. Once an object is moving, there are fewer points of contact (fewer molecules adhering), and so less force is required to keep the object moving. At pocket-sized simply nonzero speeds, friction is nearly independent of speed.

Figure six.10 Frictional forces, such as $\overrightarrow{f},$ e'er oppose motion or attempted motion between objects in contact. Friction arises in part because of the roughness of the surfaces in contact, as seen in the expanded view. For the object to move, information technology must rise to where the peaks of the top surface can skip along the bottom surface. Thus, a strength is required simply to fix the object in motion. Some of the peaks will exist broken off, also requiring a force to maintain motion. Much of the friction is actually due to attractive forces between molecules making up the two objects, then that even perfectly smooth surfaces are not friction-free. (In fact, perfectly smooth, make clean surfaces of similar materials would adhere, forming a bond called a "cold weld.")

The magnitude of the frictional force has ii forms: i for static situations (static friction), the other for situations involving motion (kinetic friction). What follows is an approximate empirical (experimentally determined) model only. These equations for static and kinetic friction are not vector equations.

Magnitude of Static Friction

The magnitude of static friction $f_{\text{s}}$ is

where ${\mu }_{\text{s}}$ is the coefficient of static friction and Northward is the magnitude of the normal strength.

The symbol $\le$ ways less than or equal to, implying that static friction can take a maximum value of ${\mu }_{\text{s}}N.$ Static friction is a responsive force that increases to exist equal and contrary to whatever strength is exerted, up to its maximum limit. Once the practical forcefulness exceeds

$f_{\text{s}}\text{(max),}$ the object moves. Thus,

$$f_{\text{s}}(\text{max})={\mu }_{\text{southward}}\mathrm{Northward}.$$

Magnitude of Kinetic Friction

The magnitude of kinetic friction $f_{\text{one thousand}}$ is given by

where ${\mu }_{\text{yard}}$ is the coefficient of kinetic friction.

A system in which $f_{\text{grand}}={\mu }_{\text{k}}N$ is described equally a arrangement in which friction behaves simply. The transition from static friction to kinetic friction is illustrated in Figure 6.11.

Figure 6.xi (a) The force of friction $\overrightarrow{f}$ betwixt the block and the rough surface opposes the management of the practical strength $\overrightarrow{F}.$ The magnitude of the static friction balances that of the applied forcefulness. This is shown in the left side of the graph in (c). (b) At some bespeak, the magnitude of the applied force is greater than the force of kinetic friction, and the block moves to the right. This is shown in the right side of the graph. (c) The graph of the frictional force versus the practical force; note that $f_{\text{southward}}(\text{max})>f_{\text{1000}}.$ This means that ${\mu }_{\text{s}}>{\mu }_{\text{k}}.$

As yous can see in Tabular array 6.ane, the coefficients of kinetic friction are less than their static counterparts. The gauge values of $\mu$ are stated to but ane or two digits to signal the estimate clarification of friction given by the preceding two equations.

System	Static Friction ${\mu }_{\text{due south}}$	Kinetic Friction ${\mu }_{\text{m}}$
Rubber on dry concrete	1.0	0.7
Prophylactic on wet concrete	0.5-0.7	0.3-0.5
Forest on forest	0.5	0.3
Waxed wood on wet snowfall	0.14	0.1
Metal on wood	0.5	0.3
Steel on steel (dry out)	0.half-dozen	0.3
Steel on steel (oiled)	0.05	0.03
Teflon on steel	0.04	0.04
Bone lubricated past synovial fluid	0.016	0.015
Shoes on wood	0.nine	0.7
Shoes on ice	0.1	0.05
Ice on ice	0.1	0.03
Steel on ice	0.iv	0.02

Tabular array six.i Approximate Coefficients of Static and Kinetic Friction

Equation 6.1 and Equation 6.two include the dependence of friction on materials and the normal force. The management of friction is e'er contrary that of motility, parallel to the surface betwixt objects, and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, and then the normal force is equal to its weight,

$$w=gg=\left(100\text{kg}\right)\left(9.80{\text{m/due south}}^2\right)=980\text{N,}$$

perpendicular to the flooring. If the coefficient of static friction is 0.45, y'all would have to exert a strength parallel to the floor greater than

$$f_{\text{south}}(\text{max})={\mu }_{\text{due south}}\mathrm{Due\; north}=(0.45)(980\text{North})=440\text{N}$$

to move the crate. Once in that location is movement, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only

$$f_{\text{k}}={\mu }_{\text{k}}N=(\mathrm{0.thirty})(980\text{N})=290\text{N}$$

keeps it moving at a constant speed. If the flooring is lubricated, both coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unitless quantity with a magnitude commonly between 0 and ane.0. The bodily value depends on the two surfaces that are in contact.

Many people accept experienced the slipperiness of walking on ice. However, many parts of the body, particularly the joints, have much smaller coefficients of friction—oftentimes iii or four times less than ice. A joint is formed past the ends of two bones, which are connected past thick tissues. The knee articulation is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a brawl (at the end of the femur) and socket (office of the pelvis) joint. The ends of the basic in the joint are covered by cartilage, which provides a smooth, almost-burnished surface. The joints also produce a fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint can exist replaced by an bogus joint (Figure 6.12). These replacements can be made of metals (stainless steel or titanium) or plastic (polyethylene), besides with very pocket-sized coefficients of friction.

Figure 6.12 Bogus articulatio genus replacement is a procedure that has been performed for more than than 20 years. These post-operative Ten-rays show a right human knee articulation replacement. (credit: modification of work by Mike Baird)

Natural lubricants include saliva produced in our mouths to assistance in the swallowing procedure, and the glace fungus found between organs in the trunk, assuasive them to move freely past each other during heartbeats, during breathing, and when a person moves. Hospitals and md's clinics normally utilise artificial lubricants, such as gels, to reduce friction.

The equations given for static and kinetic friction are empirical laws that describe the behavior of the forces of friction. While these formulas are very useful for applied purposes, they do not take the status of mathematical statements that stand for general principles (e.g., Newton'south 2d law). In fact, there are cases for which these equations are not even good approximations. For instance, neither formula is accurate for lubricated surfaces or for two surfaces siding across each other at high speeds. Unless specified, we will non be concerned with these exceptions.

Example 6.10

Static and Kinetic Friction

A 20.0-kg crate is at rest on a flooring as shown in Figure 6.13. The coefficient of static friction between the crate and floor is 0.700 and the coefficient of kinetic friction is 0.600. A horizontal force $\overrightarrow{P}$ is applied to the crate. Observe the force of friction if (a) $\overrightarrow{P}=20.0\text{Northward}\widehat{i}\text{,}$ (b) $\overrightarrow{P}=30.0\text{N}\widehat{i}\text{,}$ (c) $\overrightarrow{P}=120.0\text{Northward}\widehat{i}\text{,}$ and (d) $\overrightarrow{P}=180.0\text{Northward}\widehat{i}\text{.}$

Figure 6.13 (a) A crate on a horizontal surface is pushed with a force $\overrightarrow{P}.$ (b) The forces on the crate. Here, $\overrightarrow{f}$ may represent either the static or the kinetic frictional force.

Strategy

The complimentary-body diagram of the crate is shown in Effigy half-dozen.thirteen(b). We apply Newton'due south second law in the horizontal and vertical directions, including the friction force in opposition to the direction of motion of the box.

Solution

Newton'southward second law GIVES

$$\begin{array}{cccc}{\displaystyle \sum F_{\mathrm{10}}=\mathrm{grand}{a}_x}\hfill & & & {\displaystyle \sum F_y=\mathrm{thousand}{a}_y}\hfill \\ P-f=m{a}_{\mathrm{ten}}\hfill & & & N-w=0.\hfill \end{array}$$

Here we are using the symbol f to correspond the frictional force since nosotros take not yet determined whether the crate is subject to station friction or kinetic friction. We practise this whenever we are unsure what type of friction is interim. At present the weight of the crate is

$$w=(20.0\text{kg})(9.80{\text{one thousand/s}}^2)=196\text{N,}$$

which is as well equal to N. The maximum force of static friction is therefore $\left(0.700\right)\left(196\text{N}\right)=137\text{Northward}\text{.}$ Every bit long as $\overrightarrow{P}$ is less than 137 N, the force of static friction keeps the crate stationary and $f_{\text{southward}}=\overrightarrow{P}.$ Thus, (a) $f_{\mathrm{south}}=20.0\text{N,}$ (b) $f_s=30.0\text{Due north,}$ and (c) $f_s=120.0\text{Northward}\text{.}$

(d) If $\overrightarrow{P}=180.0\text{Northward,}$ the applied strength is greater than the maximum strength of static friction (137 N), so the crate can no longer remain at remainder. In one case the crate is in motion, kinetic friction acts. And so

$$f_{\text{k}}={\mu }_{\text{one thousand}}\mathrm{North}=(0.600)(196\text{N})=118\text{N,}$$

and the dispatch is

$$a_{\mathrm{ten}}=\frac{P-f_{\text{k}}}{m}=\frac{180.0\text{N}-118\text{N}}{20.0\text{kg}}=3.10{\text{chiliad/south}}^{\mathrm{two}}\text{.}$$

Significance

This case illustrates how nosotros consider friction in a dynamics problem. Notice that static friction has a value that matches the applied force, until we attain the maximum value of static friction. Too, no move can occur until the applied force equals the strength of static friction, but the force of kinetic friction will then become smaller.

Cheque Your Understanding 6.seven

A block of mass 1.0 kg rests on a horizontal surface. The frictional coefficients for the block and surface are ${\mu }_{\mathrm{due\; south}}=\mathrm{0.fifty}$ and ${\mu }_g=0.40.$ (a) What is the minimum horizontal forcefulness required to move the block? (b) What is the cake's acceleration when this force is applied?

Friction and the Inclined Plane

One situation where friction plays an obvious part is that of an object on a slope. Information technology might exist a crate being pushed upward a ramp to a loading dock or a skateboarder coasting down a mount, merely the bones physics is the same. We usually generalize the sloping surface and telephone call it an inclined plane but then pretend that the surface is flat. Let's wait at an case of analyzing movement on an inclined plane with friction.

Case 6.xi

Downhill Skier

A skier with a mass of 62 kg is sliding downward a snowy slope at a abiding acceleration. Find the coefficient of kinetic friction for the skier if friction is known to be 45.0 North.

Strategy

The magnitude of kinetic friction is given as 45.0 N. Kinetic friction is related to the normal forcefulness $N$ by $f_{\text{k}}={\mu }_{\text{k}}\mathrm{Northward}$ ; thus, we tin find the coefficient of kinetic friction if we can observe the normal force on the skier. The normal force is always perpendicular to the surface, and since there is no motion perpendicular to the surface, the normal force should equal the component of the skier'southward weight perpendicular to the slope. (See Effigy 6.xiv, which repeats a figure from the chapter on Newton's laws of motion.)

Effigy 6.14 The motility of the skier and friction are parallel to the slope, so it is about convenient to project all forces onto a coordinate system where ane axis is parallel to the slope and the other is perpendicular (axes shown to left of skier). The normal force $\overrightarrow{\mathrm{Due\; north}}$ is perpendicular to the gradient, and friction $\overrightarrow{f}$ is parallel to the gradient, but the skier'due south weight $\overrightarrow{w}$ has components along both axes, namely ${\overrightarrow{w}}_y$ and ${\overrightarrow{\mathrm{west}}}_x.$ The normal force $\overrightarrow{N}$ is equal in magnitude to ${\overrightarrow{\mathrm{westward}}}_y,$ and then there is no movement perpendicular to the slope.

We have

$$N=w_y=\mathrm{west}\text{cos}25\text{°}=\mathrm{1000}g\text{cos}25\text{°}.$$

Substituting this into our expression for kinetic friction, nosotros obtain

$$f_{\text{1000}}={\mu }_{\text{yard}}\mathrm{1000}\mathrm{grand}\text{cos}25\text{°},$$

which can now exist solved for the coefficient of kinetic friction ${\mu }_{\text{k}}.$

Solution

Solving for ${\mu }_{\text{k}}$ gives

$${\mu }_{\text{k}}=\frac{f_{\text{k}}}{N}=\frac{f_{\text{k}}}{\mathrm{west}\text{cos}25\text{°}}=\frac{f_{\text{k}}}{m\mathrm{thousand}\text{cos}25\text{°}}.$$

Substituting known values on the right-hand side of the equation,

$${\mu }_{\text{g}}=\frac{45.0\text{N}}{(62\text{kg})(9.80{\text{m/s}}^2)(0.906)}=0.082.$$

Significance

This outcome is a little smaller than the coefficient listed in Table 6.1 for waxed woods on snow, only it is still reasonable since values of the coefficients of friction tin vary greatly. In situations like this, where an object of mass m slides downwardly a slope that makes an angle $\theta$ with the horizontal, friction is given by $f_{\text{thou}}={\mu }_{\text{k}}\mathrm{grand}\mathrm{grand}\text{cos}\theta .$ All objects slide down a slope with constant acceleration under these circumstances.

Nosotros have discussed that when an object rests on a horizontal surface, the normal strength supporting it is equal in magnitude to its weight. Furthermore, simple friction is always proportional to the normal force. When an object is not on a horizontal surface, equally with the inclined airplane, we must find the force acting on the object that is directed perpendicular to the surface; information technology is a component of the weight.

We now derive a useful human relationship for calculating coefficient of friction on an inclined plane. Notice that the result applies but for situations in which the object slides at constant speed downwardly the ramp.

An object slides down an inclined airplane at a abiding velocity if the cyberspace forcefulness on the object is null. Nosotros tin employ this fact to measure the coefficient of kinetic friction between 2 objects. As shown in Example six.xi, the kinetic friction on a slope is $f_k={\mu }_{k}m\mathrm{thousand}\text{cos}\theta$ . The component of the weight down the slope is equal to $\mathrm{thousand}g\text{sin}\theta$ (run across the complimentary-torso diagram in Figure six.14). These forces act in opposite directions, and then when they have equal magnitude, the acceleration is zero. Writing these out,

$${\mu }_{\text{k}}mg\text{cos}\theta =\mathrm{thou}g\text{sin}\theta .$$

Solving for ${\mu }_{\text{one thousand}},$ we observe that

$${\mu }_{\text{thou}}=\frac{\mathrm{one\; thousand}\mathrm{1000}\text{sin}\theta }{k\mathrm{thousand}\text{cos}\theta }=\text{tan}\theta .$$

Put a coin on a book and tilt it until the money slides at a constant velocity downward the book. Yous might demand to tap the book lightly to get the money to motion. Measure the bending of tilt relative to the horizontal and notice ${\mu }_{\text{k}}.$ Note that the coin does not beginning to slide at all until an angle greater than $\theta$ is attained, since the coefficient of static friction is larger than the coefficient of kinetic friction. Think about how this may touch the value for ${\mu }_{\text{k}}$ and its dubiousness.

Atomic-Scale Explanations of Friction

The simpler aspects of friction dealt with so far are its macroscopic (big-scale) characteristics. Great strides have been made in the atomic-calibration caption of friction during the past several decades. Researchers are finding that the atomic nature of friction seems to take several central characteristics. These characteristics non only explain some of the simpler aspects of friction—they also hold the potential for the development of nearly friction-free environments that could save hundreds of billions of dollars in energy which is currently beingness converted (unnecessarily) into heat.

Figure half-dozen.15 illustrates 1 macroscopic characteristic of friction that is explained past microscopic (pocket-size) research. We have noted that friction is proportional to the normal force, but not to the amount of expanse in contact, a somewhat counterintuitive notion. When ii crude surfaces are in contact, the actual contact expanse is a tiny fraction of the total expanse because but high spots bear on. When a greater normal force is exerted, the actual contact area increases, and we find that the friction is proportional to this area.

Figure 6.15 Two rough surfaces in contact take a much smaller area of actual contact than their total area. When the normal force is larger as a consequence of a larger practical strength, the area of actual contact increases, as does friction.

However, the atomic-calibration view promises to explain far more than the simpler features of friction. The mechanism for how heat is generated is at present being adamant. In other words, why do surfaces become warmer when rubbed? Essentially, atoms are linked with one another to grade lattices. When surfaces rub, the surface atoms adhere and cause atomic lattices to vibrate—substantially creating sound waves that penetrate the material. The sound waves diminish with distance, and their energy is converted into rut. Chemical reactions that are related to frictional wear can also occur between atoms and molecules on the surfaces. Figure 6.16 shows how the tip of a probe fatigued across another material is plain-featured past atomic-calibration friction. The forcefulness needed to elevate the tip can exist measured and is constitute to be related to shear stress, which is discussed in Static Equilibrium and Elasticity. The variation in shear stress is remarkable (more than a cistron of ${10}^{12}$ ) and difficult to predict theoretically, just shear stress is yielding a fundamental understanding of a large-scale phenomenon known since ancient times—friction.

Figure 6.16 The tip of a probe is deformed sideways by frictional force equally the probe is dragged across a surface. Measurements of how the force varies for different materials are yielding central insights into the diminutive nature of friction.

Example half dozen.12

Sliding Blocks

The ii blocks of Figure half-dozen.17 are attached to each other by a massless cord that is wrapped effectually a frictionless pulley. When the bottom 4.00-kg block is pulled to the left past the constant strength $\overrightarrow{P},$ the top 2.00-kg cake slides across information technology to the right. Find the magnitude of the force necessary to move the blocks at abiding speed. Assume that the coefficient of kinetic friction betwixt all surfaces is 0.400.

Effigy vi.17 (a) Each block moves at constant velocity. (b) Free-trunk diagrams for the blocks.

Strategy

We clarify the motions of the two blocks separately. The top block is subjected to a contact force exerted by the lesser cake. The components of this force are the normal force ${\mathrm{Northward}}_1$ and the frictional force $\mathrm{-0.400}{\mathrm{Northward}}_1.$ Other forces on the peak block are the tension $T\text{i}$ in the cord and the weight of the peak block itself, 19.6 N. The lesser block is subjected to contact forces due to the top cake and due to the flooring. The commencement contact force has components $\text{−}{N}_1$ and $0.400N_{\mathrm{i}},$ which are simply reaction forces to the contact forces that the bottom block exerts on the tiptop block. The components of the contact force of the floor are ${\mathrm{Northward}}_2$ and $0.400N_2.$ Other forces on this block are $\text{−}P,$ the tension $T\text{i},$ and the weight –39.two Due north.

Solution

Since the top block is moving horizontally to the right at constant velocity, its acceleration is zip in both the horizontal and the vertical directions. From Newton'south second police,

$$\begin{array}{cccccccc}\hfill {\displaystyle \sum F_x}& =\hfill & {\mathrm{yard}}_1{a}_{\mathrm{ten}}\hfill & & & \hfill {\displaystyle \sum F_y}& =\hfill & m_{\mathrm{one}}{a}_y\hfill \\ \hfill T-0.400N_1& =\hfill & 0\hfill & & & \hfill N_1-\mathrm{19.half\; dozen}\text{Northward}& =\hfill & 0.\hfill \end{array}$$

Solving for the two unknowns, we obtain $N_{\mathrm{ane}}=\mathrm{19.half-dozen}\text{N}$ and $T=0.40N_1=\mathrm{vii.84}\text{N}\text{.}$ The lesser block is too not accelerating, so the application of Newton's second law to this block gives

$$\begin{array}{cccc}{\displaystyle \sum F_x={\mathrm{yard}}_{\mathrm{two}}{a}_{\mathrm{10}}}\hfill & & & {\displaystyle \sum F_y={\mathrm{chiliad}}_2{a}_y}\hfill \\ T-P+0.400N_1+0.400N_2=0\hfill & & & N_2-39.2\text{N}-N_1=0.\hfill \end{array}$$

The values of $N_1$ and T were found with the starting time gear up of equations. When these values are substituted into the 2d gear up of equations, we can decide $N_2$ and P. They are

$$N_{\mathrm{ii}}=\mathrm{58.viii}\text{N}\text{and}P=39.2\text{N}\text{.}$$

Significance

Understanding what direction in which to draw the friction forcefulness is often troublesome. Detect that each friction force labeled in Figure 6.17 acts in the management opposite the move of its respective block.

Example vi.xiii

A Crate on an Accelerating Truck

A 50.0-kg crate rests on the bed of a truck as shown in Figure half dozen.18. The coefficients of friction between the surfaces are ${\mu }_{\text{k}}=0.300$ and ${\mu }_{\text{southward}}=0.400.$ Detect the frictional strength on the crate when the truck is accelerating forward relative to the ground at (a) 2.00 m/s², and (b) v.00 m/s².

Figure 6.xviii (a) A crate rests on the bed of the truck that is accelerating forwards. (b) The free-body diagram of the crate.

Strategy

The forces on the crate are its weight and the normal and frictional forces due to contact with the truck bed. We start past bold that the crate is not slipping. In this case, the static frictional force $f_{\text{due south}}$ acts on the crate. Furthermore, the accelerations of the crate and the truck are equal.

Solution

1. Awarding of Newton'south 2nd law to the crate, using the reference frame attached to the basis, yields
   

   $$\begin{array}{cccccccc}\hfill {\displaystyle \sum F_x}& =\hfill & \mathrm{yard}{a}_x\hfill & & & \hfill {\displaystyle \sum F_y}& =\hfill & m{a}_y\hfill \\ \hfill f_{\text{southward}}& =\hfill & (50.0\text{kg})(\mathrm{ii.00}{\text{m/s}}^{\mathrm{two}})\hfill & & & \hfill N-\mathrm{iv.xc}\times {10}^{\mathrm{ii}}\text{N}& =\hfill & (50.0\text{kg})(0)\hfill \\ & =\hfill & 1.00\times {10}^2\text{North}\hfill & & & \hfill N& =\hfill & \mathrm{iv.90}\times {10}^{\mathrm{ii}}\text{N}\text{.}\hfill \end{array}$$

   Nosotros tin now check the validity of our no-slip assumption. The maximum value of the forcefulness of static friction is
   

   $${\mu }_{\text{s}}\mathrm{North}=(0.400)(\mathrm{4.xc}\times {\mathrm{ten}}^2\text{N})=196\text{N,}$$

   whereas the actual force of static friction that acts when the truck accelerates forwards at $2.00{\text{m/s}}^{\mathrm{two}}$ is only $1.00\times {10}^2\text{N}\text{.}$ Thus, the assumption of no slipping is valid.
2. If the crate is to move with the truck when information technology accelerates at $\mathrm{five.0}{\text{m/s}}^{\mathrm{two}},$ the force of static friction must be
   

   $$f_{\text{due south}}=\mathrm{yard}{a}_x=(50.0\text{kg})(\mathrm{v.00}{\text{m/south}}^{\mathrm{two}})=250\text{N}\text{.}$$

   Since this exceeds the maximum of 196 N, the crate must skid. The frictional strength is therefore kinetic and is
   

   $$f_k={\mu }_{\mathrm{one\; thousand}}\mathrm{North}=(0.300)(4.90\times {10}^2\text{N})=147\text{Due north}\text{.}$$

   The horizontal dispatch of the crate relative to the basis is now found from
   

   $$\begin{array}{ccc}\hfill {\displaystyle \sum F_x}& =\hfill & \mathrm{thousand}{a}_x\hfill \\ \hfill 147\text{Due north}& =\hfill & (\mathrm{fifty.0}\text{kg})a_x,\hfill \\ \hfill \text{so}{a}_x& =\hfill & 2.94{\text{m/due south}}^2.\hfill \end{array}$$

Significance

Relative to the ground, the truck is accelerating forward at $\mathrm{five.0}{\text{k/s}}^{\mathrm{ii}}$ and the crate is accelerating frontwards at $2.94{\text{1000/due south}}^{\mathrm{two}}$ . Hence the crate is sliding backward relative to the bed of the truck with an dispatch $2.94{\text{one thousand/south}}^{\mathrm{ii}}-\mathrm{five.00}{\text{yard/s}}^{\mathrm{ii}}=\mathrm{-2.06}{\text{m/southward}}^2\text{.}$

Example 6.14

Snowboarding

Before, we analyzed the situation of a downhill skier moving at abiding velocity to determine the coefficient of kinetic friction. Now let's practice a like analysis to determine acceleration. The snowboarder of Figure 6.nineteen glides down a slope that is inclined at $\theta ={\mathrm{thirteen}}^0$ to the horizontal. The coefficient of kinetic friction between the lath and the snow is ${\mu }_{\text{one thousand}}=0.20.$ What is the acceleration of the snowboarder?

Figure 6.19 (a) A snowboarder glides downward a slope inclined at 13° to the horizontal. (b) The free-body diagram of the snowboarder.

Strategy

The forces acting on the snowboarder are her weight and the contact force of the slope, which has a component normal to the incline and a component along the incline (strength of kinetic friction). Because she moves along the slope, the near convenient reference frame for analyzing her movement is one with the x-axis along and the y-axis perpendicular to the incline. In this frame, both the normal and the frictional forces lie along coordinate axes, the components of the weight are $g\mathrm{yard}\text{sin}\theta \text{forth the gradient and}\mathrm{chiliad}g\text{cos}\theta \text{at correct angles into the slope}$ , and the only dispatch is along the x-axis $\left(a_y=0\right).$

Solution

We can now apply Newton's second police to the snowboarder:

$$\begin{array}{cccccc}\hfill {\displaystyle \sum F_x}& =\hfill & m{a}_x\hfill & & & {\displaystyle \sum F_y=m{a}_y}\hfill \\ \hfill m\mathrm{chiliad}\text{sin}\theta -{\mu }_{\mathrm{one\; thousand}}N& =\hfill & m{a}_{\mathrm{ten}}\hfill & & & \hfill N-mg\text{cos}\theta =m(0)\text{.}\end{array}$$

From the 2nd equation, $N=\mathrm{thou}g\text{cos}\theta .$ Upon substituting this into the first equation, nosotros find

$$\begin{array}{cc}\hfill a_x& =k(\text{sin}\theta -{\mu }_{\text{one thousand}}\text{cos}\theta )\hfill \\ & =m(\text{sin}\mathrm{thirteen}\text{°}-\mathrm{0.xx}\text{cos}13\text{°})=0.29{\text{m/due south}}^2\text{.}\hfill \end{array}$$

Significance

Observe from this equation that if $\theta$ is small-scale enough or ${\mu }_{\text{thou}}$ is big enough, $a_x$ is negative, that is, the snowboarder slows down.

Cheque Your Agreement 6.8

The snowboarder is now moving down a colina with incline $\mathrm{x.0}\text{°}$ . What is the snowboarder'southward dispatch?

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Source: https://openstax.org/books/university-physics-volume-1/pages/6-2-friction

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