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College Physics by Openstax Chapter 7 Problem 6


How much work is done by the boy pulling his sister 30.0 m in a wagon as shown in Figure 7.33? Assume no friction acts on the wagon.

Figure 7.33 The boy does work on the system of the wagon and the child when he pulls them as shown.

Solution:

The work WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=FdcosθW=Fd \cos \theta

In this case, we are given the following values:

F=50 Nd=30 mθ=30\begin{align*} F & = 50\ \text{N} \\ d & = 30\ \text{m} \\ \theta & = 30^{\circ} \end{align*}

Substituting these values into the equation, we have

W=FdcosθW=(50 N)(30 m)cos30W=1299.0381 NmW=1.30×103 NmW=1.30×103 J  (Answer)\begin{align*} W & = Fd \cos \theta \\ W & = \left( 50\ \text{N} \right)\left( 30\ \text{m} \right) \cos 30^{\circ } \\ W & = 1299.0381\ \text{N} \cdot \text{m} \\ W & = 1.30 \times 10^{3}\ \text{N} \cdot \text{m} \\ W & = 1.30 \times 10^{3}\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 7 Problem 5


Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º with the horizontal. (See Figure 7.32.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.

Figure 7.32 A man pushes a crate up a ramp.

Solution:

The Work Done by the Man on the Crate

The work WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=FdcosθW=Fd \cos \theta

In case where the work done by the man to the crate, the following values are given:

F= 500 Nd= 4 mθ= 0(Force is parallel to displacement)\begin{align*} F = & \ 500\ \text{N} \\ d = & \ 4\ \text{m} \\ \theta = & \ 0^{\circ} \color{Blue} \left( \text{Force is parallel to displacement} \right) \end{align*}

Substituting these values in the equation, we have

W= FdcosθW= (500 N)(4 m)cos0W= 2000 Nm\begin{align*} W = & \ Fd \cos \theta \\ W = & \ \left( 500\ \text{N} \right) \left( 4\ \text{m} \right) \cos 0^{\circ} \\ W = & \ 2000\ \text{N} \cdot \text{m} \end{align*}

The work done by the man on his body

In this case, the force exerted is counteracted by the weight of the man. This force is directed upward. The displacement is still the 4.0 m along the inclined plane. The angle between the force and the displacement is 70 degrees.

W= FdcosθW= mgdcosθW= (85.0 kg)(9.80 m/s2)(4.0 m)cos70W= 1139.6111 Nm\begin{align*} W = & \ Fd \cos \theta \\ W = & \ mg d \cos \theta \\ W = & \ \left( 85.0\ \text{kg} \right) \left( 9.80\ \text{m/s}^2 \right)\left( 4.0\ \text{m} \right) \cos 70^{\circ} \\ W = & \ 1139.6111\ \text{N} \cdot \text{m} \end{align*}

The Total Work

The total work done by the man is the sum of the work he did on the crate and on his body.

WT=2000 Nm+1139.6111 NmWT=3139.6111 NmWT=3.14×103 NmWT=3.14×103 J  (Answer)\begin{align*} W_{T} & = 2000\ \text{N}\cdot \text{m} + 1139.6111\ \text{N}\cdot \text{m} \\ W_{T} & = 3139.6111 \ \text{N}\cdot \text{m} \\ W_{T} & = 3.14 \times 10^{3} \ \text{N}\cdot \text{m} \\ W_{T} & = 3.14 \times 10^{3} \ \text{J}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 7 Problem 4


Suppose a car travels 108 km at a speed of 30.0 m/s, and uses 2.0 gal of gasoline. Only 30% of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (See Table 7.1 for the energy content of gasoline.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s?


Solution:

Part A

According to Table 7.1, the energy in 1 gallon of gasoline is 1.2×108 J1.2 \times 10^{8}\ \text{J}. Since only 30% of the gasoline goes into useful work, the work done by the friction WfW_{f} is

Wf=0.30(2.0 gal)(1.2×108 J/gal)Wf=72×106 J\begin{align*} W_{f} & =0.30 \left( 2.0\ \text{gal} \right)\left( 1.2 \times 10^{8} \ \text{J/gal}\right) \\ W_{f} & = 72 \times 10^{6}\ \text{J} \end{align*}

Now, the work done by the friction can also be calculated using the formula below, where FfF_{f} is the magnitude of the friction force that keeps the car moving at constant speed, and dd is the distance traveled by the car.

Wf=Ffd\begin{align*} W_{f}=F_{f}d \end{align*}

We can solve for FfF_{f} in terms of the other variables.

Ff=WfdF_{f} = \frac{W_{f}}{d}

Substituting the given values, we can now solve for the unknown magnitude of the force exerted to keep the car moving at constant speed.

Ff=WfdFf=72×106 J108 kmFf=72×106 Nm108×103 mFf=666.6667 NFf=6.7×102 N  (Answer)\begin{align*} F_{f} & = \frac{W_{f}}{d} \\ F_{f} & = \frac{72 \times 10^{6}\ \text{J}}{108\ \text{km}} \\ F_{f} & = \frac{72 \times 10^{6}\ \text{N}\cdot \text{m}}{108 \times 10^{3}\ \text{m}} \\ F_{f} & = 666.6667\ \text{N} \\ F_{f} & = 6.7 \times 10^{2}\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

If the required force is directly proportional to speed, then there must be a linear relationship between the required force and speed. In this situation, we can just simply used ratio and proportion to compute for the number of gallons.

2.0 gal30.0 m/s=x28.0 m/sx=(2.0 gal)(28.0 m/s)30.0 m/sx=1.8667 galx=1.9 gal  (Answer)\begin{align*} \frac{2.0\ \text{gal}}{30.0\ \text{m/s}} & = \frac{x}{28.0\ \text{m/s}} \\ x & = \frac{\left( 2.0\ \text{gal} \right)\left( 28.0\ \text{m/s} \right)}{30.0\ \text{m/s}} \\ x & = 1.8667\ \text{gal} \\ x & = 1.9\ \text{gal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

College Physics by Openstax Chapter 7 Problem 3


(a) Calculate the work done on a 1500-kg elevator car by its cable to lift it 40.0 m at constant speed, assuming friction averages 100 N. (b) What is the work done on the elevator car by the gravitational force in this process? (c) What is the total work done on the elevator car?


Solution:

The work WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=FdcosθW=Fd \cos \theta

Part A

The force in the cable is equal to the combined effect of the weight of the elevator and the friction that opposes the motion. That is

F=mg+fF=(1500 kg)(9.80 m/s2)+100 NF=14800 N\begin{align*} F & = mg + f \\ F & = \left( 1500\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)+100\ \text{N} \\ F & = 14800\ \text{N} \end{align*}

This force in the cable is directed upward. The displacement is also upward, making the angle between the two quantities equal to zero. Thus, θ=0\theta = 0.

Substituting these values in the equation, the work done by the cable is

W=FdcosθW=(14 800 N)(40.0 m)cos0W=592 000 JW=5.92×105 J  (Answer)\begin{align*} W & = Fd \cos \theta \\ W & = \left( 14\ 800\ \text{N} \right)\left( 40.0\ \text{m} \right) \cos 0^\circ \\ W & = 592\ 000\ \text{J} \\ W & = 5.92 \times 10^{5} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The force due to gravity is equal to the weight of the elevator alone. That is

Weight=mg=(1 500 kg)(9.80 m/s2)=14 700 N\begin{align*} \text{Weight} & = mg \\ & = \left( 1\ 500\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right) \\ & = 14\ 700\ \text{N} \end{align*}

This force is directed downward, whereas the displacement is directed upward. Therefore, the angle θ\theta between the two quantities is θ=180\theta = 180^\circ.

Substituting these values in the formula for work, we have

W=FdcosθW=(14 700 N)(40.0 m)cos180W=588 000 JW=5.88×105 J  (Answer)\begin{align*} W & = Fd \cos \theta \\ W & = \left( 14\ 700\ \text{N} \right)\left( 40.0\ \text{m} \right) \cos 180^\circ \\ W & = -588\ 000\ \text{J} \\ W & = -5.88 \times 10^{5}\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

Since the elevator is moving at a constant speed, it is in equilibrium. This means that the net external force experience by the elevator is zero. Therefore, the total work done on the elevator car is

WT=0 J  (Answer)W_{T} = 0\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

College Physics by Openstax Chapter 7 Problem 2


A 75.0-kg person climbs stairs, gaining 2.50 meters in height. Find the work done to accomplish this task. (Neglect friction in your calculations.)


Solution:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy is ΔPEg=mgh\Delta PE_{g} = mgh, with hh being the increase in height and gg the acceleration due to gravity.

W=mghW=mgh

We are given the following values: m=75.0 kgm=75.0\ \text{kg}, g=9.80 m/s2g=9.80\ \text{m/s}^2, and h=2.50 mh=2.50\ \text{m}.

Substitute the given in the formula.

W=mghW=(75.0 kg)(9.80 m/s2)(2.50 m)W=1837.5 NmW=1837.5 JW=1.84×103 J  (Answer)\begin{align*} W & = mgh \\ W & = \left( 75.0\ \text{kg} \right)\left( 9.80\ \text{m/s}^2 \right)\left( 2.50\ \text{m} \right)\\ W & = 1837.5\ \text{Nm} \\ W & = 1837.5\ \text{J} \\ W & = 1.84 \times 10 ^{3} \ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The work done is about 1.84×103 Joules1.84 \times 10 ^ {3}\ \text{Joules} .


College Physics by Openstax Chapter 7 Problem 1


How much work does a supermarket checkout attendant do on a can of soup he pushes 0.600 m horizontally with a force of 5.00 N? Express your answer in joules and kilocalories.


Solution:

The work WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=FdcosθW=Fd \cos \theta

We are given the following values: F=5.00 NF=5.00\ \text{N}, d=0.600 md=0.600\ \text{m}, and θ=0\theta=0^\circ.

Substitute the given values in the formula for work.

W=FdcosθW=(5.00 N)(0.600 m)cos0W=3.00 NmW=3.00 J  (Answer)\begin{align*} W & = Fd \cos \theta \\ W & = \left( 5.00\ \text{N} \right)\left( 0.600\ \text{m} \right) \cos 0^\circ \\ W & = 3.00\ \text{Nm} \\ W & = 3.00\ \text{J} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The work done is 3.00 Joules. Now, we can convert this in unit of kilocalories knowing that 1 kcal=4186 J1\ \text{kcal} = 4186\ \text{J}.

W=3.00 JW=3.00 J × 1 kcal4186 JW=0.000717 kcalW=7.17×104 kcal  (Answer)\begin{align*} W & = 3.00\ \text{J} \\ W & = 3.00\ \text{J}\ \times \ \frac{1\ \text{kcal}}{4186\ \text{J}} \\ W & = 0.000717\ \text{kcal} \\ W & = 7.17 \times 10 ^{-4} \ \text{kcal} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The work done in kilocalories is about 7.17×1047.17 \times 10 ^{-4}.


College Physics by Openstax Chapter 6 Problem 31

The Speed of the Roller Coaster at the Top of the Loop


Problem:

Modern roller coasters have vertical loops like the one shown in Figure 6.35. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g?

Figure 6.35 Teardrop-shaped loops are used in the latest roller coasters so that the radius of curvature gradually decreases to a minimum at the top. This means that the centripetal acceleration builds from zero to a maximum at the top and gradually decreases again. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered long ago in railroad curve design. With a small radius of curvature at the top, the centripetal acceleration can more easily be kept greater than g
 so that the passengers do not lose contact with their seats, nor do they need seat belts to keep them in place.

Solution:

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College Physics by Openstax Chapter 6 Problem 30

The Ideal Speed and the Minimum Coefficient of Friction in Icy Mountain Roads


Problem:

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.

(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?


Solution:

Part A

The formula for an ideally banked curve is

tanθ=v2rg\tan \theta = \frac{v^2}{rg}

Solving for vv in terms of all the other variables, we have

v=rgtanθv= \sqrt{rg \tan \theta}

For this problem, we are given

  • radius, r=100 mr=100\ \text{m}
  • acceleration due to gravity, g=9.81 m/s2g=9.81\ \text{m/s}^2
  • banking angle, θ=15.0\theta = 15.0^\circ

Substituting all these values in the formula, we have

v=rgtanθv=(100 m)(9.81 m/s2)tan15.0v=16.2129 m/sv=16.2 m/s  (Answer)\begin{align*} v & = \sqrt{rg \tan \theta} \\ v & = \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)\tan 15.0^\circ }\\ v & = 16.2129\ \text{m/s} \\ v & = 16.2\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let us draw the free-body diagram of the car.

Summing forces in the vertical direction, we have

Ncosθ+fsinθw=0Equation 1N \cos \theta + f \sin \theta -w= 0 \quad \quad \quad \color{Blue} \text{Equation 1}

Summing forces in the horizontal directions taking to the left as the positive since the centripetal force is directed this way, we have

Nsinθfcosθ=FcEquation 2N \sin \theta - f \cos \theta = F_c \quad \quad \quad \color{Blue} \text{Equation 2}

We are given the following quantities:

  • radius of curvature, r=100 metersr=100\ \text{meters}
  • banking angle, θ=15.0\theta = 15.0^\circ
  • velocity, v=20 km/h=5.5556 m/sv=20\ \text{km/h} = 5.5556\ \text{m/s}
  • We also know that the friction, f=μsNf=\mu_{s} N and weight, w=mgw=mg

We now use equation 1 to solve for N in terms of the other variables.

Ncosθ+fsinθw=0Ncosθ+μsNsinθ=mgN(cosθ+μssinθ)=mgN=mgcosθ+μssinθ  (Equation 3)\begin{align*} N \cos \theta + f \sin \theta -w & = 0 \\ N \cos \theta +\mu_s N\sin \theta & = mg \\ N \left( \cos \theta + \mu_s \sin\theta \right) & =mg \\ N & = \frac{mg}{\cos \theta + \mu_s \sin\theta} \ \qquad \ \color{Blue} \left( \text{Equation 3} \right) \end{align*}

We also solve for N in equation 2.

NsinθμsNcosθ=mv2rN(sinθμscosθ)=mv2rN=mv2r(sinθμscosθ)  (Equation 4)\begin{align*} N \sin \theta - \mu_s N \cos \theta & = m \frac{v^2}{r} \\ N \left( \sin \theta-\mu_s \cos \theta \right) & = m \frac{v^2}{r} \\ N & = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \ \qquad \ \color{Blue} \left( \text{Equation 4} \right) \end{align*}

Now, we have two equations of NN. We now equate these two equations.

mgcosθ+μssinθ=mv2r(sinθμscosθ)\frac{mg}{\cos \theta + \mu_s \sin\theta} = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)}

We can now use this equation to solve for μs\mu_s.

mgcosθ+μssinθ=mv2r(sinθμscosθ)rg(sinθμscosθ)=v2(cosθ+μssinθ)rgsinθμsrgcosθ=v2cosθ+μsv2sinθμsv2sinθ+μsrgcosθ=rgsinθv2cosθμs(v2sinθ+rgcosθ)=rgsinθv2cosθμs=rgsinθv2cosθv2sinθ+rgcosθ\begin{align*} \frac{\bcancel{m}g}{\cos \theta + \mu_s \sin\theta} & = \frac{\bcancel{m}v^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \\ rg\left( \sin \theta-\mu_s \cos \theta \right) & = v^2 \left( \cos \theta + \mu_s \sin\theta \right) \\ rg \sin \theta - \mu_s rg \cos \theta & = v^2 \cos \theta +\mu_s v^2 \sin \theta \\ \mu_s v^2 \sin \theta + \mu _s rg \cos \theta & = rg \sin \theta - v^2 \cos \theta \\ \mu _s \left( v^2 \sin \theta + rg \cos \theta \right) & = rg \sin \theta - v^2 \cos \theta \\ \mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta} \end{align*}

Now that we have an equation for μs\mu_s, we can substitute the given values.

μs=rgsinθv2cosθv2sinθ+rgcosθμs=100 m(9.81 m/s2)sin15.0(5.5556 m/s)2cos15.0(5.5556 m/s)2sin15.0+100 m(9.81 m/s2)cos15.0μs=0.2345  (Answer)\begin{align*} \mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta} \\ \mu_s & = \frac{100\ \text{m}(9.81\ \text{m/s}^2) \sin 15.0^\circ -\left( 5.5556\ \text{m/s} \right)^2 \cos 15.0^\circ }{\left( 5.5556\ \text{m/s} \right)^2 \sin 15.0^\circ +100\ \text{m}\left( 9.81\ \text{m/s}^2 \right) \cos 15.0^\circ } \\ \mu_s & = 0.2345 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 6 Problem 29

The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts


Problem:

A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration 10g10g if the rider is 15.0 m from the center of rotation?

(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 6.34(b). At what angle θ\theta below the horizontal will the cage hang when the centripetal acceleration is  10g10g? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle 10g10g should be.)

Figure 6.34 (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to always be along its axis.

Solution:

Part A

The centripetal acceleration, aca_c, is calculated using the formula ac=rω2a_c = r \omega ^2. Solving for the angular velocity, ω\omega, in terms of the other variables, we should come up with

ω=acr\omega = \sqrt{\frac{a_c}{r}}

We are given the following values:

  • centripetal acceleration, ac=10g=10(9.81 m/s2)=98.1 m/s2a_c = 10g = 10 \left( 9.81\ \text{m/s}^2 \right) = 98.1\ \text{m/s}^2
  • radius of curvature, r=15.0 mr = 15.0\ \text{m}

Substituting the given values into the equation,

ω=acrω=98.1 m/s215.0 mω=2.5573 rad/secω=2.56 rad/sec  (Answer)\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{98.1\ \text{m/s}^2}{15.0\ \text{m}}} \\ \\ \omega & = 2.5573\ \text{rad/sec} \\ \\ \omega & = 2.56\ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The free-body diagram of the force is shown

The free-body diagram of the rider’s cage that hangs on a pivot at the end of the arm of a large centrifuge. College Physics Problem 6-29
The free-body diagram of the rider’s cage hangs on a pivot at the end of the arm of a large centrifuge.

Summing forces in the vertical direction, we have

Fy=0Farmsinθw=0Farm=wsinθ Equation 1\begin{align*} \sum_{}^{} F_y & = 0 \\ \\ F_{arm} \sin \theta-w & = 0 \\ \\ F_{arm} & = \frac{w}{\sin \theta} \ \quad \quad \color{Blue} \text{Equation 1} \end{align*}

Now, summing forces in the horizontal direction, taking into account that FcF_c is the centripetal force which is the net force. That is,

Fc=mac\begin{align*} F_c & = m a_c \end{align*}

We know that FcF_c is equal to the horizontal component of the force FarmF_{arm}. That is Fc=FarmcosθF_c = F_{arm} \cos \theta. Therefore,

Farmcosθ=mac\begin{align*} F_{arm} \cos \theta & = m a_c \\ \end{align*}

Now, we can substitute equation 1 into the equation, and the value of the centripetal acceleration given at 10g10g. Also, we note that the weight ww is equal to mgmg. So, we have

Farmcosθ=macwsinθcosθ=m(10g)mgcosθsinθ=10mg\begin{align*} F_{arm} \cos \theta & = m a_c \\ \\ \frac{w}{\sin \theta} \cos \theta & = m (10g) \\ \\ \frac{mg \cos \theta}{\sin \theta} & = 10 mg \\ \\ \end{align*}

From here, we are going to use the trigonometric identity tanθ=sinθcosθ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We can also cancel mm, and g since they can be found on both sides of the equation.

1tanθ=10tanθ=110θ=tan1(110)θ=5.71  (Answer)\begin{align*} \frac{1}{\tan \theta} & = 10 \\ \\ \tan \theta & = \frac{1}{10} \\ \\ \theta & = \tan ^{-1} \left( \frac{1}{10} \right) \\ \\ \theta & = 5.71 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 6 Problem 28

Riding a Bicycle in an Ideally Banked Curve


Problem:

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.33. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that θ\theta (as defined in the figure) is related to the speed vv and radius of curvature rr of the turn in the same way as for an ideally banked roadway—that is, θ=tan1(v2/rg)\theta = \tan ^{-1} \left( v^2/rg \right)

(b) Calculate θ\theta for a 12.0 m/s turn of radius 30.0 m (as in a race).

Figure 6.33 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system, while its horizontal component must supply the centripetal force. This process produces a relationship among the angle θ, the speed v, and the radius of curvature r of the turn similar to that for the ideal banking of roadways.

Solution:

Part A

Let us redraw the given forces in a free-body diagram with their corresponding components.

The force NN and FcF_c are the vertical and horizontal components of the force FF.

If we take the equilibrium of forces in the vertical direction (since there is no motion in the vertical direction) and solve for FF, we have

Fy=0Fcosθmg=0Fcosθ=mgF=mgcosθEquation 1\begin{align*} \sum F_y & = 0 \\ \\ F \cos \theta - mg & = 0 \\ \\ F \cos \theta & = mg \\ \\ F & = \frac{mg}{\cos \theta} \quad \quad & \color{Blue} \small \text{Equation 1} \end{align*}

If we take the sum of forces in the horizontal direction and equate it to mass times the centripetal acceleration (since the centripetal acceleration is directed in this direction), we have

Fx=macFsinθ=macFsinθ=mv2rEquation 2\begin{align*} \sum F_x & = ma_c \\ \\ F \sin \theta & = m a_c \\ \\ F \sin \theta & = m \frac{v^2}{r} \quad \quad & \color{Blue} \small \text{Equation 2} \end{align*}

We substitute Equation 1 to Equation 2.

Fsinθ=mv2rmgcosθsinθ=mv2rmgsinθcosθ=mv2r\begin{align*} F \sin \theta & = m \frac{v^2}{r} \\ \\ \frac{mg}{\cos \theta} \sin \theta & = m \frac{v^2}{r} \\ \\ mg \frac{\sin \theta}{\cos \theta} & =m \frac{v^2}{r} \\ \\ \end{align*}

We can cancel mm from both sides, and we can apply the trigonometric identity tanθ=sinθcosθ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We should come up with

gtanθ=v2rtanθ=v2rgθ=tan1(v2rg)  (Answer)\begin{align*} g \tan \theta & = \frac{v^2}{r} \\ \\ \tan \theta & = \frac{v^2}{rg} \\ \\ \theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following values:

  • linear velocity, v=12.0 m/sv = 12.0\ \text{m/s}
  • radius of curvature, r=30.0 mr=30.0\ \text{m}
  • acceleration due to gravity, g=9.81 m/s2g = 9.81\ \text{m/s}^2

We substitute the given values to the formula of θ\theta we solve in Part A.

θ=tan1(v2rg)θ=tan1[(12.0 m/s)2(30.0 m)(9.81 m/s2)]θ=26.0723θ=26.1  (Answer)\begin{align*} \theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \\ \\ \theta & = \tan ^ {-1} \left[ \frac{\left( 12.0\ \text{m/s} \right)^2}{\left( 30.0\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\ \theta & = 26.0723 ^\circ \\ \\ \theta & = 26.1 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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