Frictional Force in a Knee Joint| Further Applications of Newton’s Laws: Friction, Drag, and Elasticity| College Physics| Problem 5.3

PROBLEM:

(a) What is the maximum frictional force in the knee joint of a person who supports 66.0 kg of her mass on that knee?

(b) During strenuous exercise, it is possible to exert forces to the joints that are easily ten times greater than the weight being supported. What is the maximum force of friction under such conditions? The frictional forces in joints are relatively small in all circumstances except when the joints deteriorate, such as from injury or arthritis. Increased frictional forces can cause further damage and pain.

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College Physics 4.10 – The force a powerful motorcycle exerts backward on the ground to produce its acceleration


A powerful motorcycle can produce an acceleration of 3.50m/s² while traveling at  90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous to air friction. It always opposes the motion of an object.) What force does the motorcycle exert backward on the ground to produce its acceleration if the mass of the motorcycle with the rider is 245 kg?


Solution:

From Newton’s second law of motion, we know that

\displaystyle \text{net F}=\text{ma}

Considering the horizontal direction and let f be the resisting force, the equation becomes

\displaystyle \text{F}-\text{f}=\text{ma}

We are solving for F. Substituting the given values, we have

\displaystyle \text{F}-400\:\text{N}=\left(245\:\text{kg}\right)\left(3.50\:\text{m/s}^2\right)

\displaystyle \text{F}=\left(245\:\text{kg}\right)\left(3.50\:\text{m/s}^2\right)+400\:\text{N}

\displaystyle \text{F}=857.5\:\text{N}+400\:\text{N}

\displaystyle \text{F}=1257.5\:\text{N}        ◀


College Physics 4.9 – The acceleration of a child in a wagon push by two children in opposite directions


Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0 N, the second a force of 90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg.
(a)What is the system of interest if the acceleration of the child in the wagon is to be calculated?
(b)Draw a free-body diagram, including all forces acting on the system.
(c)Calculate the acceleration.
(d)What would the acceleration be if friction were 15.0 N?


Solution:

Part A

If the acceleration of the child in the wagon is to be calculated, the system of interest is the wagon with the child in it.

Part B

Considering all the forces acting on the wagon and child with mass m, there will be 5 forces. The horizontal forces are the two forces coming from the two other children, denoted as Fr and Fl. We also have friction going against the direction of motion.

The vertical forces are the weight of the wagon and the child and the normal force. These forces are shown in the free-body diagram shown.

Considering all the forces acting on the wagon and child with mass m. there will be 5 forces. The horizontal forces are the two forces coming from the two other children, denoted as F_l: and F_r. We also have friction going against the direction of motion. The vertical forces are the weight of the wagon and the child and the normal force. These forces are shown in the free-body diagram shown.
Free-body diagram showing the 5 forces acting on the system.

Part C

From Newton’s second law of motion, we know that

\displaystyle \text{net F}=\text{ma}

Considering the horizontal direction to the right as positive, the forces involved are Fr, Fl and f. Therefore, we have

\displaystyle \text{F}_{\text{r}}-\text{F}_{\text{l}}-\text{f}=\text{ma}

Solving for acceleration, we have

\displaystyle \text{a}=\frac{\text{F}_\text{r}-\text{F}_\text{l}-\text{f}}{\text{m}}

\displaystyle \text{a}=\frac{90.0\:\text{N}-75.0\:\text{N}-12.0\:\text{N}}{23.0\:\text{kg}}

\displaystyle \text{a}=0.1304\:\text{m/s}^2       ◀

Part D

If the friction is 15 N, the acceleration would be

\displaystyle \text{a}=\frac{90.0\:\text{N}-75.0\:\text{N}-15.0\:\text{N}}{23.0\:\text{kg}}

\displaystyle \text{a}=0\:\text{m/s}^2       ◀

There would be no acceleration, meaning that the system is in equilibrium. The system is either not moving or it’s moving at a constant velocity.


College Physics 4.8 – The deceleration of a rocket sled


What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)


Solution:

We can answer this question by referring to the definition of acceleration–change in velocity over change in time. That is

\displaystyle \text{a}=\frac{\Delta \text{v}}{\Delta \text{t}}

Before substituting the given values, the units should be homogeneous. We will convert 1000 km/hr to m/s.

\displaystyle 1000\:\frac{\text{km}}{\text{hr}}\times \frac{1000\:\text{m}}{1\:\text{km}}\times \frac{1\:\text{hr}}{3600\:\text{s}}=277.7778\:\text{m/s}

We substitute the values into the equation,

\displaystyle \text{a}=\frac{277.7778\:\text{m/s}}{1.1\:\text{s}}

\displaystyle \text{a}=252.5252\:\text{m/s}^2       ◀

The deceleration is about 252.5252 m/s2


College Physics 4.7 – The acceleration of a rocket sled given its thrust


(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4×104 N, and the force of friction opposing the motion is known to be 650 N.
(b) Why is the acceleration not one-fourth of what it is with all rockets burning?

(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4×104 N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?
Figure 4.31

Solution:

Part A

If we look at the given figure, there are two forces responsible for the motion of the object in the horizontal direction–the thrust and the friction. We can state that the net force is equal to thrust minus friction. We are also given that T=25,900 Newtons. Based from Newton’s second law of motion, we can calculate for the acceleration as

\displaystyle \text{net F}=\text{ma}

\displaystyle \text{a}=\frac{\text{net F}}{\text{m}}

\displaystyle \text{a}=\frac{2.4\times 10^4\:\text{N}-650\:\text{N}}{2100\:\text{kg}}

\displaystyle \text{a}=11.1180\:\text{m/s}^2       ◀

Part B

The magnitude of the acceleration depends on two forces–the thrust and the friction. Since the magnitude of the friction is the same no matter how many rockets are burning, then the acceleration can not be 1/4 when only 1 of the four rockets are burning.


College Physics 4.6 – Force necessary for a rocket sled deceleration


The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s². What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg. 

The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s2 . What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.
Figure 4.30: The rocket sled

Solution:

From the problem, we know that the acceleration a is 196 m/s² and the mass is 2100 kg. We are asked to solve for the net force F. Basically, we can compute for the net force using the Second Law of Motion by Newton. 

\displaystyle \text{net F}=\text{ma}

\displaystyle \text{net F}=\left(2100\:\text{kg}\right)\left(196\:\text{m/s}^2\right)

\displaystyle \text{net F}=411,600\:\text{N}       ◀

The net force opposing motion should be 411,600 Newtons to produce the given deceleration.


College Physics 4.5 – The force exerted on the mower


In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

Figure 4.7 The net force on a lawn mower is 51 N to the right

Solution:

From the problem, we know that the net external force is 51 N. This net force is the difference between the force exerted by the man and the opposing friction force. That is

\text{net F}=\text{F}-\text{f}

So, we can solve for the force by the person to the mower. 

\text{F}=\text{net F}+\text{f}

\text{F}=51\:\text{N}\:+24\:\text{N}

\text{F}=75\:\text{N}       ◀

We shall use kinematical equations to solve for the distance traveled by the mower after force F is removed. When F is removed, only f is now the only present force in the mower, so we can compute for the acceleration using the negative force. 

\displaystyle \text{a}=\frac{\text{F}}{\text{m}}

\displaystyle \text{a}=\frac{-24\:\text{N}}{24\:\text{kg}}

\text{a}=-1\:\text{m/s}^2

We shall use this acceleration to compute for the distance travelled. 

\text{v}^2=\left(\text{v}_0\right)^2+2\text{ax}

\left(0\right)^2=\left(1.5\:\text{m/s}\right)^2+2\left(-1.0\:\text{m/s}^2\right)\text{x}

\text{x}=1.125\:\text{m}       ◀

The mower will go 1.125 meters before stopping when the force F is already removed.


Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

SOLUTION:

Draw the free-body diagram of the car

week 2 problem 8

Consider the vertical direction

\sum F_y=ma_y

F_N-mg=0

F_N=mg

Consider the motion in the horizontal direction

Solve for the acceleration of the car.

v^2=\left(v_0\right)^2+2a_x\Delta x

a_x=\frac{v^2-\left(v_0\right)^2}{2\Delta x}=\frac{0-50^2}{2\left(300\right)}=-4.17\:m/s^2

Solve for the coefficient of kinetic friction

\sum F_x=ma_x

-F_{fr}=ma_x

-\mu _kF_N=ma_x

\mu _k=\frac{ma_x}{-F_N}=\frac{m\:\left(-4.17\right)}{-m\left(9.80\right)}=\frac{4.17}{9.80}

\mu _k=0.43