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.

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

Two children pushing on a third child in a wagon| Newton’s Second Law of Motion: Concept of a System| Dynamics| College Physics| Problem 4.9

PROBLEM:

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?

Continue reading “Two children pushing on a third child in a wagon| Newton’s Second Law of Motion: Concept of a System| Dynamics| College Physics| Problem 4.9”

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