College Physics 3.9 – Sum of two vectors


Show that the sum of the vectors discussed in the example Subtracting Vectors Graphically: A Woman Sailing a Boat gives the result shown in Figure 3.24.

The resultant vector R' is shown together with vectors A and B. Vector A is directed 66° from the positive x-axis and vector B is directed 112° from the positive x-axis
Figure 3.24

Solution:

So, we are given the two vectors shown below.

The figure shows vectors A and B. Vector A has magnitude 27.5 meters and is directed 66° from the positive x-axis, while a separate vector B has magnitude 30.0 meters is directed 112° from the positive x-axis.
Figure 3.9A: Vectors A and B

If we use the graphical method of adding vectors, we can join the two vectors using head-tail addition and come up with the following:

The figure shows vector A is drawn from the origin, and then vector B starts from the head of vector A
Figure 3.9B: Vectors A and B added graphically

The resultant is drawn from the tail of the first vectors (the origin) to the head of the last vector. The resultant is shown in red in the figure below.

The figure shows the vectors A and B together with their resultant in red
Figure 3.9B: The resultant of vectors A and B using graphical method

The result is in conformity with that in figure 3.24 shown on the question shown above.


College Physics 3.8 – Order of addition of 3 vectors does not affect their sum


Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors A, B, and C, all having different lengths and directions. Find the sum A + B + C then find their sum when added in a different order and show the result is the same. (There are five other orders in which A, B, and C can be added; choose only one.)


Solution:

Consider the three vectors shown in the figure below:

Three vectors are given: Vector A is directed to the right, vector B is directed upward, and vector C is directed to the left.
Figure 3.8A: The 3 given vectors

First, we shall add them A+B+C. Using the head-tail or graphical method of vector addition, we have the figure shown below.

The 3 vectors A, B, and C are added and the resultant force is also shown in red color directed upward.
Figure 3.8B: The resultant force of A+B+C

Now, let us try to find the sum of the three vectors by reordering vectors A, B, and C. Let us try to find the sum of C+B+A in that order. The result is shown below.

The sum of vectors A, B and C is shown with their resultant in red.
Figure 3.8C: The resultant of 3 vectors added in different order.

We can see that the resultant is the same directed from the origin upward. This proves that the resultant must be the same even if the vectors are added in different order.


College Physics 2.66 – Solving for velocity and acceleration from position graph


Figure 2.68 shows the position graph for a particle for 6 s.

(a) Draw the corresponding Velocity vs. Time graph.

(b) What is the acceleration between 0 s and 2 s?

(c) What happens to the acceleration at exactly 2 s?

position graph for a particle for 6 s.
Figure 2.68

Solution:

Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is compose of straight lines, we can say that the velocity is constant for several time ranges.

Time RangesSlope of the position Graph/Velocity
0 to 2 seconds=\frac{2-0}{2-0}=1\:\text{m/s}
2 to 3 seconds=\frac{-3-2}{3-2}=\frac{-5}{1}=-5\:\text{m/s}
3 to 5 seconds0 \text{m/s}
5 to 5 seconds=\frac{-2-\left(-3\right)}{6-5}=\frac{1}{1}=1\:\text{m/s}

Based from the data in the table, we can draw the velocity diagram

velocity vs time graph

Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.


College Physics 2.65 – Velocity graph of a world-class sprinter


A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67).

(a) What is his average velocity for the first 4 s?

(b) What is his instantaneous velocity at t=5 s?

(c) What is his average acceleration between 0 and 4 s?

(d) What is his time for the race?

A graph of  v(t)  is shown for a world-class track sprinter in a 100-m race.
Figure 2.67

Solution:

Part A

To find for the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is

\text{v}_{\text{ave}}=6\:\text{m/s}

Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is

\text{a}=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2

Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve.

\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by

\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec

Therefore, the total time of the sprint is

\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec


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”