<< Chapter < Page Chapter >> Page >
  • Apply problem-solving techniques to solve for quantities in more complex systems of forces.
  • Integrate concepts from kinematics to solve problems using Newton's laws of motion.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills.

Drag force on a barge

Suppose two tugboats push on a barge at different angles, as shown in [link] . The first tugboat exerts a force of 2.7 × 10 5 N size 12{2 "." 7 times "10" rSup { size 8{5} } " N"} {} in the x -direction, and the second tugboat exerts a force of 3.6 × 10 5 N size 12{3 "." 6 times "10" rSup { size 8{5} } " N"} {} in the y -direction.

(a) A view from above two tugboats pushing on a barge. One tugboat is pushing with the force F sub x equal to two point seven multiplied by ten to the power five newtons, shown by a vector arrow acting toward the right in the x direction. Another tugboat is pushing with a force F sub y equal to three point six multiplied by ten to the power five newtons acting upward in the positive y direction. Acceleration of the barge, a, is shown by a vector arrow directed fifty-three point one degree angle above the x axis. In the free-body diagram, F sub y is acting on a point upward, F sub x is acting toward the right, and F sub D is acting approximately southwest. (b) A right triangle is made by the vectors F sub x and F sub y. The base vector is shown by the force vector F sub x. and the perpendicular vector is shown by the force vector F sub y. The resultant is the hypotenuse of this triangle, making a fifty-three point one degree angle from the base, shown by the vector force F sub net pointing up the inclination. A vector F sub D points down the incline.
(a) A view from above of two tugboats pushing on a barge. (b) The free-body diagram for the ship contains only forces acting in the plane of the water. It omits the two vertical forces—the weight of the barge and the buoyant force of the water supporting it cancel and are not shown. Since the applied forces are perpendicular, the x - and y -axes are in the same direction as F x size 12{F rSub { size 8{x} } } {} and F y size 12{F rSub { size 8{y} } } {} . The problem quickly becomes a one-dimensional problem along the direction of F app size 12{F rSub { size 8{"app"} } } {} , since friction is in the direction opposite to F app size 12{F rSub { size 8{"app"} } } {} .

If the mass of the barge is 5.0 × 10 6 kg size 12{5 times "10" rSup { size 8{6} } " kg"} {} and its acceleration is observed to be 7 . 5 × 10 2 m/s 2 size 12{7 "." "52" times "10" rSup { size 8{ - 2} } " m/s" rSup { size 8{2} } } {} in the direction shown, what is the drag force of the water on the barge resisting the motion? (Note: drag force is a frictional force exerted by fluids, such as air or water. The drag force opposes the motion of the object.)

Strategy

The directions and magnitudes of acceleration and the applied forces are given in [link] (a) . We will define the total force of the tugboats on the barge as F app size 12{F rSub { size 8{"app"} } } {} so that:

F app = F x + F y size 12{F rSub { size 8{ ital "app"} } ital "= F" rSub { size 8{x} } ital "+ F" rSub { size 8{y} } } {}

Since the barge is flat bottomed, the drag of the water F D size 12{F rSub { size 8{D} } } {} will be in the direction opposite to F app size 12{F rSub { size 8{"app"} } } {} , as shown in the free-body diagram in [link] (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Our strategy is to find the magnitude and direction of the net applied force F app size 12{F rSub { size 8{"app"} } } {} , and then apply Newton’s second law to solve for the drag force F D size 12{F rSub { size 8{D} } } {} .

Solution

Since F x size 12{F rSub { size 8{x} } } {} and F y size 12{F rSub { size 8{y} } } {} are perpendicular, the magnitude and direction of F app size 12{F rSub { size 8{"app"} } } {} are easily found. First, the resultant magnitude is given by the Pythagorean theorem:

F app = F x 2 + F y 2 F app = ( 2.7 × 10 5 N ) 2 + ( 3.6 × 10 5 N ) 2 = 4.5 × 10 5 N. alignl { stack { size 12{F rSub { size 8{ ital "app"} } = \( F rSub { size 8{x} rSup { size 8{2} } } + F rSub { size 8{y} rSup { size 8{2} } } \) rSup { size 8{1/2} } } {} #F rSub { size 8{ ital "app"} } = \( \( 2 "." 7 times "10" rSup { size 8{5} } " N" \) rSup { size 8{2} } + \( 3 "." 6 times "10" rSup { size 8{5} } " N" \) rSup { size 8{2} } \) rSup { size 8{1/2} } =4 "." "50" times "10" rSup { size 8{5} } " N" "." {} } } {}

The angle is given by

θ = tan 1 F y F x θ = tan 1 3.6 × 10 5 N 2.7 × 10 5 N = 53º , alignl { stack { size 12{θ="tan" rSup { size 8{ - 1} } left ( { {F rSub { size 8{y} } } over {F rSub { size 8{x} } } } right )} {} #θ="tan" rSup { size 8{ - 1} } left ( { { \( 2 "." 7 times "10" rSup { size 8{5} } " N" \) } over { \( 3 "." 6 times "10" rSup { size 8{5} } " N" \) } } right )="53" "." 1°, {} } } {}

which we know, because of Newton’s first law, is the same direction as the acceleration. F D size 12{F rSub { size 8{D} } } {} is in the opposite direction of F app size 12{F rSub { size 8{"app"} } } {} , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F app size 12{F rSub { size 8{"app"} } } {} , but its magnitude is slightly less than F app size 12{F rSub { size 8{"app"} } } {} . The problem is now one-dimensional. From [link] (b) , we can see that

F net = F app F D size 12{F rSub { size 8{"net"} } =F rSub { size 8{"app"} } - F rSub { size 8{D} } } {} .

But Newton’s second law states that

F net = ma size 12{F rSub { size 8{"net"} } = ital "ma"} {} .

Thus,

F app F D = ma size 12{F rSub { size 8{"app"} } - F rSub { size 8{D} } = ital "ma"} {} .

This can be solved for the magnitude of the drag force of the water F D size 12{F rSub { size 8{D} } } {} in terms of known quantities:

F D = F app ma size 12{F rSub { size 8{D} } =F rSub { size 8{"app"} } - ital "ma"} {} .

Substituting known values gives

F D = ( 4 . 5 × 10 5 N ) ( 5 . 0 × 10 6 kg ) ( 7 . 5 × 10 –2 m/s 2 ) = 7 . 5 × 10 4 N size 12{F rSub { size 8{D} } = \( 4 "." "50" times "10" rSup { size 8{5} } " N" \) - \( 5 "." "00" times "10" rSup { size 8{6} } " kg" \) \( 7 "." "50" times "10" rSup { size 8{"-2"} } " m/s" rSup { size 8{2} } \) =7 "." "50" times "10" rSup { size 8{4} } " N"} {} .

The direction of F D size 12{F rSub { size 8{D} } } {} has already been determined to be in the direction opposite to F app size 12{F rSub { size 8{"app"} } } {} , or at an angle of 53º size 12{"53" "." 1°} {} south of west.

Questions & Answers

how do you get the 2/50
Abba Reply
number of sport play by 50 student construct discrete data
Aminu Reply
width of the frangebany leaves on how to write a introduction
Theresa Reply
Solve the mean of variance
Veronica Reply
Step 1: Find the mean. To find the mean, add up all the scores, then divide them by the number of scores. ... Step 2: Find each score's deviation from the mean. ... Step 3: Square each deviation from the mean. ... Step 4: Find the sum of squares. ... Step 5: Divide the sum of squares by n – 1 or N.
kenneth
what is error
Yakuba Reply
Is mistake done to something
Vutshila
Hy
anas
hy
What is the life teble
anas
hy
Jibrin
statistics is the analyzing of data
Tajudeen Reply
what is statics?
Zelalem Reply
how do you calculate mean
Gloria Reply
diveving the sum if all values
Shaynaynay
let A1,A2 and A3 events be independent,show that (A1)^c, (A2)^c and (A3)^c are independent?
Fisaye Reply
what is statistics
Akhisani Reply
data collected all over the world
Shaynaynay
construct a less than and more than table
Imad Reply
The sample of 16 students is taken. The average age in the sample was 22 years with astandard deviation of 6 years. Construct a 95% confidence interval for the age of the population.
Aschalew Reply
Bhartdarshan' is an internet-based travel agency wherein customer can see videos of the cities they plant to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400 a. what is the probability of getting more than 12,000 hits? b. what is the probability of getting fewer than 9,000 hits?
Akshay Reply
Bhartdarshan'is an internet-based travel agency wherein customer can see videos of the cities they plan to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400. a. What is the probability of getting more than 12,000 hits
Akshay
1
Bright
Sorry i want to learn more about this question
Bright
Someone help
Bright
a= 0.20233 b=0.3384
Sufiyan
a
Shaynaynay
How do I interpret level of significance?
Mohd Reply
It depends on your business problem or in Machine Learning you could use ROC- AUC cruve to decide the threshold value
Shivam
how skewness and kurtosis are used in statistics
Owen Reply
yes what is it
Taneeya
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask