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  • 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

where we get a research paper on Nano chemistry....?
Maira Reply
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
Maira Reply
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
Google
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
Hafiz Reply
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
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Source:  OpenStax, Physics 110 at une. OpenStax CNX. Aug 29, 2013 Download for free at http://legacy.cnx.org/content/col11566/1.1
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