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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

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In Inelastic collision cunculate the vilocity
Anshu Reply
elucidate
Usman
explain how a body becomes electrically charged based on the presence of charged particles
Kym Reply
induction
babar
induction
DEMGUE
definitely by induction
Raymond
induction
Raymond
induction
Shah
induction
Korodhso
please why does a needle sinks in water
DEMGUE
induction
Korodhso
induction
Auwal
what are the calculations of Newton's third law of motiow
Murtala Reply
what is dark matter
apex Reply
(in some cosmological theories) non-luminous material which is postulated to exist in space and which could take either of two forms: weakly interacting particles ( cold dark matter ) or high-energy randomly moving particles created soon after the Big Bang ( hot dark matter ).
Usman
if the mass of a trolley is 0.1kg. calculate the weight of plasticine that is needed to compensate friction. (take g=10m/s and u=0.2)
Declan Reply
what is a galaxy
Maduka Reply
a galaxy is a type of phone e.g samsung galaxy there are diff types of samsung galaxy there is s5 s6 s7 s8 s9
lasisi
what isflow rate of volume
Abcd Reply
flow rate is the volume of fluid which passes per unit time;
Rev
flow rate or discharge represnts the flow passing in unit volume per unit time
bhat
When two charges q1 and q2 are 6 and 5 coulomb what is ratio of force
Mian Reply
incomplete question
lasisi
When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?
bimo Reply
is that the answer
nehemiah
why is it proportional
nehemiah Reply
i don't know
Adah
y
nehemiah
what are the relationship between distance and displacement
Usman Reply
They are interchangeable.
Shii
Distance is scalar, displacement is vector because it must involve a direction as well as a magnitude. distance is the measurement of where you are and where you were displacement is a measurement of the change in position
Shii
Thanks a lot
Usman
I'm beginner in physics so I can't reason why v=u+at change to v2=u2+2as and vice versa
Usman
what is kinematics
praveen
kinematics is study of motion without considering the causes of the motion
Theo
The study of motion without considering the cause 0f it
Usman
why electrons close to the nucleus have less energy and why do electrons far from the nucleus have more energy
Theo
thank you frds
praveen
plz what is the third law of thermodynamics
Chidera Reply
third law of thermodynamics states that at 0k the particles will collalse its also known as death of universe it was framed at that time when it waa nt posible to reach 0k but it was proved wrong
bhat
I have not try that experiment but I think it will magnet....
Rev Reply
Hey Rev. it will
Jeff
I do think so, it will
Chidera
yes it will
lasisi
If a magnet is in a pool of water, would it be able to have a magnetic field?.
Stella Reply
yes Stella it would
Jeff

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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