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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

how did they solve for "t" after getting 67.6=.5(Voy + 0)t
Martin Reply
Find the following for path D in [link] : (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.
David Reply
the topic is kinematics
David
can i get notes of solid state physics
Lohitha
just check the chpt. 13 kinetic theory of matter it's there
David
is acceleration a fundamental unit.
David Reply
no it is derived
Abdul
no
Nisha
K thanks
David
hi guys can you teach me how to solve a logarithm?
Villaflor Reply
how about a conceptual framework can you simplify for me? needed please
Villaflor
Hello what happens when electrone stops its rotation around its nucleus if it possible how
Afzal
I think they are constantly moving
Villaflor
yep what is problem you are stuck into context?
S.M
not possible to fix electron position in space,
S.M
Physics
Beatriz
yes of course Villa flor
David
equations of kinematics for constant acceleration
Sagcurse Reply
A bottle full of water weighs 45g when full of mercury,it weighs 360g.if the empty bottle weighs 20g.calculate the relative density of mercury and the density of mercury....pls I need help
Lila Reply
well You know the density of water is 1000kg/m^3.And formula for density is density=mass/volume Then we must calculate volume of bottle and mass of mercury: Volume of bottle is (45-20)/1000000=1/40000 mass of mercury is:(360-20)/1000 kg density of mercury:(340/1000):1/50000=(340•40000):1000=13600
Sobirjon
the latter is true
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture...take density of water as 1g/cm3 and density of liquid 1.2g/cm3
Lila
plz hu can explain Heisenberg's uncertainty principle
Emmanuel Reply
who can help me with my problem about acceleration?
Vann Reply
ok
Nicholas
how to solve this... a car is heading north then smoothly made a westward turn during the travel the speed of the car remains constant at 1.5km/h what is the acceleration of the car? the total travel time of the car as it smoothly changed its direction is 15 minutes
Vann
i think the acceleration is 0 since the car does not change its speed unless there are other conditions
Ben
yes I have to agree, the key phrase is, "the speed of the car remains constant...," all other information is not needed to conclude that acceleration remains at 0 during the entire time
Luis
who can help me with a relative density question
Lila
1cm3 sample of tin lead alloy has mass 8.5g.the relative density of tin is 7.3 and that of lead is 11.3.calculate the percentage by weight of tin in the alloy. assuming that there is no change of volume when the metals formed the alloy
Lila
morning, what will happen to the volume of an ice block when heat is added from -200°c to 0°c... Will it volume increase or decrease?
adefenwa Reply
no
Emmanuel
hi what is physical education?
Kate
BPED..is my course.
Kate
No
Emmanuel
I think it is neither decreases nor increases ,it remains in the same volume because of its crystal structure
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture. take density of water as 1g/cm3 and density of liquid as 1.2g/cm3
Lila
Sorry what does it means"no changes in volume occured"?
Sobirjon
volume can be the amount of space occupied by an object. But when an object does not change in shape it will still occupy the same space. Thats why the volume will still remain the same
Ben
Most soilds expand when heated but if it changes state at 0C it will have less volume. Ice floats because it is less dense ie a larger mass per unit volume.
Richard
how to calculate velocity
Okwethu Reply
v=d/t
Emeka
his about the speed?
Villaflor
how about speed
Villaflor
v=d/t
Nisha
hello bro hw is life with you
Jacob Reply
Mine is good. How about you?
Chase
Hi room of engineers
lawan Reply
yes,hi sir
Okwethu
hello
akinmeji
Hello
Mishael
hello
Jerry
hi
Sakhi
hi
H.C
so, what is going on here
akinmeji
u are all wlc just ask your question anybody. can answer
Ajayi
good morning ppl
ABDUL
If someone has not studied Mathematics enough yet, should theu study it first then study Phusics or Study Basics of Physics whilst srudying Math as well?
Riaz Reply
whether u studied maths or not, it is advisable to start from d basics cuz it is essential to know dem
Nuru
yea you are right
Badmus
wow, you got this w/o knowing math
Thomas
I guess that's it
Thomas
later people
Thomas
mathematics is everywhere
Anand
thanks but dat doesn't mean it is good without maths @Riaz....... Maths is essential in sciences particularly wen it comes to PHYSICS but PHYSICS must be started from the basic which may also help in ur mathematical ability
Nuru
A hydrometer of mass 0.15kg and uniform cross sectional area of 0.0025m2 displaced in water of density 1000kg/m3.what depth will the hydrometer sink
Lila
16.66 meters?
Darshik
16.71m2
aways
,i have a question of let me give answer
aways
the mass is stretched a distance of 8cm and held what is the potential energy? quick answer
aways
oscillation is a to and fro movement, it can also be referred to as vibration. e.g loaded string, loaded test tube or an hinged door
Olatunji Reply
what property makes the magnet to break?
Akshaya Reply

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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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