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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

why is it dat when using double pan balance the known and unknown mass are the same
Victor Reply
is there more then 4 dimensions
Miguel Reply
hii
princy
hi
Miguel
hello I kinda need help in physics... a lot
Brown
Brown. what kind of help
Jeff
when it comes to physics stick with the basics don't overthink things
Jeff
yes
ayesha
sticking to the basics will take you farther than overwhelming yourself with more than you need to physics is simple keep it simple
Jeff
thk u Ayesha
Jeff
for real....? so I've got to know the fundamentals and use the formula to solve any problem
Brown
read Stephan hawkings a brief history of time
ayesha
it'll help you understand better than summing up formulas or ready textbooks
ayesha
physics isn't hard it's just understanding and applying the formulas if u need help ask any question
ayesha
okay...because I've got an exam next year February a Computer based exam
Brown
start with a pace a plan and stick to it
ayesha
well best of luck can't help you much there contact your teachers for tips and helpful notes
ayesha
how can we find absolute uncertainty
ayesha Reply
it what?
Luke
in physics
ayesha
the basic formula is uncertainty in momentum multiplied buy uncertainty In position is greater than or equal to 4×pi/2. same formula for energy and time
Luke
I have this one question can you please look it up it's 9702/22/O/N/17 Question 1 B 3
ayesha
what
uma
would you like physics?
Suthar
yes
farooq
precision or absolute uncertainty is always equal to least count of that instrument
Iram
how do I unlock the MCQ and the Essay?
Ojeh Reply
what is the dimension of strain
Joy Reply
Is there a formula for time of free fall given that the body has initial velocity? In other words, formula for time that takes a downward-shot projectile to hit the ground. Thanks!
Cyclone Reply
hi
Agboro
hiii
Chandan
Hi
Sahim
hi
Jeff
hey
Priscilla
sup guys
Bile
Hy
Kulsum
What is unit of watt?
Kulsum
watt is the unit of power
Rahul
p=f.v
Rahul
watt can also be expressed as Nm/s
Rahul
what s i unit of mass
Maxamed
SI unit of mass is Kg(kilogram).
Robel
what is formula of distance
Maxamed
Formula for for the falling body with initial velocity is:v^2=v(initial)^2+2*g*h
Mateo
i can't understand
Maxamed
we can't do this calculation without knowing the height of the initial position of the particle
Chathu
sorry but no more in science
Imoreh
2 forces whose resultant is 100N, are at right angle to each other .if one of them makes an angle of 30 degree with the resultant determine it's magnitude
Victor Reply
50 N... (50 *1.732)N
Sahim
Plz cheak the ans and give reply..
Sahim
50 N...(50 *1.732)N
Ibrahim
show the working
usiomon
what is the value of f1 and f2
Syed
what is the value of force 1 and force 2.
Syed
.
muhammad
Is earth is an inertial frame?
Sahim Reply
The abacus (plural abaci or abacuses), also called a counting frame, is a calculating tool that was in use in Europe, China and Russia, centuries before the adoption of the written Hindu–Arabic numeral system
Sahim
thanks
Irungu
Most welcome
Sahim
Hey.. I've a question.
Sahim Reply
Is earth inertia frame?
Sahim
only the center
Shii
What is an abucus?
Irungu
what would be the correct interrogation "what is time?" or "how much has your watch ticked?"
prakash Reply
someone please give answer to this.
prakash
a load of 20N on a wire of cross sectional area 8×10^-7m produces an extension of 10.4m. calculate the young modules of the material of the wire is of length 5m
Ebenezer Reply
Young's modulus = stress/strain strain = extension/length (x/l) stress = force/area (F/A) stress/strain is F l/A x
El
so solve it
Ebenezer
please
Ebenezer
two bodies x and y start from rest and move with uniform acceleration of a and 4a respectively. if the bodies cover the same distance in terms of tx and ty what is the ratio of tx to ty
Oluwatola Reply
what is cesium atoms?
prakash Reply
The atoms which form the element Cesium are known as Cesium atoms.
Naman
A material that combines with and removes trace gases from vacuum tubes.
Shankar
what is difference between entropy and heat capacity
Varun
Heat capacity can be defined as the amount of thermal energy required to warm the sample by 1°C. entropy is the disorder of the system. heat capacity is high when the disorder is high.
Chathu
I want learn physics
Vinodhini Reply
sir how to understanding clearly
Vinodhini
try to imagine everything you study in 3d
revolutionary
pls give me one title
Vinodhini
displacement acceleration how understand
Vinodhini
vernier caliper usage practically
Vinodhini
karthik sir is there
Vinodhini
what are the solution to all the exercise..?
What is realm
Vinodhini Reply
The quantum realm, also called the quantum scale, is a term of art inphysics referring to scales where quantum mechanical effects become important when studied as an isolated system. Typically, this means distances of 100 nanometers (10−9meters) or less or at very low temperature.
revolutionary

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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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