# 4.7 Further applications of newton’s laws of motion

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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 $2.7×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}$ in the x -direction, and the second tugboat exerts a force of $3.6×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}$ in the y -direction.

If the mass of the barge is $5.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{kg}$ and its acceleration is observed to be $7\text{.}\text{5}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$ 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 ${\mathbf{\text{F}}}_{\text{app}}$ so that:

${\mathbf{\text{F}}}_{\text{app}}\text{=}{\mathbf{\text{F}}}_{\mathit{x}}+{\mathbf{\text{F}}}_{\mathit{y}}$

Since the barge is flat bottomed, the drag of the water ${\mathbf{\text{F}}}_{\text{D}}$ will be in the direction opposite to ${\mathbf{\text{F}}}_{\text{app}}$ , 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 ${\mathbf{\text{F}}}_{\text{app}}$ , and then apply Newton’s second law to solve for the drag force ${\mathbf{\text{F}}}_{\text{D}}$ .

Solution

Since ${\mathbf{\text{F}}}_{x}$ and ${\mathbf{\text{F}}}_{y}$ are perpendicular, the magnitude and direction of ${\mathbf{\text{F}}}_{\text{app}}$ are easily found. First, the resultant magnitude is given by the Pythagorean theorem:

$\begin{array}{lll}{F}_{\text{app}}& =& \sqrt{{\text{F}}_{x}^{2}+{\text{F}}_{y}^{2}}\\ {F}_{\text{app}}& =& \sqrt{\left(2.7×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}{\right)}^{2}+\left(3.6×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}{\right)}^{2}}& =& 4.5×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N.}\end{array}$

The angle is given by

$\begin{array}{lll}\theta & =& {\text{tan}}^{-1}\left(\frac{{F}_{y}}{{F}_{x}}\right)\\ \theta & =& {\text{tan}}^{-1}\left(\frac{3.6×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}}{2.7×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}}\right)=\text{53º},\end{array}$

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

${F}_{\text{net}}={F}_{\text{app}}-{F}_{\text{D}}.$

But Newton’s second law states that

${F}_{\text{net}}=\text{ma}.$

Thus,

${F}_{\text{app}}-{F}_{\text{D}}=\text{ma}.$

This can be solved for the magnitude of the drag force of the water ${F}_{\text{D}}$ in terms of known quantities:

${F}_{\text{D}}={F}_{\text{app}}-\text{ma}.$

Substituting known values gives

${\text{F}}_{\text{D}}=\left(4\text{.}\text{5}×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}\right)-\left(5\text{.}\text{0}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(7\text{.}\text{5}×{\text{10}}^{\text{–2}}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}\right)=7\text{.}\text{5}×{\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\text{N}.$

The direction of ${\mathbf{\text{F}}}_{\text{D}}$ has already been determined to be in the direction opposite to ${\mathbf{\text{F}}}_{\text{app}}$ , or at an angle of $\text{53º}$ south of west.

how did they solve for "t" after getting 67.6=.5(Voy + 0)t
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.
the topic is kinematics
David
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Lohitha
just check the chpt. 13 kinetic theory of matter it's there
David
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no it is derived
Abdul
no
Nisha
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David
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S.M
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S.M
Physics
Beatriz
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David
equations of kinematics for constant acceleration
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
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
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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
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who can help me with my problem about acceleration?
ok
Nicholas
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Vann
i think the acceleration is 0 since the car does not change its speed unless there are other conditions
Ben
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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?
no
Emmanuel
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No
Emmanuel
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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
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v=d/t
Emeka
Villaflor
Villaflor
v=d/t
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ABDUL
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Thomas
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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
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Darshik
16.71m2
aways
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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
what property makes the magnet to break?