4.7 Further applications of newton’s laws of motion  (Page 7/6)

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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.

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