Chapter 4: Linear Kinetics, Force and Newton’s Laws of Motion
Chapter 4: Linear Kinetics, Force and Newton’s Laws of Motion
4.11 Further Applications of Newton’s Laws of Motion
Authors: Paul Peter Urone, Roger Hinrichs
Adapted by: Rob Pryce, Alix Blacklin
Learning Objectives
By the end of this section, you will be able to:
- 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.
Example: Drag Force on a Barge
Suppose two tugboats push on a barge at different angles, as shown in Figure 4.22. The first tugboat exerts a force of [latex]2.7×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}[/latex] in the x-direction, and the second tugboat exerts a force of [latex]3.6×{\text{10}}^{5}\phantom{\rule{0.25em}{0ex}}\text{N}[/latex] in the y-direction.
If the mass of the barge is [latex]5.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{kg}[/latex] and its acceleration is observed to be [latex]7\text{.}\text{5}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}[/latex] 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 Figure 4.22(a). We will define the total force of the tugboats on the barge as [latex]{\mathbf{\text{F}}}_{\text{app}}[/latex] so that:
[latex]F_{app} = F_x+F_y[/latex]
Since the barge is flat bottomed, the drag of the water [latex]{\mathbf{\text{F}}}_{\text{D}}[/latex] will be in the direction opposite to [latex]{\mathbf{\text{F}}}_{\text{app}}[/latex], as shown in the free-body diagram in Figure 4.22(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 [latex]{\mathbf{\text{F}}}_{\text{app}}[/latex], and then apply Newton’s second law to solve for the drag force [latex]{\mathbf{\text{F}}}_{\text{D}}[/latex].
Solution
Since [latex]{\mathbf{\text{F}}}_{x}[/latex] and [latex]{\mathbf{\text{F}}}_{y}[/latex] are perpendicular, the magnitude and direction of [latex]{\mathbf{\text{F}}}_{\text{app}}[/latex] are easily found. First, the resultant magnitude is given by the Pythagorean theorem:
[latex]F_{app} =\sqrt{F_x^2+F_y^2} = \sqrt{\left(2.7×10^5\text{N}\right)^{2}+\left(3.6×10^5\text{N}\right)^{2}} =4.5×10^5\text{N.}[/latex]
The angle is given by
[latex]\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}\phan