31 Further Applications of Newton’s Laws of Motion
31 Further Applications of Newton’s Laws of Motion
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 1. Drag Force on a Barge
Suppose two tugboats push on a barge at different angles, as shown in Figure 1. The first tugboat exerts a force of 2.7 × 105 N in the x-direction, and the second tugboat exerts a force of 3.6 × 105 in the y-direction.
If the mass of the barge is 5.10 × 106 kg and its acceleration is observed to be 7.5 × 10-2 m/s2 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 1(a). We will define the total force of the tugboats on the barge as Fapp so that:
Since the barge is flat bottomed, the drag of the water FD will be in the direction opposite to Fapp, as shown in the free-body diagram in Figure 1(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 Fapp, and then apply Newton’s second law to solve for the drag force FD.
Solution
Since Fx and Fy are perpendicular, the magnitude and direction of Fapp are easily found. First, the resultant magnitude is given by the Pythagorean theorem:
[latex]\begin{array}{lll}{F}_{\text{app}}& =& \sqrt{{{\text{F}}_{x}}^{2}+{{{\text{F}}_{y}}^{2}}} \\ {F}_{\text{app}}& =& \sqrt{\left(2.7\times {10}^{5}\text{ N}\right)^{2}+\left(3.6\times {10}^{5}\text{ N}\right)^{2}}& =& 4.5\times {10}^{5}\text{ N}\end{array}\\[/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\times {\text{10}}^{5}\text{ N}}{2.7\times {\text{10}}^{5}\text{ N}}\right)=53^{\circ},\end{array}\\[/latex]
which we know, because of Newton’s first law, is the same direction as the acceleration. FD is in the opposite direction of Fapp, since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as Fapp, but its magnitude is slightly less than Fapp. The problem is now one-dimensional. From Figure 1(b), we can see that
But Newton’s second law states that
Fnet = ma
Thus,
Fapp – FD = ma
This can be solved for the magnitude of the drag force of the water FD in terms of known quantities:
FD = Fapp – ma
Substituting known values gives
FD = (4.5 × 105 N) – (5.0 × 106