The task of determining the amount of influence of a single
vector in a given direction involves the use of trigonometric functions. The
use of these functions to determine the components of a single vector was also
discussed in Lesson 1 of this unit. As a quick review, let's consider the use of SOH CAH
TOA to determine the components of force acting upon Fido. Assume that the
chain is exerting a 60 N force upon Fido at an angle of 40 degrees above the
horizontal. A quick sketch of the situation reveals that to determine the
vertical component of force, the sine function can be used and to determine the
horizontal component of force, the cosine function can be used. The solution to
this problem is shown below.
As another example of the use of SOH CAH TOA to resolve a single vector
into its two components, consider the diagram at the right. A 400-N force is
exerted at a 60-degree angle (a direction of 300 degrees) to move a railroad car eastward along a railroad track. A top view of the
situation is depicted in the diagram. The force applied to the car has both a
vertical (southward) and a horizontal component (eastward). To determine the
magnitudes of these two components, the sine and cosine function will have to
be used. The task is made clearer by beginning with a diagram of the situation
with a labeled angle and a labeled hypotenuse. Once a triangle is constructed, it
becomes obvious that the sine function will have to be used to determine the vertical (southward)
component and the cosine function will have to be used to determine the horizontal (eastward)
component. The triangle and accompanying work is shown below.
Anytime a force vector is directed at an angle to the horizontal, the
trigonometric functions can be used to determine the components of that force
vector. To assure that you understand the use of SOH CAH TOA to determine the
components of a vector, try the following three practice problems. To view the
answers, click on the button.
cos (30 degrees) = Fx / (100N)
Fx = 100 N • cos (30
degrees) = 86.6 N
sin (30 degrees) = (Fy / (100N)
Fy = 100 N * sin (30
degrees) = 50.0
N
cos (45 degrees) = Fx / (100N)
Fx = 100 N • cos (45
degrees) = 70.7 N
sin (45 degrees) = Fy / (100N)
Fy = 100 N • sin (45
degrees) = 70.7 N
cos (60 degrees) = Fx / (100N)
Fx = 100 N • cos (60
degrees) = 50.0 N
sin (60 degrees) = Fy/ (100N)
Fy = 100 N • sin (60
degrees) = 86.6 N
An important concept is revealed by the above three diagrams.
Observe that the force is the same magnitude in each diagram; only the angle
with the horizontal is changing. As the angle that a force makes with the
horizontal increases, the component of force in the horizontal direction (Fx) decreases. The principle makes some sense;
the more that a force is directed upwards (the angle with the horizontal
increases), the less that the force is able to exert an influence in the
horizontal direction. If you wish to drag Fido horizontally, then you would
make an effort to pull in as close to a horizontal direction as possible; you
would not pull vertically on Fido's chain if you wish to pull him horizontally.
One important application of this principle is
in the recreational sport of sail boating. Sailboats encounter a force of wind
resistance due to the impact of the moving air molecules against the sail. This
force of wind resistance is directed perpendicular to the face of the sail, and
as such is often directed at an angle to the direction of the sailboat's
motion. The actual direction of this force is dependent upon the orientation of
the sail. To determine the influence of the wind resistance force in the direction
of motion, that force will have to be resolved into two components - one in the
direction that the sailboat is moving and the other in a direction
perpendicular to the sailboat's motion. See diagram at right. In the diagram
below, three different sail orientations are shown. Assuming that the wind
resistance force is the same in each case, which case would produce the
greatest influence in the direction of the sailboat's motion? That is, which
case has the greatest component of force in the direction parallel to the
boats' heading?
Case C will provide the
greatest force of propulsion. In Case C, the component of the wind resistance
force parallel to the direction of the boat's motion is greatest. The diagrams
depict the two components of force; and clearly the parallel component of force
is longest (i.e., greatest magnitude) in Case C.
Many people believe that a sailboat cannot travel "upwind." It
is their perception that if the wind blows from north to south, then there is
no possible way for a sailboat to travel from south to north. This is simply
not true. Sailboats can travel "upwind" and commonly do so by a
method known as tacking into the wind. It is true to
say that a sailboat can never travel upwind by heading its boat directlyinto the wind. As seen in the diagram at the right, if the boat heads
directly into the wind, then the wind force is directed due opposite its heading.
In such a case, there is no component of force in the direction that the
sailboat is heading. That is, there is no "propelling force." On the
other hand, if the boat heads at an angle into the wind, then the wind force
can be resolved into two components. In the two orientations of the sailboat
shown below, the component of force in the direction parallel to the sailboat's
heading will propel the boat at an angle into the wind. When tacking into the
wind, a sailboat will typically travel at 45-degree angles, tacking back and
forth into the wind.
The following problems focus on concepts discussed in this
lesson. Answer each question and then click the button to view the answer.
1. The diagram at the right depicts a force that makes an angle to the
horizontal. This force will have horizontal and vertical components. Which one
of the choices below best depicts the direction of the horizontal and vertical
components of this force?
The answer is D.
The force is
directed downwards and leftwards. Thus, this force will have a downward
vertical component and a leftward horizontal component. This would be
consistent with choice d.
2. Three sailboats are shown below. Each sailboat experiences
the same amount of force, yet has different sail orientations.
In which case (A, B or C) is the sailboat most likely to tip over sideways?
Explain.
The answer is Case A.
While it is the
parallel component of force which propels the boat forward, it is the
perpendicular component of force which tips the boat over. This component of
force is greatest in Case A as seen in the diagram.
3. Consider the tow truck at the right. If the tensional force in the
cable is 1000 N and if the cable makes a 60-degree angle with the horizontal,
then what is the vertical component of force that lifts the car off the ground?
The
answer is 866 N, upward.
The
vertical component can be found if a triangle is constructed with the 1000 N diagonal
force as the hypotenuse. The vertical component is the length of the side
opposite the hypotenuse. Thus,
sin (60 degrees) = (Fvert) / (1000 N)
Solving
for Fvert will give the answer 866 N.