Download Example Problem - Resolving a Velocity Vector into Its Components

Chapter 6 Notes 6.2 ANALYTICAL METHOD OF VECTOR ADDITION
You have seen how an arrow can represent a vector quantity. Two numbers can also represent a vector quantity in
two dimensions. One gives its length and the second gives the angle between that vector and a reference direction.
The sum of any two vectors can be determined using trigonometry. Trigonometry deals with the relationships
among angles and sides of triangles.
Trigonometric functions are defined in terms of one of the non - 900 angles of a right triangle. The common
trigonometric functions of an angle are the sine (sin), cosine (cos), and tangent (tan). For the triangle in Figure 66a,
sin θ = opposite side / hypotenuse
cos θ = adjacent side / hypotenuse
tan θ = opposite side / adjacent side
The numerical value of the trigonometry functions can be found with a calculator or from Table D-6. For example,
suppose you need to build a triangular form with the dimensions shown in Figure 6-6b. You know that the
hypotenuse of the triangle is 7.6 m and the side adjacent to angle θ is 3.1 m. Therefore, by using the cosine function, you can determine what size you should make θ.
cos θ =
adjacent side
hypotenuse
3.1 m = 0.41
7.6 m
On my small calculators – 3.1/7.6 – enter - 2nd – cos = answer 65.9 or 660
TI 83 – 2nd – cos – (3.1/7.6) – enter – answer 65.9 or 660
Therefore, from a calculator or trigonometry table, θ = 66o.
Adding Perpendicular Vectors
If two vectors are perpendicular, a right angle is formed when the tail of the second vector is placed at the head of
the first. The resultant vector, drawn from the tail of the first to the head of the second, is the hypotenuse of the right
triangle formed by the three vectors. The length of the resultant can be calculated using the Pythagorean theorem
c2=a2+ b2 which means c or h can be found by taking the square root of a 2+ b2. You can find the interior angle, θ, by
using the trigonometric tangent function. In Figure 6-6, tan θ = A/B. When solving vector addition analytically, draw
a careful sketch of the vectors and check that your answer agrees with the sketch.
Example Problem - Finding a Resultant Velocity
An airplane flying toward 00 at 90.0 km/h is being blown toward 900 at 50.0 km/h, Figure 6-7. What is the resultant
velocity of the plane?
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Do practice problems 6-3
Components of Vectors (Vector Resolution)
We have seen that two or more vectors acting in different directions from the same point may be replaced by a
single vector, the resultant. The resultant has the same effect as the original vectors.
It is also possible to begin with a single vector and think of it as the resultant of two vectors. We usually choose two
new vectors in directions that are perpendicular to each other. These two vectors are called the components of the
vector.
The process of finding the magnitude of a component in a given direction is called vector resolution. Consider a
man pulling a sled through the snow. A 58-N force is exerted on a rope held at an angle of 300 with the horizontal.
The rope pulls both forward and upward on the sled. The only force that pulls the sled forward is the horizontal
component, Fh or Fx. The vertical component, Fv or Fy, pulls the sled upward.
The magnitudes of the horizontal and vertical components of F are found by first drawing a set of perpendicular
axes. The x-axis is the horizontal axis. And the y-axis is vertical axis. The vector that represents the force in the
rope, F, is then drawn to scale at the proper angle. To resolve the rope force into the components F h or Fx and Fv or
Fy , draw perpendicular lines from each axis to the tip of the force vector. The magnitudes of the two components
can then be measured using the scale used for F.
Note that we can reverse the process, and show that the resultant of F x and Fy is F, the original force.
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Fh or Fx and Fv or Fy, can be found using trigonometry. In this case,
cos θ =(Fh or Fx )/ F so (Fh or Fx ) = F cos θ
sin θ = Fv or Fy / F so (Fv or Fy) = F sin θ
The signs of Fx and Fy can be found from the vector diagram. If F x is positive it acts to the right if it is negative it
acts to the left; when Fy is positive it acts upward, if it is negative it acts downward.
Look at quads – Quad I x+y+
Quad II x-y+
Quad III x-yQuad IV x+ySuppose the person pulling the sled keeps the force constant but lowers the rope. The angle, θ, is decreased, and
the horizontal component of the force is increased. The vertical component decreases. On the other hand, if the
angle between the rope and the horizontal is increased, the horizontal component decreases, and the vertical
component increases. Thus, the magnitude of the components change as the direction of the pulling force changes.
PROBLEM SOLVING STRATEGY
In resolving force, velocity, or displacement vectors, choose the direction of the component axes according to the
specifics of the problem. Since the choice of axes will not affect the results, choose the set that simplifies the
solution. To simplify the problem choose the x-axis (right left or east west)1st and the y-axis (up down or north
south) second. Be sure to pay attention to direction and work from the horizontal axis.
Example Problem - Resolving a Velocity Vector into Its Components
A wind with a velocity of 40.0 km/h blows toward 30.0o , Figure 6-10.
a. What is the component of the wind's velocity toward 900?
b. What is the component of the wind's velocity toward 00?
Do practice problems 6- 4
Adding Vectors at Any Angle
Vector resolution can also be used to add two or more vectors that are not perpendicular to each other.
 Each vector is resolved into its perpendicular components x’s and y’s.
 Horizontal x-axis components of the vectors are added together to produce a single horizontal vector. F x =
F cos θ
 Vertical y-axis components of all the vectors are added together to produce a single vector that acts in the
vertical direction. Fy = F sin θ
 The sum of resulting horizontal and vertical components can be used in the Pythagorean theorem to obtain
the final resultant.
 The tangent which is equal to y/x is used to find the angle of the resultant θ
To find the resultant from horizontal - Back of notes 4 quads
 Quad I angle looking for = θ
 Quad II angle looking for = 180 - θ
 Quad III angle looking for = 180 + θ
 Quad IV angle looking for = 360 - θ
If two vectors are acting at angles other than 90 0 another way to determine the resultant mathematically is by using
the law of cosines and the law of sines.
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Example Problem - Adding Non-Perpendicular Vectors
Two men are using two ropes to pull a log up to the top of a cliff. The force applied by one rope is 12.0N @ 10.0 0
the force applied by the second rope is 8.00N @ 120 0. What is the net force on the log? The log weighs 6.00N,
what is its acceleration?
Do Practice Problems 6-5 and concept review 6-2
6.3
Applications Of Vectors
Forces and accelerations are vector quantities, so it seems natural to think of Newton's laws in terms of vector
equations. Newton, however, never used vectors. It was not until almost two hundred years later, in the 1880s, that
Sir Oliver Heaviside, a British electrician, convinced skeptical physicists that Newton's laws were best understood
using vectors. We can use the techniques for adding and resolving vectors to analyze the acceleration of objects
that have several forces exerted on them.
Equilibrium
When the net force-the net sum of all the forces acting-is zero, the object is in equilibrium. According to Newton's
laws, the object will not be accelerated because there is no net force on it. A simple example of equilibrium is the
case in which two equal forces act in opposite directions on an object, as shown in Figure 6-13. The resultant force
is zero.
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Figure a shows an object at point P with three forces acting on it. The 3-N force and the 4-N force are at right
angles to each other. When the three vectors are added head-to-tail, they form a closed triangle, as in Figure b.
The distance from the tail of the first vector to the head of the last vector is then zero, so the length of the resultant
is zero. The length of the resultant is zero, so the vector sum is zero. Therefore, the three forces produce zero net
force on point P. The object at point P is in equilibrium.
When the vector sum of forces acting at one point is not zero, a force can be applied that will produce equilibrium.
This force is called the equilibrant force. The equilibrant force is the single additional force that, if applied at the
same point as the other forces, will produce equilibrium.
To find the equilibrant of two or more forces, first find the resultant force. The equilibrant force is equal in magnitude
to the resultant, but opposite in direction. In Figure 6-15, the equilibrant force is a 10-N force whose direction is
opposite to the direction of the resultant.
Sometimes the equilibrant is exerted by two or more forces. For example, when a high wire tight-rope walker
performs in a circus, the upward force that balances the weight of the performer is provided by the cable. The force
exerted by the two ends of the cable can be found by following the method in the example below.
Example Problem - Finding Forces When Components Are Known
A sign that weighs 168 N is supported by ropes a and b, that make 22.50 angles with the horizontal. The
sign is not moving. What forces do the ropes exert on the sign?
The force in each rope is larger than the weight of the sign. Try different angles. As the angle between the ropes
and the horizontal increases, the horizontal components of the forces become smaller. When the ropes are vertical,
the horizontal components are zero, and each rope exerts only half the weight of the sign, 84 N. Note that since the
forces exerted by the ropes must always have vertical components adding to 168 N, the ropes can never be totally
horizontal.
Do practice problems 6-6
Gravitational Force and Inclined Planes
The gravitational force acting on an object is directed toward the center of Earth. The object's weight, W, can be
represented by a vector directed down. Down is the direction an object falls.
Picture a trunk resting on an inclined plane. To analyze the forces acting on the trunk, the weight of the
trunk, W, is resolved into two components perpendicular to each other. When doing inclined plane problems, we
may choose one axis along the incline, and the second axis perpendicular to it. Thus, we have one component, F II,
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the parallel force, that acts parallel to the incline. The second component, F ┴ called the perpendicular force, acts
perpendicular to the incline.
Look at the right triangle formed by the surface of the incline and its horizontal and vertical sides. It is similar to the
right triangle formed by W, F┴ , and FII because the corresponding sides are mutually perpendicular. Therefore the
angles are equal. If θ and W are both known, a vector diagram similar to Figure 6-18 can be drawn, and the force
W resolved into the components F┴ , and FII . You can calculate the values of F┴ , and FII using either graphical or
trigonometric methods. Hand out diagram of incline.
From this we come to the conclusion that:
F┴ = W cos θ and FII = W sin θ
Example Problem - Finding F┴ , and FII
A trunk weighing 562 N is resting on a plane inclined at 300 from the horizontal, Figure 6-18. Find the components
of the weight parallel and perpendicular to the plane.
Move F┴ so that it is tail is at the end of FII and its head is at the end of W – 90
becomes the intersection of F┴ and FII and the angle were W and F┴ is equal to
the angle of the incline plane.
The directions of the two components are shown on Figure 6-18. Note that as the incline in Figure 6-19 becomes
steeper, FII becomes greater and F┴ becomes smaller.
What other forces act on the trunk besides the force of gravity? The inclined plane exerts an upward force
perpendicular to its surface, the normal force, F N. Since the trunk has no acceleration perpendicular to the plane, all
forces in that direction must be balanced. Therefore,
FN + F┴ = 0 and F┴ = W cos θ so FN – W cos θ = 0 ; which means FN = W cos θ
If there is no friction between the trunk and the plane, the only force on the trunk along the plane is the parallel
component of its weight, FII = W sin θ. According to Newton’s second law, the acceleration
a = F/m = the F is FII so a = W sin θ/m
But, W/m =g, so the acceleration of the trunk is a = g sin θ. As the plane becomes more horizontal, the acceleration
approaches zero. When the plane is tilted more vertically, the acceleration approaches closer to g = 9.80 m/s2.
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Example Problem -Finding Acceleration Down a Plane
The 562-N trunk is on a frictionless plane inclined at 30.00 from the horizontal. Find the acceleration of the trunk.
What is its direction?
Usually there is friction between the inclined plane and the object on it. The resolution of weight on an inclined
plane can be used to measure the force of friction or the coefficient of static friction between the two surfaces. Put a
coin on one of your textbooks. Now slowly lift the cover. The coin will remain at rest until the cover reaches a
certain angle. Then the coin begins to accelerate down the book.
The downward force along the cover is the parallel component of the coin's weight, proportional to sin θ. As the
cover angle is increased, the parallel component of the weight increases. It is balanced by the static friction force.
Static friction can never be larger than μFN , so when the force along the cover exceeds the frictional force, the coin
starts to move. Once the coin starts to move, static friction is replaced by sliding friction, which is a smaller force.
The coin, with a non-zero net force on it, accelerates down the book. If you measure the angle of the cover when
the coin first begins to move, you can determine the coefficient of static friction. Try different objects and books.
Which combination has the greatest coefficient of static friction? the least?
Example Problem - Finding the Coefficient of Static Friction
A coin placed on the cover of a book just begins to move when the cover makes an angle of 380 with the horizontal,
Figure 6-21. What is the coefficient of static friction between the cover and coin?
Practice Problems 6-7
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