PHYS2001 Recitation 1 – Friday, September 1, 2006
1. Consider the vectors A and B shown in the figure below. Which of the other four
vectors in the figure (C, D, E, and F) best represents the direction of (a) A +B, (b) A – B,
and (c) B – A?
y
A
C
D
B
F
E
x
2. A vector A has a magnitude of 50.0 m and points in a direction 20.0o below the
positive x-axis. A second vector, B, has a magnitude of 70.0 m and points in a direction
50.0o above the positive x-axis. (a) Sketch the vectors A, B, and C, where C = A + B. (b)
Use the component method of vector addition to find the magnitude and direction of C.
3. Vector A points in the negative x-direction. Vector B points at an angle of 30.0o above
the positive x-axis. Vector C has a magnitude of 15 m and points in a direction 40.0o
below the positive x-axis. Use the component method of vector addition and the fact that
A + B + C = 0 to find the magnitudes of vectors A and B.
4. Quiz: Vector A lies in the xy-plane. Its scalar components are Ax = 15 m and Ay = 20 m. What is the magnitude of vector A, and what angle does vector A make with the
negative y-axis?
PHYS 2001 Recitations
Practice Exercises & Quiz ANSWERS
Fall 2006
1.
Friday, September 1st
• Problem #1:
– The vector E best represents the direction of A + B.
– The vector F best represents the direction of A − B = A + (−B).
– The vector D best represents the direction of B − A = B + (−A).
• Problem #2:
The top figure on the following page (Fig. 2a) displays a sketch of vectors A and B, according to their
definitions in the problem. As illustrated in the bottom figure on the same page (Fig. 2b), the vector
C is obtained by first sliding vector B over and down (without changing its angular orientation!) until
the “tail” of B connects to the “head” of A; the vector drawn from the “tail” of A to the “head” of
B is then the desired vector C.
Here is how to use the “component method” of vector addition to find the magnitude and direction of
the vector C. First, break the two vectors A and B into their respective x and y components:
Ax
=
(50.0 m) × cos(−20.0◦ ) = 47.0 m ;
Ay
=
(50.0 m) × sin(−20.0◦ ) = −17.1 m ;
Bx
=
(70.0 m) × cos(50.0◦ ) = 45.0 m ;
By
=
(70.0 m) × sin(50.0◦ ) = 53.6 m .
Therefore, the x and y components of the resultant vector C are,
Cx
=
Ax + Bx = 92.0 m ;
Cy
=
Ay + By = 36.5 m .
The magnitude of the vector C is, then,
C =
q
Cx 2 + Cy 2 = 99.0 m ;
and it is oriented at an angle θC above the positive x axis such that,
µ ¶
Cy
θC = tan−1
= 21.6◦ .
Cx
SUMMARY ANSWER (expressed to the same degree of precision as the quoted magnitudes and angles
of vectors A and B): C = 99.0 m and θC = 21.6◦ .
SELF-CONSISTENCY CHECK: As a check, compare these numerical results to the sketch of vector
C in Fig. 2b. We see that vector C is a bit longer than vector B, which is consistent with its calculated
magnitude C = 99.0 m; and we see that vector C points above the positive x-axis at an angle that is
comparable in size to the angle at which vector A points below the x-axis, which is consistent with our
calculation that, for vector C, θ = 21.6◦ .
Figures Associated with Problem #2
Figure 2a: Sketch of Vectors A and B
[Illustrated angles are only approximate.]
Figure 2b: Sketch of Vector C
[Illustrated angles are only approximate.]
1
FRIDAY, SEPTEMBER 1ST
2
• Problem #3:
Begin by carefully writing down everything that we know about the vectors A, B, and C, based on
the statement of the problem. (It is useful to note, as well, certain key unknown properties of these
vectors.) The magnitudes of the three vectors A, B, and C are:
A
=
unknown ;
B
=
unknown ;
C
=
15 m ;
and the angles that they each make with the positive x-axis are:
θA
=
180◦ ;
θB
=
30◦ ;
θC
=
−40◦ .
In the statement of the problem, we are also told that the sum of the three vectors is zero. In an effort
to reexpress this information in terms that are more useful, let’s define a new vector,
D = A + B + C.
We are told that D = 0, which means that both the x and y components of D must be zero, that is,
Dx = 0 and Dy = 0.
Given the above information, we can now use the “component method” of vector addtion to determine
the two unknown vector magnitudes, A and B. First, let’s determine the x and y components of the
three vectors A, B, and C:
Ax
=
A cos(θA ) = A cos(180◦ ) = A(−1) = −A ;
Ay
=
Bx
=
By
=
Cx
=
A sin(θA ) = A sin(180◦ ) = 0 ;
√
3
◦
B cos(θB ) = B cos(30 ) =
B = 0.866B ;
2
1
B sin(θB ) = B sin(30◦ ) = B = 0.5B ;
2
C cos(θC ) = (15 m) cos(−40◦ ) = (15 m) × 0.766 = 11.5 m ;
Cy
=
C sin(θC ) = (15 m) sin(−40◦ ) = (15 m) × (−0.643) = −9.64 m .
Because the sum of the three x-components must produce Dx = 0, and the sum of the three ycomponents must produce Dy = 0, we can write the following two algebraic expressions:
Dx
=
Ax + Bx + Cx = −A + 0.866B + (11.5 m) = 0 ;
Dy
=
Ay + By + Cy = 0 + 0.5B + (−9.64 m) = 0 .
We now can calculate the value of B from the second of these expressions, that is,
0.5B
=⇒ B
= −(−9.64 m) = +9.64 m
9.64 m
=
= 19.3 m .
0.5
1
FRIDAY, SEPTEMBER 1ST
3
Finally, having deduced the value of B, we can use the first of the algebraic expressions to determine
A, namely,
A
=
0.866B + (11.5 m) = 0.866(19.3 m) + (11.5 m) = 28.2 m .
SUMMARY ANSWER (expressed to the same degree of precision as the quoted magnitude of vector
C): A = 28 m and B = 19 m.
SELF-CONSISTENCY CHECK: The top figure on the following page (Fig. 3a) illustrates what is
known about the three vectors A, B, and C initially. The vector C has been drawn as a solid arrow
with a definite length because both its magnitude and angle of orientation are known initially. However,
the lengths of vectors A and B are unknown initially, so they have been drawn as dashed “arrows”
without the location of the “head” of either arrow being specified.
In the bottom figure (Fig. 3b), the vectors have been moved around in the xy-plane (without changing
their angular orientations!) in such a way that they illustrate the expression,
A + B + C = 0.
Because it is the first vector in the summation, the “tail” of A touches the origin of the coordinate
axes. Because the sum of the three vectors is zero, the “head” of the last vector in the summation
(vector C) must also touch the origin. Vector B is then positioned so that its “head” touches the
“tail” of vector C. Finally, we realize that the point at which the two dashed lines cross one another
identifies both the “head” of A and the “tail” of B; it is in this manner that the magnitudes of both
A and B can be determined graphically. A is clearly the longest of the three vectors while C is clearly
the shortest of the three. This is entirely consistent with the numerical values determined above, that
is, A = 28.2 m, B = 19.3 m, and C = 15 m.
Figures Associated with Problem #3
Figure 3a: Sketch of Vectors A, B, and C (as far as they are determined initially)
[Illustrated angles are only approximate.]
Figure 3b: Sketch of Vectors A, B, and C (after shifting them to form a closed triangle)
[Illustrated angles are only approximate.]
1
FRIDAY, SEPTEMBER 1ST
4
• QUIZ during Week #1:
RESTATEMENT OF QUIZ: Vector A lies in the xy-plane. Its scalar components are Ax = 15 m
and Ay = −20 m. What is the magnitude of vector A, and what angle does vector A make with the
negative y-axis?
NOTE: This question is slightly tricky because you’re asked to specify the angle with respect to the
negative y-axis instead of the customary positive x-axis!
QUICK DETERMINATION: A straightforward way to work this problem is as follows:
q
√
p
A = A2x + A2y = (15 m)2 + (−20 m)2 = 625 m2 = 25 m ;
and with respect to the negative y-axis, this vector makes an angle,
¶
µ
¶
µ
15 m
Ax
= tan−1
= tan−1 (0.75) = 36.9◦ .
θ = tan−1
−Ay
+20 m
USEFUL RELATED INFORMATION: The figure on the following page (Fig. 4a) illustrates graphically
how this vector A is oriented.
Figure Associated with QUIZ (Week #1)
Figure 4a: Sketch of Vector A and its components