2.1 adding forces by the parallelogram la adding forces by the parallelogram law example 1, page 1...
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2.1 Adding Forces by the Parallelogram Law
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2.1 Adding Forces by the Parallelogram Law Example 1, page 1 of 4
1. Determine the magnitude and direction
of the resultant of the forces shown.
30°
150 N
200 N
30°
200 N
150 N
30°
Construct a parallelogram by drawing two
lines. Each line starts at the tip of one vector
and is parallel to the other vector.Tip
Parallel
Tip
1
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2.1 Adding Forces by the Parallelogram Law Example 1, page 2 of 4
30°
200 N
150 N
30°
Since opposite sides of a parallelogram are
equal in length, the length of each line
represents the magnitude of the vector opposite.
200 N
150 N
2
30°
R
The resultant R is drawn from the tails of the
vectors to the opposite vertex of the parallelogram.
Tails
150 N
200 N
150 N
200 N
3
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2.1 Adding Forces by the Parallelogram Law Example 1, page 3 of 4
200 N
150 N
30°
R150 N
200 N
150 N
200 N
Tails Head
Heads
4
30°
200 N
R150 N
To calculate the magnitude and direction of R, consider
the triangle formed by one half of the parallelogram.5
200 N
150 N
This is not the resultant because it is not
drawn from the intersection of the tails.
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2.1 Adding Forces by the Parallelogram Law Example 1, page 4 of 4
30°
200 N
R 150 N
Use trigonometry to calculate the magnitude and direction
of the resultant.
R2 = (200 N)2 + (150 N)2 2(200 N)(150 N) cos 30°
The result is
R = 102.66 N Ans.
=
Solving gives
= 46.9° Ans.
b
a
c
C
A
B
Law of cosines
C2 = A2 + B2 2AB cos c
Law of sines
= =
6
sin a A B
sin b C
sin c
sin 150 N
sin 30° R
= 102.66 N
Trigonometric formulas for a
general triangle are given below.
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2.1 Adding Forces by the Parallelogram Law Example 2, page 1 of 3
2. Determine the magnitude and direction
of the resultant of the forces shown.
3 kN
2 kN
Construct a parallelogram by drawing
two lines parallel to the forces.
1
y
x
3 kN
2 kN
y
x
3 kN
2 kN
10°
20°
10°
20°
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2.1 Adding Forces by the Parallelogram Law Example 2, page 2 of 3
Draw the resultant R from the tails of the vectors to the
opposite vertex of the parallelogram.
3 kN
2 kN
y
x
3 kN
2 kN
2
Tails
R
2 kN2 kN
R
3 kN
3 kN
y
x
To calculate the magnitude and direction of R, consider
the triangle formed by one half of the parallelogram.
3
R
3 kN
2 kN
10°
20°
20°
10°
10°
20° 20°
10°
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2.1 Adding Forces by the Parallelogram Law Example 2, page 3 of 3
R
3 kN
2 kN
Use trigonometry to calculate the magnitude
and direction of the resultant.
4
total angle= 10° + 90° + 20° = 120°
Law of cosines
R2 = (3 kN)2 + (2 kN)2 2(3 kN)(2 kN) cos 120°
R = 4.359 kN Ans.
sin 120° R
Law of sines
=
Solving gives
= 36.6°
sin 3 kN = 4.359 kN
5
6
R = 4.36 kN
x
y
36.6° + 20° = 56.6° Ans.
= 36.6°
7 Angle measured with
respect to the vertical axis
20°
10°
20°20°
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2.1 Adding Forces by the Parallelogram Law Example 3, page 1 of 3
110°
100 N
80 N
Construct a parallelogram1
100 N
80 N
40°
110°
y
40°
3. Determine the magnitude and direction of the resultant force.
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2.1 Adding Forces by the Parallelogram Law Example 3, page 2 of 3
Draw the resultant R from the tails of the vectors to
the opposite vertex of the parallelogram.
100 N
80 N
40°
110°
y
2
100 N
80 N
R
40°
110°
80 N
100 N
y
R
30°
To calculate R, consider the triangle formed
by the lower half of the parallelogram.
3
Calculate angle
180° 110° 40° = 30°
Parallel lines make 30° angle
with vertical direction
4
5
Calculate angle
30° + 40° = 70°
6
40°
Law of cosines
R2 = (80 N)2 + (100 N)2 2(80 N)(100 N) cos 70°
R = 104.54 N Ans.
7
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2.1 Adding Forces by the Parallelogram Law Example 3, page 3 of 3
y
100 N
R = 104.54 N
70°
Calculate the angle that the
resultant makes with the vertical.
Law of sines
=
Solving gives
= 64.0°
R sin 70°
100 N sin
104.54 N
8
R = 104.54 N
y
Angle measured from the vertical9
40° + 64.0° = 104.0° Ans.
40°
64°
40°
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2.1 Adding Forces by the Parallelogram Law Example 4, page 1 of 4
x
40 lb
60 lb
30°
y
4. The resultant of the two forces acting on the screw eye is known to
be vertical. Determine the angle and the magnitude of the resultant.
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2.1 Adding Forces by the Parallelogram Law Example 4, page 2 of 4
40 lb
30°
x
y
30°
40 lb
x
y
30°
40 lb
y
x
To determine what needs to be calculated, make some
sketches of several possible parallelograms.
1
R
R
R
Each parallelogram is based on two facts that are given:
1) One side of the parallelogram is known (40 lb at 30°), and
2) The resultant R lies on the y axis.
2
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2.1 Adding Forces by the Parallelogram Law Example 4, page 3 of 4
30°
40 lb
x
y
How do we determine the actual parallelogram? We have
to use the additional fact that one of the forces is 60 lb.
3
The point of the intersection of the arc and the
vertical axis must be the vertex of the
parallelogram since it lies on the vertical axis and
also lies a "distance" of 60 lb from the tip of the
40-lb vector.
4
Now the parallelogram is completely defined.
60 lb
40 lb
30°
x
y
5
40 lb
60 lb
Radius of circular
arc = 60 lb
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2.1 Adding Forces by the Parallelogram Law Example 4, page 4 of 4
To calculate the resultant R and and the
angle (see below), analyze the triangle
formed by the left half of the parallelogram.
Angle = 90° 30° = 60°
Corresponding angles are equal
y30°
x
40 lb
60 lb
6
8
7
30°
Parallel
60 lb
40 lb
60°
+ 30°
R
sin 60° 60 lb
Law of sines
=
Solving gives
= 35.26°
sin 40 lb
Law of sines
=
Solving gives
R = 69.0 lb Ans.
sin 60° 60 lb
sin( + 30°)
R
9
11 54.74°
The sum of the angles of the triangle is 180°:
+ ( + 30°) + 60° = 180°
Solving gives
= 54.74° Ans.
10
35.26°
R
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2.1 Adding Forces by the Parallelogram Law Example 5, page 1 of 2
5. Determine the magnitude F and the angle , if the
resultant of the two forces acting on the block is to be a
horizontal 80-N force directed to the right.
50 NF
Draw the parts of parallelogram that are known:
Two sides are of length 50 N and make
an angle of 25° with the horizontal axis.
The diagonal of the parallelogram (the
resultant) is 80 N long and horizontal.
50 N
50 N
80 N
1
2
3
25°
25°
25°
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2.1 Adding Forces by the Parallelogram Law Example 5, page 2 of 2
Complete the parallelogram.
25°
50 N
50 N
80 N
4
25°
25°
F
F
Analyze the triangle forming the lower half of the parallelogram.
6
25°
F
80 N
50 N Calculate F from the law of cosines.
F2 = (50 N)2 + (80 N)2 2(50 N)(80 N) cos 25°
The result is
F = 40.61 N Ans. sin
50 N
Calculate from the law of sines.
=
Solving gives
= 31.4° Ans.
sin 25°
F 40.61 N
7
5
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2.1 Adding Forces by the Parallelogram Law Example 6, page 1 of 2
A
B
30 N25 N
6. To support the 2-kg flower pot shown, the resultant of the two
wires must point upwards and be equal in magnitude to the
weight of the flower pot. Determine the angles and , if the
forces in the wires are known to be 25 N and 30 N.
B
Weight of flower pot
mg = (2 kg)(9.81 m/s2 )
= 19.62 N
1
19.62 N 19.62 N
B
R = 19.62 N
Resultant, R, of forces in wires
balances the weight.2
C
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2.1 Adding Forces by the Parallelogram Law Example 6, page 2 of 2
B
25 N
30 N
30 N
25 N
The resultant R = 19.62 N must be the diagonal of a
parallelogram with sides 25 N and 30 N long.
3 Analyze the triangle forming the left-hand
half of the parallelogram.
4
25 N
30 N
19.62 N 19.62 N
Law of cosines to calculate
(25 N)2 = (30 N)2 + (19.62 N)2 2(30 N)(19.62 N) cos
Solving gives
= 55.90° Ans.
Law of sines to calculate
=
Solving gives
= 83.6° Ans.
30 N sin
5
sin
25 N 55.90°
6
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2.1 Adding Forces by the Parallelogram Law Example 7, page 1 of 2
25°
40°
v
u
120 lb
7. Resolve the 120-lb force into components
acting in the u and v directions.
Construct a parallelogram with
the 120-lb force as a diagonal.
Draw another line from the
tip but parallel to u.
Draw a line from the tip of the
force vector parallel to v.120 lb
v
40°
u
40°
25°
1
2
3 R u
R v
Label the components R u and R
v.4
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2.1 Adding Forces by the Parallelogram Law Example 7, page 2 of 2
120 lb40°
25°R v
R u
Analyze the triangle forming the left-hand half of the
parallelogram.
180° 40° 25° = 115°
5
Calculate R u from the law of sines.
=
Solving gives
R u = 78.9 lb Ans.
120 lb
sin 40° sin 25°
R u
6
Calculate R v from the law of sines.
=
Solving gives
R v = 169.2 lb Ans.
sin 40° sin 115°
R v 120 lb
7
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2.1 Adding Forces by the Parallelogram Law Example 8, page 1 of 2
8. Resolve the 4-kN
horizontal force into
components along truss
members AB and AC.
35°
5
12
A
B C
4 kN
Extend line AC.
35°
Construct a parallelogram with the
4-kN force as the diagonal.
12
5
4 kN
2
B C
Draw another line from
the tip but parallel to AC.
4
5
12
ADraw a line from the tip of the
force vector parallel to AB.
3
1
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2.1 Adding Forces by the Parallelogram Law Example 8, page 2 of 2
35°
12
5
4 kN
B C5
12
A
Label the components R B and R
C.5
R B
R C
4 kN
12
5
R C
R B
Analyze the triangle forming the upper half of the
parallelogram (The drawing has been enlarged for clarity).6
35°
12 5
Geometry
= tan-1 = 67.38°
= 180° 35° 67.38° = 77.62°
7
77.62°
Law of sines to calculate R C
=
Solving gives
R C = 3.78 kN Ans.
sin sin R
C 4 kN
67.38°
Law of sines to calculate R B
=
Solving gives
R B = 2.35 kN Ans.
R B
sin 35° 4 kN
sin 77.62°
8
9
35°
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2.1 Adding Forces by the Parallelogram Law Example 9, page 1 of 3
9. Find two forces, one acting along rod AB and one
along rod CB, which when added, are equivalent to the
200-N vertical force.
200 N
A
B
C
30° 40°
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2.1 Adding Forces by the Parallelogram Law Example 9, page 2 of 3
30°
40° 30°
40°
30°
40° 30°
200 N
A
B
C
30°
1 Construct a parallelogram with
sides parallel to AB and BC and
with the 200-N force as a
diagonal.Draw a line from the tail of the
force vector parallel to AB.
Draw another line from the
tail but parallel to BC.
Extend AB.Extend BC.
4
5
32
200 N
B
R C
R A
Label the sides of the
parallelogram R A and R
C.
6
R C
R A
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2.1 Adding Forces by the Parallelogram Law Example 9, page 3 of 3
200 N
R C
30°
40°
180° 50° (30° + 40°) = 60°
200 N
Law of sines to calculate R A
=
Solving gives
R A = 163.0 N Ans.
R A
sin 50°
90° 40° = 50°
Analyze the triangle
formed by the
left-hand side of the
parallelogram.
Angles are equal
7
11
sin (30° + 40°)
Law of sines to calculate R C
=
Solving gives
R C = 184.3 N Ans.
R C
sin 60° sin (30° + 40°)
200 N
12
89
10
R A
40°
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