2.1 adding forces by the parallelogram la adding forces by the parallelogram law example 1, page 1...

2.1 Adding Forces by the Parallelogram Law

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


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


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.


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.


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°


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°


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°


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.


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


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°


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.


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


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


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


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°


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


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


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


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


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


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


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°


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°


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


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°

2.1 adding forces by the parallelogram la adding forces by the parallelogram law example 1, page 1...

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