UNIVERSITY OF SOUTH AFRICA
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
EME1501: ENGINEERING MECHANICS I
ASSIGNMENT 5
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
EME1501
Module Code:
Engineering Mechanics I
Module Name:
Assignment 5
Assessment:
Dr. MJ Sithole
Module Leader:
Dr. SJ Mofokeng
Internal Moderator:
100
Total Marks:
Submitted in partial fulfilment of the requirements for Engineering Mechanics I — UNISA
,QUESTION 1 [10]
1.1 Resolve force 𝑃 shown in the Figure below
into components in the x- and y-directions for the given x-y-
axes orientation as shown in the Figure below. (6)
1.2 Use the trigonometric functions of sine, cosine, and tangent to
determine the resultant, 𝑅, of the vectors shown in the Figure
below. (4)
,UNISA | EME1501 Engineering Mechanics I — Assignment 5
Question 1.1: Resolution of Force P into Rectangular Components
Resolve force P into components in the x and y directions for the given x–y axes orientation
shown in the figure. Given: P = 20 lb.
Step 1: Determine the Direction of the Force
The force is 120◦ measured from the downward vertical. The downward vertical is 270◦ from
the positive horizontal. Therefore,
θP = 270◦ − 120◦ = 150◦
The positive x-axis is inclined at 20◦ above the horizontal, so
θx = 20◦
The angle between the force and the positive x-axis is
150◦ − 20◦ = 130◦
Since the x and y axes are perpendicular,
θy = 130◦ − 90◦ = 40◦
The force acts opposite the positive x-direction and toward the positive y-direction.
Step 2: Calculate the x-Component
Px = P cos 130◦
Substituting the values:
Px = 20 cos 130◦ = 20(−0.6428) = −12.86 lb
Step 3: Calculate the y-Component
Py = P sin 130◦
Substituting the values:
Py = 20 sin 130◦ = 20(0.7660) = 15.32 lb
Page 2 of 21
,UNISA | EME1501 Engineering Mechanics I — Assignment 5
Final Answer
Px = −12.86 lb Py = 15.32 lb
Page 3 of 21
, UNISA | EME1501 Engineering Mechanics I — Assignment 5
Question 1.2: Resultant of Two Perpendicular Vectors
Use the trigonometric functions of sine, cosine, and tangent to determine the resultant, R, of
the vectors shown in the figure. Given: A = 12 N, B = 5 N.
Step 1: Calculate the Magnitude of the Resultant
Using Pythagoras’ theorem:
p
R= A2 + B 2
Substituting the values:
p √ √
R= 122 + 52 = 144 + 25 = 169 = 13 N
Step 2: Calculate the Direction of the Resultant
Using the tangent function:
B
tan θ =
A
Substituting the values:
5
tan θ = = 0.4167
12
Taking the inverse tangent:
θ = tan−1 (0.4167) = 22.62◦
Final Answer
Magnitude of the resultant: R = 13 N
Direction of the resultant: θ = 22.62◦ measured above the horizontal (along vector A).
Page 4 of 21
College of Science, Engineering and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
EME1501: ENGINEERING MECHANICS I
ASSIGNMENT 5
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
EME1501
Module Code:
Engineering Mechanics I
Module Name:
Assignment 5
Assessment:
Dr. MJ Sithole
Module Leader:
Dr. SJ Mofokeng
Internal Moderator:
100
Total Marks:
Submitted in partial fulfilment of the requirements for Engineering Mechanics I — UNISA
,QUESTION 1 [10]
1.1 Resolve force 𝑃 shown in the Figure below
into components in the x- and y-directions for the given x-y-
axes orientation as shown in the Figure below. (6)
1.2 Use the trigonometric functions of sine, cosine, and tangent to
determine the resultant, 𝑅, of the vectors shown in the Figure
below. (4)
,UNISA | EME1501 Engineering Mechanics I — Assignment 5
Question 1.1: Resolution of Force P into Rectangular Components
Resolve force P into components in the x and y directions for the given x–y axes orientation
shown in the figure. Given: P = 20 lb.
Step 1: Determine the Direction of the Force
The force is 120◦ measured from the downward vertical. The downward vertical is 270◦ from
the positive horizontal. Therefore,
θP = 270◦ − 120◦ = 150◦
The positive x-axis is inclined at 20◦ above the horizontal, so
θx = 20◦
The angle between the force and the positive x-axis is
150◦ − 20◦ = 130◦
Since the x and y axes are perpendicular,
θy = 130◦ − 90◦ = 40◦
The force acts opposite the positive x-direction and toward the positive y-direction.
Step 2: Calculate the x-Component
Px = P cos 130◦
Substituting the values:
Px = 20 cos 130◦ = 20(−0.6428) = −12.86 lb
Step 3: Calculate the y-Component
Py = P sin 130◦
Substituting the values:
Py = 20 sin 130◦ = 20(0.7660) = 15.32 lb
Page 2 of 21
,UNISA | EME1501 Engineering Mechanics I — Assignment 5
Final Answer
Px = −12.86 lb Py = 15.32 lb
Page 3 of 21
, UNISA | EME1501 Engineering Mechanics I — Assignment 5
Question 1.2: Resultant of Two Perpendicular Vectors
Use the trigonometric functions of sine, cosine, and tangent to determine the resultant, R, of
the vectors shown in the figure. Given: A = 12 N, B = 5 N.
Step 1: Calculate the Magnitude of the Resultant
Using Pythagoras’ theorem:
p
R= A2 + B 2
Substituting the values:
p √ √
R= 122 + 52 = 144 + 25 = 169 = 13 N
Step 2: Calculate the Direction of the Resultant
Using the tangent function:
B
tan θ =
A
Substituting the values:
5
tan θ = = 0.4167
12
Taking the inverse tangent:
θ = tan−1 (0.4167) = 22.62◦
Final Answer
Magnitude of the resultant: R = 13 N
Direction of the resultant: θ = 22.62◦ measured above the horizontal (along vector A).
Page 4 of 21
