EME1501 Assignment 2 solutions 2026

EME1501 Assignment 2 solutions 2026
UNISA
Module code: EME1501

Module title: ENGINEERING MECHANICS I

Assessment: ASSIGNMENT 2

Total marks: 100

Module leader: Dr. MJ SITHOLE

Internal moderator: Dr. SJ MOFOKENG

,QUESTION 1

1.1 Resolve force 𝑷 = 𝟐𝟎 lb

Understand angle

From the diagram:

• Force makes 120° with vertical

• Convert to standard angle from x-axis:

𝜃 = 120∘ − 90∘ = 30∘ above negative x-axis


So effectively:

𝜃 = 150∘ from +x axis
Resolve into components

𝑃𝑥 = 𝑃cos⁡ 𝜃, 𝑃𝑦 = 𝑃sin⁡ 𝜃


𝑃𝑥 = 𝑃cos⁡ 𝜃, 𝑃𝑦 = 𝑃sin⁡ 𝜃



Substitute values

𝑃𝑥 = 20cos⁡(150∘ ) = 20(−0.866) = −17.32 lb
𝑃𝑦 = 20sin⁡(150∘ ) = 20(0.5) = 10 lb




Final Answer:

𝑃𝑥 = −17.32 lb, 𝑃𝑦 = 10 lb




1.2 Resultant of vectors (A = 12 N, B = 5 N)

From diagram:

• A = 12 N (horizontal)

• B = 5 N (vertical)



Use Pythagoras

, 𝑅 = √𝐴2 + 𝐵 2

𝑅 = √122 + 52 = √144 + 25 = √169 = 13 N




Direction
5
tan⁡ 𝜃 =
12
𝜃 = 22.6∘


Final Answer:

𝑅 = 13 N, 𝜃 = 22.6∘


QUESTION 2

Given:

• 𝑃 = 20 Nwith direction triangle 3–4–5

• 𝑄 = 26 Nwith direction triangle 5–12–13

We must find the resultant 𝑅using a vector triangle (components method is equivalent
and clearer).

: Resolve Force 𝑷into Components

From the 3–4–5 triangle:
4
• Horizontal ratio = 5
3
• Vertical ratio = 5

4
𝑃𝑥 = 20 ×
5
𝑃𝑥 = 16 N
3
𝑃𝑦 = 20 ×
5
𝑃𝑦 = 12 N