PHY1505 ASSIGNMENT 2 Solutions 2026
UNISA
Module Code: PHY1505
Module Title: Elementary Mechanics
Assessment Format: Take-Home Assignment
Total Marks: 80
Duration: 8 -19 June 2026
First Examiner: Dr SJ Mofokeng
Second Examiner: Dr M Ramantswana
, Question 1
Given
• 𝑻𝟏 = 𝟒𝟔 N(horizontal to the left)
• 𝑻𝟐 = 𝟕𝟒 N(vertical upward)
• The loop is in equilibrium.
For equilibrium:
∑𝑭𝒙 = 𝟎, ∑𝑭𝒚 = 𝟎
Therefore, 𝑻𝟑 must balance both 𝑻𝟏 and 𝑻𝟐 .
Components of 𝑻𝟑
𝑻𝟑𝒙 = 𝟒𝟔 N
𝑻𝟑𝒚 = 𝟕𝟒 N
(The rope pulls downwards and to the right.)
Magnitude of 𝑻𝟑
Using Pythagoras:
𝑻𝟑 = √𝑻𝟐𝟑𝒙 + 𝑻𝟐𝟑𝒚
𝑻𝟑 = √𝟒𝟔𝟐 + 𝟕𝟒𝟐
𝑻𝟑 = √𝟐𝟏𝟏𝟔 + 𝟓𝟒𝟕𝟔
𝑻𝟑 = √𝟕𝟓𝟗𝟐
𝑻𝟑 = 𝟖𝟕. 𝟏𝟑 N
Direction of 𝑻𝟑
𝟕𝟒
𝐭𝐚𝐧 𝜽 =
𝟒𝟔
𝟕𝟒
𝜽 = 𝐭𝐚𝐧−𝟏 ( )
𝟒𝟔
𝜽 = 𝟓𝟖. 𝟏𝟑∘
Answer
𝑻𝟑 = 𝟖𝟕. 𝟏 N
𝜽 = 𝟓𝟖. 𝟏∘
Direction: 58.1° below the horizontal toward the right.
UNISA
Module Code: PHY1505
Module Title: Elementary Mechanics
Assessment Format: Take-Home Assignment
Total Marks: 80
Duration: 8 -19 June 2026
First Examiner: Dr SJ Mofokeng
Second Examiner: Dr M Ramantswana
, Question 1
Given
• 𝑻𝟏 = 𝟒𝟔 N(horizontal to the left)
• 𝑻𝟐 = 𝟕𝟒 N(vertical upward)
• The loop is in equilibrium.
For equilibrium:
∑𝑭𝒙 = 𝟎, ∑𝑭𝒚 = 𝟎
Therefore, 𝑻𝟑 must balance both 𝑻𝟏 and 𝑻𝟐 .
Components of 𝑻𝟑
𝑻𝟑𝒙 = 𝟒𝟔 N
𝑻𝟑𝒚 = 𝟕𝟒 N
(The rope pulls downwards and to the right.)
Magnitude of 𝑻𝟑
Using Pythagoras:
𝑻𝟑 = √𝑻𝟐𝟑𝒙 + 𝑻𝟐𝟑𝒚
𝑻𝟑 = √𝟒𝟔𝟐 + 𝟕𝟒𝟐
𝑻𝟑 = √𝟐𝟏𝟏𝟔 + 𝟓𝟒𝟕𝟔
𝑻𝟑 = √𝟕𝟓𝟗𝟐
𝑻𝟑 = 𝟖𝟕. 𝟏𝟑 N
Direction of 𝑻𝟑
𝟕𝟒
𝐭𝐚𝐧 𝜽 =
𝟒𝟔
𝟕𝟒
𝜽 = 𝐭𝐚𝐧−𝟏 ( )
𝟒𝟔
𝜽 = 𝟓𝟖. 𝟏𝟑∘
Answer
𝑻𝟑 = 𝟖𝟕. 𝟏 N
𝜽 = 𝟓𝟖. 𝟏∘
Direction: 58.1° below the horizontal toward the right.
