EME1501 Assignment 6 Solutions 2026

EME1501 Assignment 6 Solutions 2026
MECHANICS
UNISA

, Question 1
Given:
• Mass of bar: m = 1 slug
• Coefficient of friction: μ = 0.4
• Inclination: θ = 40°
• Bar length: L = 40 in
𝑊𝑒𝑖𝑔ℎ𝑡: 𝑊 = 𝑚𝑔 = (1)(32.2) = 32.2 𝑙𝑏

Normal reaction
𝑁 = 𝑊 𝑐𝑜𝑠 40°
𝑁 = 32.2 𝑐𝑜𝑠 40°
𝑁 = 24.66 𝑙𝑏

Friction force
𝐹 = 𝜇𝑁
𝐹 = 0.4(24.66)
𝐹 = 9.86 𝑙𝑏
At impending motion downward, friction acts up the incline.

Force equilibrium along the incline
𝛴𝐹 ∥ = 0
𝑃 + 𝐹 − 𝑊 𝑠𝑖𝑛 40° = 0
𝑃 = 𝑊 𝑠𝑖𝑛 40° − 𝐹
𝑃 = 32.2 𝑠𝑖𝑛 40° − 9.86
𝑃 = 20.70 − 9.86
𝑃 = 10.84 𝑙𝑏

Moment equilibrium about the contact point
40
= 20 𝑖𝑛
2
𝑃𝑑 = 𝑊(20 𝑠𝑖𝑛 40°)
𝑊(20 𝑠𝑖𝑛 40°)
= 𝑑
𝑃
32.2(20) 𝑠𝑖𝑛 40°
= 38.2 𝑖𝑛
10.84

𝑃𝑚𝑖𝑛 = 10.84 𝑙𝑏
𝑑 = 38.2 𝑖𝑛 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑜𝑤𝑒𝑟 𝑐𝑜𝑛𝑡𝑎𝑐𝑡 𝑝𝑜𝑖𝑛𝑡