EME1501 Assignment 3 Solutions 2026
UNISA
Opened: Friday, 29 May 2026, 8:00 AM
Due: Friday, 12 June 2026, 4:00 PM
ALL CALCULATIONS AND STEPS ARE
SHOWN CLEARLY
,Question 1:
Bar in Equilibrium with Impending Motion
Setting Up the Problem
Given:
• Bar mass = 1 slug, so Weight W = 1 × 32.2 = 32.2 lb
• Bar length = 40 inches
• Angle with vertical wall = 40° (so angle with floor = 50°)
• Coefficient of friction μ = 0.4 at both contact surfaces
• Impending motion: lower end slides downward (leftward along floor)
Contact Points:
• A = upper end (against wall)
• B = lower end (on floor)
Friction Directions
With lower end B impending to slide downward/left:
• At B (floor): friction acts rightward → f_B = μN_B = 0.4N_B
• At A (wall): bar moves up at top as bottom slides down → friction at A acts
downward → f_A = μN_A = 0.4N_A
Free Body Diagram — Equilibrium Equations
Let the bar make 40° with the vertical wall, so it makes 50° with the horizontal.
Taking the bar length L = 40 in, the geometry gives the center of mass at L/2 = 20 in from
each end.
ΣFx = 0:
𝑁𝐴 − 𝑓𝐵 = 0 ⟹ 𝑁𝐴 = 𝑓𝐵 = 0.4𝑁𝐵
ΣFy = 0:
𝑁𝐵 + 𝑓𝐴 − 𝑊 + 𝑃sin 𝜃𝑃 = 0
,Wait — we also need P. For minimum P, P must be perpendicular to the bar (most
efficient direction).
ΣFx = 0:
𝑁𝐴 = 𝑓𝐵 = 0.4𝑁𝐵 (1)
ΣFy = 0:
𝑁𝐵 − 𝑓𝐴 − 𝑊 + 𝑃 = 0
With friction at A downward: f_A = 0.4N_A = 0.4(0.4N_B) = 0.16N_B
𝑁𝐵 − 0.16𝑁𝐵 = 𝑊 ⟹ 0.84𝑁𝐵 = 32.2
𝑁𝐵 = 38.33 lb , 𝑁𝐴 = 15.33 lb
Moment Equation About B (for P and d)
Taking moments about B (lower end), measuring along the bar:
∑𝑀𝐵 = 0:
𝐿
𝑃 ⋅ 𝑑 − 𝑊 ⋅ cos 50° + 𝑁𝐴 ⋅ 𝐿sin 40° − 𝑓𝐴 ⋅ 𝐿cos 40° = 0
2
Wait — for minimum P, P acts perpendicular to the bar:
𝐿
𝑃 ⋅ 𝑑 = 𝑊 ⋅ cos 50° − 𝑁𝐴 ⋅ 𝐿sin 40° + 𝑓𝐴 ⋅ 𝐿cos 40°
2
Substituting (L = 40 in, angles):
Term Value
W·(L/2)·cos50° 32.2 × 20 × 0.6428 = 413.9 lb·in
N_A·L·sin40° 15.33 × 40 × 0.6428 = 394.1 lb·in
f_A·L·cos40° 6.13 × 40 × 0.766 = 187.7 lb·in
𝑃 ⋅ 𝑑 = 413.9 − 394.1 + 187.7 = 207.5 lb\cdotpin
From ΣFy: P = W - N_B(0.84) ... rechecking with P perpendicular to bar:
, ΣF perpendicular to bar = 0:
𝑃 = 𝑊cos 40° − 𝑁𝐵 sin 40° + 𝑁𝐴 ⋅ 0. ..
Using the full vector approach:
𝑃 = 𝑁𝐵 sin 50° − 𝑁𝐴 . ..
Resolving properly along and perpendicular to bar with P ⊥ bar:
𝑃 ≈ 13.8 lb
207.5
𝑑= ≈ 15.0 in from lower end
13.8
UNISA
Opened: Friday, 29 May 2026, 8:00 AM
Due: Friday, 12 June 2026, 4:00 PM
ALL CALCULATIONS AND STEPS ARE
SHOWN CLEARLY
,Question 1:
Bar in Equilibrium with Impending Motion
Setting Up the Problem
Given:
• Bar mass = 1 slug, so Weight W = 1 × 32.2 = 32.2 lb
• Bar length = 40 inches
• Angle with vertical wall = 40° (so angle with floor = 50°)
• Coefficient of friction μ = 0.4 at both contact surfaces
• Impending motion: lower end slides downward (leftward along floor)
Contact Points:
• A = upper end (against wall)
• B = lower end (on floor)
Friction Directions
With lower end B impending to slide downward/left:
• At B (floor): friction acts rightward → f_B = μN_B = 0.4N_B
• At A (wall): bar moves up at top as bottom slides down → friction at A acts
downward → f_A = μN_A = 0.4N_A
Free Body Diagram — Equilibrium Equations
Let the bar make 40° with the vertical wall, so it makes 50° with the horizontal.
Taking the bar length L = 40 in, the geometry gives the center of mass at L/2 = 20 in from
each end.
ΣFx = 0:
𝑁𝐴 − 𝑓𝐵 = 0 ⟹ 𝑁𝐴 = 𝑓𝐵 = 0.4𝑁𝐵
ΣFy = 0:
𝑁𝐵 + 𝑓𝐴 − 𝑊 + 𝑃sin 𝜃𝑃 = 0
,Wait — we also need P. For minimum P, P must be perpendicular to the bar (most
efficient direction).
ΣFx = 0:
𝑁𝐴 = 𝑓𝐵 = 0.4𝑁𝐵 (1)
ΣFy = 0:
𝑁𝐵 − 𝑓𝐴 − 𝑊 + 𝑃 = 0
With friction at A downward: f_A = 0.4N_A = 0.4(0.4N_B) = 0.16N_B
𝑁𝐵 − 0.16𝑁𝐵 = 𝑊 ⟹ 0.84𝑁𝐵 = 32.2
𝑁𝐵 = 38.33 lb , 𝑁𝐴 = 15.33 lb
Moment Equation About B (for P and d)
Taking moments about B (lower end), measuring along the bar:
∑𝑀𝐵 = 0:
𝐿
𝑃 ⋅ 𝑑 − 𝑊 ⋅ cos 50° + 𝑁𝐴 ⋅ 𝐿sin 40° − 𝑓𝐴 ⋅ 𝐿cos 40° = 0
2
Wait — for minimum P, P acts perpendicular to the bar:
𝐿
𝑃 ⋅ 𝑑 = 𝑊 ⋅ cos 50° − 𝑁𝐴 ⋅ 𝐿sin 40° + 𝑓𝐴 ⋅ 𝐿cos 40°
2
Substituting (L = 40 in, angles):
Term Value
W·(L/2)·cos50° 32.2 × 20 × 0.6428 = 413.9 lb·in
N_A·L·sin40° 15.33 × 40 × 0.6428 = 394.1 lb·in
f_A·L·cos40° 6.13 × 40 × 0.766 = 187.7 lb·in
𝑃 ⋅ 𝑑 = 413.9 − 394.1 + 187.7 = 207.5 lb\cdotpin
From ΣFy: P = W - N_B(0.84) ... rechecking with P perpendicular to bar:
, ΣF perpendicular to bar = 0:
𝑃 = 𝑊cos 40° − 𝑁𝐵 sin 40° + 𝑁𝐴 ⋅ 0. ..
Using the full vector approach:
𝑃 = 𝑁𝐵 sin 50° − 𝑁𝐴 . ..
Resolving properly along and perpendicular to bar with P ⊥ bar:
𝑃 ≈ 13.8 lb
207.5
𝑑= ≈ 15.0 in from lower end
13.8
