RBD CYU 19

Solution to Question 19

Let unit vector $\hat{i}$ be horizontal to the right and $\hat{j}$ be vertical upwards. The container has width $l$ and height $h$. The Center of Mass (CM) is at $(l/2, h/2)$ relative to the bottom-left corner.

Part 1: Determining Coefficient of Friction ($\mu$)

Scenario: Plank accelerates right with $a_0$.
Pseudo Force: $\vec{F}_p = m a_0$ acting to the Left.
Tendency: Container tends to slide Left.
Friction: Acts to the Right at the Wedge (Right edge).

Torque Balance about CM:
Let $N_{wh}$ be normal at the wheel (Left) and $N_{we}$ be normal at the wedge (Right).

• $N_{wh}$ (at left) produces Clockwise (CW) torque: $N_{wh} \frac{l}{2}$.
• $N_{we}$ (at right) produces Counter-Clockwise (CCW) torque: $N_{we} \frac{l}{2}$.
• Friction $f = m a_0$ (at right, acting Right) produces Counter-Clockwise (CCW) torque about CM: $f \frac{h}{2}$.

Equation: $$ N_{wh} \frac{l}{2} = N_{we} \frac{l}{2} + f \frac{h}{2} $$ Substitute $N_{wh} = mg – N_{we}$ and $f = m a_0$: $$ (mg – N_{we}) \frac{l}{2} = N_{we} \frac{l}{2} + m a_0 \frac{h}{2} $$ $$ mg l – 2 N_{we} l = m a_0 h \implies N_{we} = \frac{m}{2} \left( g – a_0 \frac{h}{l} \right) $$

Friction Limit: $$ f = \mu N_{we} \implies m a_0 = \mu \left[ \frac{m}{2} \left( g – a_0 \frac{h}{l} \right) \right] $$ $$ \mu = \frac{2 a_0}{g – a_0 \frac{h}{l}} \quad \text{…(i)} $$

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Part 2: Max Retardation to Prevent Sliding

Scenario: Plank moves right but decelerates with magnitude $a$.
Pseudo Force: $\vec{F}_p = m a$ acting to the Right.
Tendency: Container tends to slide Right.
Friction: Acts to the Left at the Wedge (Right edge).

Torque Balance about CM:

• $N_{wh}$ (Left): CW torque.
• $N_{we}$ (Right): CCW torque.
• Friction $f = ma$ (at right, acting Left): Produces Clockwise (CW) torque about CM. (Pulling the “feet” left rotates the body CW). Torque magnitude: $f \frac{h}{2}$.

Equation: $$ N_{we} \frac{l}{2} = N_{wh} \frac{l}{2} + f \frac{h}{2} $$ Substitute $N_{wh} = mg – N_{we}$: $$ N_{we} \frac{l}{2} = (mg – N_{we}) \frac{l}{2} + ma \frac{h}{2} $$ $$ 2 N_{we} \frac{l}{2} = mg \frac{l}{2} + ma \frac{h}{2} \implies N_{we} = \frac{m}{2} \left( g + a \frac{h}{l} \right) $$

Sliding Condition: $$ ma = \mu N_{we} = \mu \left[ \frac{m}{2} \left( g + a \frac{h}{l} \right) \right] $$ $$ 2a = \mu g + \mu a \frac{h}{l} $$ $$ a \left( 2 – \mu \frac{h}{l} \right) = \mu g \implies a = \frac{\mu g}{2 – \mu \frac{h}{l}} $$

Substituting $\mu$ from (i): $$ a = \frac{ g \left( \frac{2 a_0}{g – a_0 \frac{h}{l}} \right) }{ 2 – \frac{h}{l} \left( \frac{2 a_0}{g – a_0 \frac{h}{l}} \right) } $$ Multiply numerator and denominator by $(g – a_0 \frac{h}{l})$: $$ a = \frac{ 2 a_0 g }{ 2(g – a_0 \frac{h}{l}) – 2 a_0 \frac{h}{l} } = \frac{ 2 a_0 g }{ 2g – 2a_0 \frac{h}{l} – 2a_0 \frac{h}{l} } $$ $$ a = \frac{ 2 a_0 g }{ 2g – 4 a_0 \frac{h}{l} } = \frac{ a_0 g }{ g – 2 a_0 \frac{h}{l} } $$ $$ \mathbf{a_{sliding}} = \frac{ a_0 g l }{ g l – 2 a_0 h } $$

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Part 3: Max Retardation to Prevent Toppling

The pseudo force acts to the Right. The container will tend to topple over the leading edge, which is the Wedge (Right edge).

Condition: Toppling begins when the Normal reaction on the rear side (Wheels, Left) becomes zero ($N_{wh} = 0$). Taking moments about the Wedge (Right bottom corner):
Torque from Pseudo Force (CW) $\ge$ Torque from Weight (CCW) $$ (ma) \frac{h}{2} = (mg) \frac{l}{2} $$ $$ a = g \frac{l}{h} $$ $$ \mathbf{a_{toppling}} = \frac{g l}{h} $$