Solution

1. Kinematic Setup and Constraint

Let's establish a coordinate system where the wall is the y-axis and the ground is the x-axis. The triangle slides down the wall with its side AC on the wall. For the hypotenuse AB to push the block rightward as the triangle moves downward, the triangle must be oriented such that vertex C is above vertex A. This causes the hypotenuse AB to slope upwards and to the right.

The core physical principle is the constraint of contact: the components of the velocities of the triangle and the block normal to the contact surface (the hypotenuse) must be equal for them to remain in contact without penetrating each other.

$$ \vec{v}_{\text{triangle}} \cdot \hat{n} = \vec{v}_{\text{block}} \cdot \hat{n} $$

The same relationship holds for their accelerations, as the geometry is constant.

$$ \vec{a}_{\text{triangle}} \cdot \hat{n} = \vec{a}_{\text{block}} \cdot \hat{n} $$

2. Geometric Analysis

• The triangle's velocity is purely vertical and downwards: $\vec{v}_{T} = -v \hat{j}$.
• The block's velocity is purely horizontal and rightward: $\vec{v}_{B} = v_{B} \hat{i}$.
• In the chosen orientation, angle A ($\alpha$) is the angle between the vertical side AC and the hypotenuse AB. The hypotenuse AB makes an angle of $90^\circ - \alpha$ with the horizontal.
• The normal vector $\hat{n}$ to the hypotenuse is perpendicular to it. It makes an angle of $-\alpha$ with the horizontal axis. Thus, the unit normal vector pointing from the hypotenuse towards the block is: $$ \hat{n} = \cos(-\alpha) \hat{i} + \sin(-\alpha) \hat{j} = \cos(\alpha) \hat{i} - \sin(\alpha) \hat{j} $$

3. Derivation of Speed

Applying the velocity constraint:

$$ (-v \hat{j}) \cdot (\cos(\alpha) \hat{i} - \sin(\alpha) \hat{j}) = (v_{B} \hat{i}) \cdot (\cos(\alpha) \hat{i} - \sin(\alpha) \hat{j}) $$ $$ (-v)(-\sin(\alpha)) = v_{B}(\cos(\alpha)) $$ $$ v \sin(\alpha) = v_{B} \cos(\alpha) $$

Solving for the speed of the block, $v_{B}$:

$$ v_{B} = v \frac{\sin(\alpha)}{\cos(\alpha)} = v \tan(\alpha) $$

4. Derivation of Acceleration

The relationship between the motions is linear, so the same derivation applies to acceleration.

• The triangle's acceleration is $\vec{a}_{T} = -a \hat{j}$.
• The block's acceleration is $\vec{a}_{B} = a_{B} \hat{i}$.

Applying the acceleration constraint:

$$ (-a \hat{j}) \cdot (\cos(\alpha) \hat{i} - \sin(\alpha) \hat{j}) = (a_{B} \hat{i}) \cdot (\cos(\alpha) \hat{i} - \sin(\alpha) \hat{j}) $$ $$ a \sin(\alpha) = a_{B} \cos(\alpha) $$

Solving for the acceleration of the block, $a_{B}$:

$$ a_{B} = a \tan(\alpha) $$

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Alternative Derivation (Calculus): Let vertex A be at $(0, y)$ and the block's top-left corner be at $(x, h)$. The hypotenuse passes through A and has a slope of $m = \cot(\alpha)$. The line equation is $Y - y = (\cot\alpha)X$. The corner $(x, h)$ lies on this line: $h - y = x \cot(\alpha)$. This gives the constraint equation: $x = (h - y)\tan(\alpha)$. Differentiating with respect to time gives the velocity relation:

$\frac{dx}{dt} = -\tan(\alpha)\frac{dy}{dt}$. With $v_B = \frac{dx}{dt}$ and $-v = \frac{dy}{dt}$, we get $v_B = -\tan(\alpha)(-v) = v\tan(\alpha)$.

Differentiating again yields the acceleration: $a_B = a\tan(\alpha)$.