Classical-Mechanics · Unit 9 · Video 4 · Interactive Practice

When the Ramp Runs Away: A Block on an Accelerating Wedge

IKey Formulas

Formula	Name	Key idea
$a_{b,y} = -(a_{b,x} - A)\tan\phi$	Acceleration constraint	Block stays on the moving face
$a_{b,x} = \dfrac{g + A\tan\phi}{\cot\phi + \tan\phi}$	Horizontal acceleration	Newton's 2nd law $+$ constraint
$a_{b,y} = \dfrac{A - g\tan\phi}{\cot\phi + \tan\phi}$	Vertical acceleration	Fed back through the constraint
$a = \sqrt{\dfrac{g^2 + A^2}{2}}$	Magnitude at $\phi = 45\degree$	From $a_{b,x}=\tfrac{g+A}{2},\ a_{b,y}=\tfrac{A-g}{2}$

Key Insight: The drive $A$ decides the block's fate — it slides down when $A < g\tan\phi$ , rides along when $A = g\tan\phi$ , and climbs when $A > g\tan\phi$ .

IIThe Ramp's Constraint

On a ramp of angle $\phi$ , the block's height and horizontal offset stay locked in the fixed ratio $\tan\phi$ .

💡 Differentiating $y_b = (\ell - (x_b - x_w))\tan\phi$ twice in time turns this locked ratio into the acceleration constraint $a_{b,y} = -(a_{b,x} - A)\tan\phi$ .

IIIDriving the Wedge

Only one drive $A$ makes the block ride along without sliding at $\phi = 45\degree$ — which one?

Wedge drive (A) 0.5 g

IVAcceleration Magnitude

At $\phi = 45\degree$ , how does the block's acceleration magnitude grow as the wedge drive $A$ increases?

VQuiz Questions

Problem 1 · Horizontal Acceleration

Given: a frictionless wedge at $\phi = 45\degree$ driven at $A = g$ . Find the block's horizontal acceleration $a_{b,x}$ .

✅ Correct! At

45\degree

a_{b,x} = \tfrac{g+A}{2} = \tfrac{g+g}{2} = g

— the block matches the wedge and rides along.

❌ Close, but…

\tfrac{g}{2}

is the fixed-wedge (

A=0

) value; here

A=g

, so the numerator is

g+g

❌ Not quite.

2g = g+A

is only the numerator — divide by

\cot 45\degree + \tan 45\degree = 2

❌ Not quite. Use

a_{b,x} = \tfrac{g + A\tan\phi}{\cot\phi + \tan\phi}

with

\tan 45\degree = \cot 45\degree = 1

Show solution

At $\phi = 45\degree$ we have $\tan\phi = \cot\phi = 1$ , so the denominator is $2$ :

a_{b,x} = \frac{g + A\tan\phi}{\cot\phi + \tan\phi} = \frac{g + g}{1 + 1} = g

With $A = g$ the block's horizontal acceleration equals the wedge's own acceleration $A$ — it is carried along.

Problem 2 · Vertical Acceleration on a Fixed Wedge

Given: $\phi = 45\degree$ and the wedge held fixed ( $A = 0$ ). Find the block's vertical acceleration $a_{b,y}$ . It is not $-g$ — the ramp pushes back.

✅ Correct!

a_{b,y} = \tfrac{A - g}{2} = \tfrac{0 - g}{2} = -\tfrac{g}{2}

— the block drops at half of

g

❌ Not quite. The normal force from the ramp carries part of the weight, so the drop is below

g

❌ Check the sign. The block moves downward, so

a_{b,y} < 0

❌ Not quite.

a_{b,y} = 0

requires the driven case

A = g

; here

A = 0

❌ Not quite. Use

a_{b,y} = \tfrac{A - g\tan\phi}{\cot\phi + \tan\phi}

\phi = 45\degree

Show solution

With $\phi = 45\degree$ (so the denominator is $2$ ) and $A = 0$ :

a_{b,y} = \frac{A - g\tan\phi}{\cot\phi + \tan\phi} = \frac{0 - g}{2} = -\frac{g}{2}

The block does not free-fall at $g$ : the incline's normal force supports part of the weight, halving the downward acceleration.

Problem 3 · Components and Magnitude (Fixed Wedge)

Given: $\phi = 45\degree$ and $A = 0$ . Find the horizontal component $a_{b,x}$ and the magnitude $|\vec{a}|$ .

What is $a_{b,x}$ ?

What is the magnitude $|\vec{a}|$ ?

✅ Correct!

a_{b,x} = \tfrac{g}{2}

a_{b,y} = -\tfrac{g}{2}

, so

|\vec{a}| = \tfrac{g}{\sqrt{2}} = g\sin 45\degree

— the fixed-ramp result.

❌ Check $a_{b,x}$ .

a_{b,x} = \tfrac{g + A}{2} = \tfrac{g}{2}

A = 0

❌ Check the magnitude. Combine both components:

\sqrt{(g/2)^2 + (g/2)^2} = \tfrac{g}{\sqrt{2}}

Show solution

Components at $\phi = 45\degree$ , $A = 0$ :

a_{b,x} = \frac{g+0}{2} = \frac{g}{2}, \qquad a_{b,y} = \frac{0-g}{2} = -\frac{g}{2}

Magnitude:

|\vec{a}| = \sqrt{a_{b,x}^2 + a_{b,y}^2} = \sqrt{\frac{g^2}{4} + \frac{g^2}{4}} = \sqrt{\frac{g^2}{2}} = \frac{g}{\sqrt{2}}

This equals $g\sin 45\degree$ — exactly the acceleration down a frictionless fixed ramp.

Problem 4 · The No-Slide Drive (General Angle)

Given: a wedge at general angle $\phi$ . For what drive $A$ does the block ride along with the wedge — no sliding, $a_{b,y} = 0$ ?

✅ Correct! Setting the numerator

A - g\tan\phi = 0

gives

A = g\tan\phi

— and at

\phi = 45\degree

this is

A = g

❌ Close, but…

A = g

works only at

\phi = 45\degree

where

\tan\phi = 1

; in general set

A - g\tan\phi = 0

❌ Not quite.

A = g\cot\phi

leaves

a_{b,y} \propto g\cot\phi - g\tan\phi \ne 0

unless

\phi = 45\degree

❌ Not quite. The block rides along when its vertical acceleration vanishes:

a_{b,y} = 0

Show solution

The block stops drifting vertically when $a_{b,y} = 0$ :

a_{b,y} = \frac{A - g\tan\phi}{\cot\phi + \tan\phi} = 0 \;\Longrightarrow\; A = g\tan\phi

At $\phi = 45\degree$ this becomes $A = g\tan 45\degree = g$ — matching the driven-wedge limit from the video, where the block rides motionless relative to the ramp.

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