In short

When a block of mass m sits on a smooth wedge of mass M that is free to slide on a smooth floor, both bodies accelerate. The wedge's acceleration is A = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta} and the block's acceleration relative to the wedge along the incline is a_{\text{rel}} = \frac{(M+m)g\sin\theta}{M + m\sin^2\theta}. You solve the system by drawing separate FBDs, writing Newton's second law for each body, and using the constraint that the block stays on the wedge surface.

Picture a loading ramp at a godown in a factory compound. A heavy crate is placed at the top of a steel ramp, but today there are no chocks — nothing pins the ramp to the concrete floor. The moment you release the crate, something surprising happens. The crate does not just slide down the ramp. The ramp itself starts skidding backward across the smooth floor. Two bodies, both accelerating, their motions tangled together by the simple fact that the crate stays in contact with the ramp surface.

This is the wedge-and-block problem — one of the most important two-body problems in Newtonian mechanics. It appears in every competitive physics exam from KVPY to JEE Advanced, and it teaches you a skill that no single-body problem can: how to handle two free body diagrams simultaneously, linked by a geometric constraint.

Why the wedge moves

If the wedge were bolted to the floor, the problem would be straightforward: resolve forces along and perpendicular to the incline, and the block slides down with acceleration g\sin\theta. The wedge does not move because the floor provides whatever horizontal force is needed to hold it in place.

But remove those bolts. Now the floor is smooth, and the wedge is free. The block presses on the wedge's incline surface — Newton's third law guarantees that. That press has a horizontal component, and since nothing resists it (the floor is frictionless), the wedge accelerates.

Here is the key subtlety: once the wedge starts moving, the block is no longer sliding on a stationary surface. The block's absolute motion is a combination of sliding along the incline and moving with the wedge. Their accelerations are coupled, and you cannot solve for one without the other.

Setting up the problem

Take a wedge of mass M with incline angle \theta, resting on a smooth horizontal floor. A block of mass m is placed on the smooth incline. Both start from rest.

The geometry matters. The wedge has its vertical face on the left, its horizontal base on the floor, and the incline rising from the lower-left corner up to the upper-right. When the block slides down the slope, it moves toward the lower-left. By Newton's third law, the block pushes the wedge toward the right.

Setup: block on a movable wedge on a smooth floor A triangular wedge of mass M sits on a horizontal floor. The incline rises from lower-left to upper-right. A small block of mass m sits on the inclined surface near the top. The angle between the incline and the horizontal is theta. An arrow shows the block tending to slide down-left along the incline, and another arrow shows the wedge tending to slide right along the floor. M θ m slides down wedge slides smooth floor
A block of mass $m$ on a wedge of mass $M$. Both surfaces are smooth. The block tends to slide down the incline (toward the lower-left), and the wedge slides to the right.

To solve this system, you need three things:

  1. FBD of the block — forces acting on the block alone
  2. FBD of the wedge — forces acting on the wedge alone
  3. The constraint equation — the geometric fact that the block stays on the incline surface

Free body diagrams — one body at a time

The most common mistake in wedge problems is drawing all forces on both bodies in the same diagram. That mixes up action-reaction pairs and leads to sign errors. Always draw each body separately.

FBD of the block

Isolate the block. Replace every contact with the force it produces. The block touches only the incline surface, so exactly two forces act on it:

  1. Weight mg, vertically downward
  2. Normal force N, perpendicular to the incline surface, pointing away from the wedge

Since the incline rises from lower-left to upper-right, the outward normal (away from the wedge body) points to the upper-left.

Free body diagram of the block A square representing the block, with two force arrows: weight mg pointing straight down, and normal force N pointing perpendicular to the incline surface (upper-left direction). Coordinate axes show x to the right and y upward. x y m mg N θ incline FBD of the block
FBD of the block. Only two forces: weight $mg$ downward and normal force $N$ perpendicular to the incline. The angle between $N$ and the vertical is $\theta$. There is no friction (smooth surface).

Resolve these forces in the ground frame (x to the right, y upward):

Horizontal (x): The normal force has a horizontal component N\sin\theta pointing to the left.

ma_{bx} = -N\sin\theta \tag{1}

Why: the only horizontal force on the block is the horizontal component of N. Since N points to the upper-left, its horizontal projection points left (negative x).

Vertical (y): The normal force has a vertical component N\cos\theta pointing upward, and the weight mg pulls downward.

ma_{by} = N\cos\theta - mg \tag{2}

Why: the two vertical forces are N\cos\theta (up) and mg (down). The block accelerates downward, so a_{by} will be negative.

FBD of the wedge

Now isolate the wedge. Three forces act on it:

  1. Weight Mg, vertically downward
  2. Normal from the floor N_f, vertically upward
  3. Reaction from the block N', perpendicular to the incline, pointing into the wedge

By Newton's third law, the block pushes on the wedge with force N' = N (same magnitude as the normal the wedge exerts on the block) in the opposite direction — into the incline, pointing to the lower-right.

Free body diagram of the wedge A triangular wedge with three force arrows: weight Mg pointing straight down from the centre, normal from floor Nf pointing straight up from the base, and reaction from block N-prime pointing into the incline surface (lower-right direction, perpendicular to the incline). x y M Mg Nf N' = N θ FBD of the wedge
FBD of the wedge. Three forces: weight $Mg$ downward, floor normal $N_f$ upward, and the reaction $N'$ from the block pushing into the incline (lower-right). By Newton's third law, $N' = N$.

The wedge does not accelerate vertically (it stays on the floor), so only the horizontal equation matters:

MA = N\sin\theta \tag{3}

Why: the horizontal component of N' is N\sin\theta to the right (the block pushes the wedge to the right). The floor is frictionless, so this is the only horizontal force on the wedge. The wedge accelerates to the right with acceleration A.

The constraint — the block stays on the wedge surface

This is the equation that ties the two FBDs together. The block does not fly off the incline or burrow into it — it stays on the surface. So the block's acceleration relative to the wedge must be directed purely along the incline surface.

The block's absolute acceleration is (a_{bx}, a_{by}). The wedge's acceleration is (A, 0) — purely horizontal. The relative acceleration of the block with respect to the wedge is:

\vec{a}_{\text{rel}} = (a_{bx} - A,\; a_{by})

For the block to stay on the incline, this relative acceleration must point along the slope. The down-slope direction on this wedge (from upper-right to lower-left) has direction (-\cos\theta, -\sin\theta). So the ratio of vertical to horizontal components of the relative acceleration must match:

\frac{a_{by}}{a_{bx} - A} = \frac{-\sin\theta}{-\cos\theta} = \tan\theta
a_{by} = (a_{bx} - A)\tan\theta \tag{4}

Why: the relative acceleration must be parallel to the incline surface. Since the block slides down-slope (toward lower-left), both the horizontal component (a_{bx} - A) and the vertical component a_{by} of the relative acceleration are negative. Their ratio equals \tan\theta.

Solving the system — the non-inertial frame approach

You have four equations (1)–(4) in four unknowns (a_{bx}, a_{by}, A, N). You can grind through the algebra in the ground frame, but there is a more elegant method: work in the accelerating frame of the wedge.

In the wedge's frame, the wedge is stationary, and the block slides along the incline with some acceleration a_{\text{rel}}. Since this is a non-inertial frame (the wedge accelerates at A to the right), you must add a pseudo-force mA on the block, pointing to the left — opposite to the wedge's acceleration.

Assumptions: Both surfaces (floor and incline) are smooth (frictionless). The block stays in contact with the wedge. All motion is in the vertical plane.

Now resolve all forces on the block along and perpendicular to the incline surface.

Along the incline (down-slope positive):

The down-slope direction is from upper-right to lower-left, making angle \theta with the horizontal.

ma_{\text{rel}} = mg\sin\theta + mA\cos\theta \tag{I}

Why: both gravity's along-slope component and the pseudo-force's along-slope component drive the block down the slope. The pseudo-force helps the block slide faster — the wedge accelerating to the right is equivalent to the block feeling an extra push to the left, which has a component down the slope.

Perpendicular to the incline (outward positive, block stays on surface so net force is zero):

N - mg\cos\theta + mA\sin\theta = 0
N = mg\cos\theta - mA\sin\theta \tag{II}

Why: the perpendicular acceleration is zero (the block stays on the surface). The pseudo-force pushes the block away from the incline — it reduces the normal force. This makes physical sense: a block presses less firmly into a movable wedge than into a fixed one, because the wedge is "giving way."

For the wedge (ground frame, horizontal):

MA = N\sin\theta \tag{III}

Now solve. Substitute A = N\sin\theta / M from (III) into (II):

N = mg\cos\theta - m\cdot\frac{N\sin\theta}{M}\cdot\sin\theta
N = mg\cos\theta - \frac{mN\sin^2\theta}{M}
N + \frac{mN\sin^2\theta}{M} = mg\cos\theta
N\left(\frac{M + m\sin^2\theta}{M}\right) = mg\cos\theta
\boxed{N = \frac{Mmg\cos\theta}{M + m\sin^2\theta}} \tag{5}

Why: this is the normal force between the block and the wedge. Check the limits: when M \to \infty (fixed wedge), N \to mg\cos\theta — the standard result for a block on a fixed incline. When \theta = 0 (flat surface), N = mg — the block just sits on a horizontal platform. When \theta = 90°, N = 0 — the block is in free fall beside a vertical wall. All correct.

Acceleration of the wedge

From equation (III) and the expression for N:

A = \frac{N\sin\theta}{M} = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta}

Using \sin\theta\cos\theta = \tfrac{1}{2}\sin 2\theta:

\boxed{A = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta}} \tag{6}

Why: the wedge acceleration depends on the mass ratio and the angle. When M is very large, A \approx 0 — a heavy wedge barely moves. When \theta = 0 or \theta = 90°, A = 0 — a flat wedge has no horizontal push, and a vertical wedge has no normal force. The maximum wedge acceleration occurs at some intermediate angle.

Acceleration of the block relative to the wedge

From equation (I):

a_{\text{rel}} = g\sin\theta + A\cos\theta = g\sin\theta + \frac{mg\sin\theta\cos^2\theta}{M + m\sin^2\theta}
= g\sin\theta\left(1 + \frac{m\cos^2\theta}{M + m\sin^2\theta}\right) = g\sin\theta\cdot\frac{M + m\sin^2\theta + m\cos^2\theta}{M + m\sin^2\theta}

Since \sin^2\theta + \cos^2\theta = 1:

\boxed{a_{\text{rel}} = \frac{(M + m)g\sin\theta}{M + m\sin^2\theta}} \tag{7}

Why: the block accelerates faster down a movable wedge than down a fixed one. On a fixed wedge, the acceleration is g\sin\theta. On a movable wedge, the denominator M + m\sin^2\theta < M + m (for \theta < 90°), so a_{\text{rel}} > g\sin\theta. This makes physical sense — the wedge retreating from under the block lets it fall faster.

Sanity check: fixed wedge limit

When M \to \infty: A \to 0 (wedge does not move) and a_{\text{rel}} \to g\sin\theta (the standard inclined-plane result). ✓

When M \to 0: A \to g\cos\theta/\sin\theta = g\cot\theta... but actually, A = mg\sin\theta\cos\theta/(m\sin^2\theta) = g\cos\theta/\sin\theta = g\cot\theta. And a_{\text{rel}} \to mg\sin\theta/(m\sin^2\theta) = g/\sin\theta. In this limit, the block is essentially in free fall — the massless wedge offers no resistance. You can verify that the block's absolute downward acceleration equals g in this limit (as it should for a free-falling body with nothing to support it).

Worked examples

Example 1: Smooth wedge on smooth floor

A smooth wedge of mass M = 10 kg with incline angle \theta = 45° rests on a smooth horizontal floor. A block of mass m = 5 kg is released from rest on the incline. Find the acceleration of the wedge, the acceleration of the block relative to the wedge, and the normal force between them. Take g = 10 m/s².

Example 1 setup: 5 kg block on 10 kg wedge at 45 degrees A 10 kg wedge at 45 degrees on a smooth floor with a 5 kg block on the incline. Arrows show the block sliding down-left and the wedge sliding right. 10 kg 45° 5 kg smooth floor
A 5 kg block on a 10 kg wedge with $\theta = 45°$, both surfaces smooth.

Step 1. Identify the knowns and the formulas.

M = 10 kg, m = 5 kg, \theta = 45°, g = 10 m/s².

\sin 45° = \cos 45° = \frac{1}{\sqrt{2}}, \sin^2 45° = \frac{1}{2}.

Why: at 45°, the sine and cosine are equal, which simplifies the algebra considerably.

Step 2. Compute the denominator M + m\sin^2\theta.

M + m\sin^2\theta = 10 + 5 \times \frac{1}{2} = 10 + 2.5 = 12.5 \text{ kg}

Why: this denominator appears in all three results. Compute it once.

Step 3. Find the acceleration of the wedge.

A = \frac{mg\sin\theta\cos\theta}{M + m\sin^2\theta} = \frac{5 \times 10 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}}{12.5} = \frac{5 \times 10 \times \frac{1}{2}}{12.5} = \frac{25}{12.5} = 2 \text{ m/s}^2

Why: the wedge accelerates at 2 m/s² to the right. For a 10 kg wedge, this corresponds to a net horizontal force of 20 N — the horizontal component of the block's push.

Step 4. Find the block's acceleration relative to the wedge.

a_{\text{rel}} = \frac{(M + m)g\sin\theta}{M + m\sin^2\theta} = \frac{(10 + 5) \times 10 \times \frac{1}{\sqrt{2}}}{12.5} = \frac{150/\sqrt{2}}{12.5} = \frac{106.07}{12.5} \approx 8.49 \text{ m/s}^2

Why: compare this to the fixed-wedge result g\sin 45° = 10/\sqrt{2} \approx 7.07 m/s². The block slides about 20% faster on a movable wedge because the wedge retreats from under it.

Step 5. Find the normal force.

N = \frac{Mmg\cos\theta}{M + m\sin^2\theta} = \frac{10 \times 5 \times 10 \times \frac{1}{\sqrt{2}}}{12.5} = \frac{500/\sqrt{2}}{12.5} = \frac{353.6}{12.5} \approx 28.3 \text{ N}

Why: on a fixed wedge, the normal force would be mg\cos 45° = 5 \times 10/\sqrt{2} \approx 35.4 N. The movable wedge gives a normal force of 28.3 N — about 20% less. The block presses less firmly into a surface that is accelerating away from it.

Now let's draw the dual FBD showing the solved forces and accelerations:

Dual FBD for Example 1 with computed values Side by side FBDs of the block and wedge with numerical force values. Block has mg = 50 N down and N = 28.3 N upper-left. Wedge has Mg = 100 N down, Nf up, and N = 28.3 N into the incline, with A = 2 m/s² to the right. m 50 N 28.3 N arel ≈ 8.5 m/s² Block M 100 N Nf 28.3 N A = 2 m/s² Wedge
Dual FBD with computed values. The block feels 50 N downward and 28.3 N normal force. The wedge receives the 28.3 N reaction and accelerates at 2 m/s² to the right. The block slides down the incline at 8.49 m/s² relative to the wedge.

Step 6. Verify conservation of horizontal momentum.

The system starts at rest, so total horizontal momentum should remain zero at all times. The horizontal acceleration of the block (ground frame) is a_{bx} = -N\sin\theta/m = -28.3 \times (1/\sqrt{2})/5 = -4 m/s². The horizontal acceleration of the wedge is A = 2 m/s².

Check: m \cdot a_{bx} + M \cdot A = 5 \times (-4) + 10 \times 2 = -20 + 20 = 0 ✓

Why: no external horizontal force acts on the system (the floor is smooth), so horizontal momentum is conserved. The block accelerates left at 4 m/s² and the wedge accelerates right at 2 m/s², and their momenta cancel exactly.

Result: The wedge accelerates at A = 2 m/s² to the right. The block accelerates at a_{\text{rel}} \approx 8.49 m/s² down the incline relative to the wedge. The normal force is N \approx 28.3 N.

What this shows: On a movable wedge, the block slides faster and presses less firmly than on a fixed wedge. The system conserves horizontal momentum, so the heavier wedge moves slower while the lighter block moves faster.

Example 2: Wedge with friction — when does the block not slide?

A wedge of mass M = 8 kg with incline angle \theta = 30° sits on a smooth horizontal floor. A block of mass m = 2 kg is placed on the incline. The coefficient of static friction between the block and the wedge surface is \mu_s = 0.4. Does the block slide on the wedge, or do both move together as a single unit?

Example 2 setup: block on rough wedge on smooth floor A 2 kg block on a wedge with incline angle 30 degrees and coefficient of friction 0.4 between block and wedge. The floor is smooth. 8 kg 30° 2 kg μs = 0.4 smooth floor
A 2 kg block on an 8 kg wedge with $\theta = 30°$. Friction exists between block and wedge ($\mu_s = 0.4$), but the floor is smooth.

The idea: If friction is strong enough, the block does not slide on the wedge. Both move together as a single body of mass (M + m). But the block still tends to slide — gravity pulls it down the incline — and static friction must prevent this. The question is whether the maximum static friction is large enough.

Step 1. Assume the block does NOT slide. Then the block and wedge move as one body.

The system accelerates horizontally. But what drives the horizontal acceleration? There is no external horizontal force on the (block + wedge) system — the floor is smooth. So the horizontal acceleration of the combined system is zero!

Wait — that means both bodies remain stationary? That would imply the block is sitting on the incline without sliding, held by friction, with the entire system at rest on a frictionless floor.

Why: if the block and wedge move together, they form a single system on a smooth floor. The only external forces are gravity (vertical) and the floor normal (vertical). No horizontal force exists. So they cannot accelerate horizontally. The system stays at rest.

Step 2. Draw the FBD of the block assuming it is at rest.

FBD of the block with friction, assuming static equilibrium Block with three forces: weight mg downward, normal force N perpendicular to incline (upper-left), and static friction f along the incline (up-slope, preventing the block from sliding down). m mg N f incline FBD of block (static case)
FBD of the block assuming it is in equilibrium on the incline. Static friction $f$ acts up the slope, preventing the block from sliding down.

If the block is stationary on the incline, the forces along and perpendicular to the incline must balance:

Along incline: f = mg\sin\theta = 2 \times 10 \times \sin 30° = 2 \times 10 \times 0.5 = 10 N

Perpendicular to incline: N = mg\cos\theta = 2 \times 10 \times \cos 30° = 2 \times 10 \times 0.866 = 17.32 N

Why: for the block to remain stationary, friction must exactly balance the gravity component pulling it down the slope, and the normal force must balance the component pushing it into the surface.

Step 3. Check whether static friction can provide this force.

Maximum static friction: f_{\max} = \mu_s N = 0.4 \times 17.32 = 6.93 N.

Required friction: f = 10 N.

Since f_{\text{required}} = 10 N > f_{\max} = 6.93 N, the friction is not sufficient. The block DOES slide on the wedge.

Why: the incline angle \theta = 30° is too steep for \mu_s = 0.4. On a fixed incline, the block would slide when \tan\theta > \mu_s. Here, \tan 30° = 0.577 > 0.4, confirming the slide.

Step 4. Find the condition for the block NOT to slide.

For the block to remain stationary on a movable wedge that is on a smooth floor (and hence the system is at rest), you need:

mg\sin\theta \leq \mu_s \cdot mg\cos\theta
\tan\theta \leq \mu_s
\boxed{\theta \leq \arctan(\mu_s)}

Why: this is the same condition as for a fixed incline — because when the system is at rest, there is no difference between a fixed and a movable wedge. The wedge has no reason to move unless the block slides. If friction holds the block, nothing moves.

For \mu_s = 0.4: \theta_{\max} = \arctan(0.4) \approx 21.8°. Since 30° > 21.8°, the block slides.

Result: The block slides. The maximum angle for the block to remain stationary is \arctan(\mu_s) \approx 21.8°. Since \theta = 30° exceeds this, friction cannot prevent the slide.

What this shows: On a smooth floor, the no-sliding condition for a block on a wedge is identical to the no-sliding condition on a fixed wedge: \tan\theta \leq \mu_s. This is because when neither body moves, the "movable" wedge is indistinguishable from a fixed one. The movable wedge adds complexity only when the block actually slides — then you must use the full two-body analysis from the derivation above.

Common confusions

If you came here to solve wedge-block problems, you have everything you need above. What follows is for readers preparing for JEE Advanced who want the general case with friction, and a deeper look at the constraint geometry.

Wedge-block system with friction on the incline (block sliding)

When kinetic friction \mu_k acts between the block and the wedge (but the floor remains smooth), the equations change. In the wedge's non-inertial frame:

Along the incline (down-slope positive, block sliding down):

ma_{\text{rel}} = mg\sin\theta + mA\cos\theta - \mu_k N

Why: friction opposes relative motion. Since the block slides down the incline, kinetic friction acts up the slope, opposing the slide.

Perpendicular to incline:

N = mg\cos\theta - mA\sin\theta

Wedge horizontal (ground frame):

MA = N\sin\theta - \mu_k N\cos\theta = N(\sin\theta - \mu_k\cos\theta)

Why: the friction from the block on the wedge acts along the incline, and by Newton's third law, it points down the slope (opposite to the friction on the block). This friction has a horizontal component \mu_k N\cos\theta pointing to the LEFT, which partially opposes the rightward push of the normal force's horizontal component.

Solving these simultaneously (substituting A from the wedge equation into the perpendicular equation):

N = mg\cos\theta - m\sin\theta \cdot \frac{N(\sin\theta - \mu_k\cos\theta)}{M}
N\left[1 + \frac{m\sin\theta(\sin\theta - \mu_k\cos\theta)}{M}\right] = mg\cos\theta
\boxed{N = \frac{Mmg\cos\theta}{M + m\sin\theta(\sin\theta - \mu_k\cos\theta)}}
\boxed{A = \frac{mg\cos\theta(\sin\theta - \mu_k\cos\theta)}{M + m\sin\theta(\sin\theta - \mu_k\cos\theta)}}

When \mu_k = 0, these reduce to the smooth-incline formulas derived earlier. When \mu_k is large enough that \sin\theta - \mu_k\cos\theta < 0 (i.e., \tan\theta < \mu_k), the wedge acceleration A becomes negative — the formula breaks down because the block would not slide at all in that case, so kinetic friction would not apply.

Why the constraint works geometrically

The constraint "block stays on the incline" can be understood as a requirement on the position, not just the acceleration.

Let the wedge move a distance X to the right. The block, relative to the wedge, slides a distance s down the incline. In the ground frame, the block's horizontal displacement is:

x_{\text{block}} = X - s\cos\theta

And the block's vertical displacement (downward):

y_{\text{block}} = -s\sin\theta

Taking two time derivatives: a_{bx} = A - a_{\text{rel}}\cos\theta and a_{by} = -a_{\text{rel}}\sin\theta.

This gives the same constraint as equation (4): \frac{a_{by}}{a_{bx} - A} = \frac{-a_{\text{rel}}\sin\theta}{-a_{\text{rel}}\cos\theta} = \tan\theta.

The position-based derivation makes the constraint's geometric meaning transparent: the block's vertical drop per unit horizontal displacement relative to the wedge is fixed at \tan\theta.

Energy conservation as a check

For a smooth wedge on a smooth floor, the only force doing work is gravity (normals are perpendicular to displacement). After the block has descended a height h:

mgh = \frac{1}{2}mv_{\text{block}}^2 + \frac{1}{2}MV_{\text{wedge}}^2

Combined with conservation of horizontal momentum (mv_{bx} + MV = 0), this gives two equations in two unknowns — providing an independent route to the speeds at any instant. This is particularly useful for finding the speed of the block after it slides off the bottom of the wedge.

Where this leads next