A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surfac

Q: Why does the wedge accelerate to the left?

As the block slides down, gravity pulls it against the wedge. The normal force between them acts perpendicular to the surface. Because the surface is slanted, this normal force has a horizontal component that pushes the wedge to the left.

Q: Why do we need a pseudo force?

We are analyzing the block's motion relative to the wedge. Since the wedge is accelerating, it's a non-inertial frame of reference. Newton's laws only work in inertial frames, unless we introduce a pseudo force to account for the frame's acceleration.

Q: Why does the pseudo force point to the right?

The pseudo force always points in the exact opposite direction of the non-inertial frame's acceleration. Since the wedge accelerates to the left, the pseudo force on the block points to the right.

Q: Why does the pseudo force have a component pointing away from the wedge?

The pseudo force is horizontal to the right. The incline slopes downwards to the right. If you push horizontally to the right against a right-facing slope, your force actually tends to lift the object off the slope! That's why its perpendicular component is $ma_2 \sin 30^\circ$ outwards.

Q: Can the block lose contact with the wedge?

Yes! If the wedge accelerates fast enough to the left, the outward component of the pseudo force could exceed the inward component of gravity. The normal force would become zero, and the block would be in free fall. However, in this problem, the masses are such that contact is maintained.

Question

A block of mass $m$ slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is [Given, $m=8	ext{ kg}, M=16	ext{ kg}$]
Assume all the surfaces shown in the figure to be frictionless.

Options:

(A) $\frac{4}{3}g$

(B) $\frac{6}{5}g$

(C) $\frac{3}{5}g$

(D) $\frac{2}{3}g$

Sigma Percentile · Accepted Answer

### The Setup: A Dance of Two Masses

Imagine a block of mass $m$ resting on a smooth wooden wedge of mass $M$. The wedge itself is free to slide on a frictionless horizontal floor.

As gravity pulls the block down the incline, the block pushes against the wedge. This normal force has a horizontal component that drives the wedge backward.

Let's call the wedge's leftward acceleration $a_2$. If we isolate the wedge, the equation of motion is simply $N \sin 30^\circ = M a_2$, where $N$ is the normal force from the block.

### Entering the Matrix: The Non-Inertial Frame

To analyze the block's motion, it is incredibly powerful to shift our perspective and "ride" the wedge.

However, because the wedge is accelerating, it is a non-inertial frame of reference. Newton's laws don't work here unless we introduce a mathematical correction: the **pseudo force**.

This pseudo force is equal to the block's mass times the frame's acceleration ($m a_2$). It always acts in the exact opposite direction of the frame's acceleration. Since the wedge accelerates left, the pseudo force pushes the block to the right.

### The Perpendicular Balance: Finding the Normal Force

Now, let's look at the forces acting on the block perpendicular to the incline. The block remains in contact with the wedge, so the net force in this direction must be zero.

Gravity pulls the block *into* the wedge with a component $mg \cos 30^\circ$.

But here is the catch! The horizontal pseudo force has a component that pushes the block *away* from the wedge, equal to $m a_2 \sin 30^\circ$.

Therefore, the normal force $N$ is the difference between these two: $N = mg \cos 30^\circ - m a_2 \sin 30^\circ$.

### Calculating the Wedge's Acceleration

We now have a beautiful system of two equations. We can substitute our expression for $N$ back into the wedge's equation of motion.

This gives us $(mg \cos 30^\circ - m a_2 \sin 30^\circ) \sin 30^\circ = M a_2$.

By plugging in the given masses ($m=8	ext{ kg}$ and $M=16	ext{ kg}$) and the trigonometric values, we can solve for $a_2$.

After some careful algebra, we find that $a_2 = \frac{\sqrt{3}}{9}g$.

### The Parallel Push: Accelerating Down the Slope

With the wedge's acceleration known, we can finally determine the block's acceleration down the incline, let's call it $a_1$.

What forces are driving the block down the slope? Both gravity and the pseudo force contribute!

Gravity provides a downward pull of $mg \sin 30^\circ$. The pseudo force adds an extra push down the slope equal to $m a_2 \cos 30^\circ$.

So, the equation of motion is $m a_1 = mg \sin 30^\circ + m a_2 \cos 30^\circ$.

### The Final Assembly

We are in the endgame now. We substitute our calculated value of $a_2$ into the block's equation.

Notice how the mass $m$ cancels out entirely from the equation. We are left with $a_1 = g \sin 30^\circ + a_2 \cos 30^\circ$.

Substituting the values, we get $a_1 = \frac{g}{2} + \left(\frac{\sqrt{3}}{9}gight)\left(\frac{\sqrt{3}}{2}ight)$.

This simplifies elegantly to $\frac{g}{2} + \frac{g}{6}$, which equals **$\frac{2g}{3}$**. This is our final answer!

Select Answer:

Visualized Solution

$F B D of Wedge$

$Pseudo Force$

$Perpendicular Force Balance$

$Calculating a_{2}$

$Parallel Equation of Motion$

$Calculating a_{1}$

$Friction Scenario$

The Sigma Insight: Pseudo Force

The Setup

Entering the Matrix

The Perpendicular Balance

Calculating the Wedge's Acceleration

The Parallel Push

The Final Assembly

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FBD of Wedge

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pseudo Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Perpendicular Force Balance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating a2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Parallel Equation of Motion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating a1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Friction Scenario

The Sigma Insight: Pseudo Force

The Setup

Entering the Matrix

The Perpendicular Balance

Calculating the Wedge's Acceleration

The Parallel Push

The Final Assembly

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