Block A of mass m and block B of mass 2m are placed on a fixed triangular wedge by means of a massle

Q: Wait, why does Block B's friction reach its maximum before Block A's?

Think of Block B as the 'aggressor' in this setup. Because it has a larger driving force ($2mg \sin 45^\circ$), it is the one actively trying to slide down and pull the string. Its own surface friction will act as the first line of defense. Only when Block B's friction is completely overwhelmed does it pull the string hard enough to threaten Block A's equilibrium, thereby engaging Block A's friction.

Q: Why is the friction on Block A acting downwards?

Friction always opposes the *tendency* of relative motion. Because Block B is pulling the string, Block A has a tendency to be dragged *up* the incline. To oppose this upward tendency, the static friction on Block A must point *down* the incline, acting in the same direction as gravity.

Q: Can the tension be different on both sides of the pulley?

No. The problem states that the pulley is frictionless and the string is massless. Under these ideal conditions, the tension remains uniform throughout the entire length of the string.

Q: What would happen if the net pulling force was greater than the total maximum friction?

If $F_{\text{net}} > (f_{\text{max}})_A + (f_{\text{max}})_B$, the static friction would be broken. Both blocks would accelerate, and you would have to use kinetic friction (if given) and Newton's Second Law ($F_{\text{net}} - f_k = m_{\text{total}} a$) to find the acceleration.

Q: Why do we use $\cos 45^\circ$ for the normal force instead of $\sin 45^\circ$?

The normal force acts perpendicular to the inclined surface. Gravity acts straight down. When you resolve the gravity vector into components, the component perpendicular to the incline is $mg \cos \theta$, which balances the normal force. The component parallel to the incline is $mg \sin \theta$, which drives the motion.

Question

Block $A$ of mass $m$ and block $B$ of mass $2m$ are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure. The wedge is inclined at $45^\circ$ to the horizontal on both sides. The coefficient of friction between block $A$ and the wedge is $2/3$ and that between block $B$ and the wedge is $1/3$. If the blocks $A$ and $B$ are released from rest, find (1997)
(a) the acceleration of $A$,
(b) tension in the string and
(c) the magnitude and direction of the force of friction acting on $A$.

Sigma Percentile · Accepted Answer

### The Setup: A Triangular Tug of War

Imagine a fixed triangular wedge with both of its base angles perfectly symmetrical at $45^\circ$. On the left incline rests Block A with a mass of $m$, and on the right incline sits Block B, which is twice as heavy with a mass of $2m$. These two blocks are locked in a mechanical tug of war, connected by a massless, inextensible string that runs over a frictionless pulley at the very apex of the wedge.

Before we dive into the math, we must understand the physical reality. Block B is heavier, so intuitively, it wants to slide down its side of the wedge, dragging Block A up the other side. But the surfaces are rough. The coefficient of friction for Block A is $\mu_A = \frac{2}{3}$, and for Block B, it is $\mu_B = \frac{1}{3}$. The central question is: will the system move, and if not, how do these frictional forces distribute themselves to maintain peace?

### The Invisible Tug of War: Driving Forces

Before we even think about friction, we must determine the 'net pulling force'—the raw driving force that exists purely due to gravity. Gravity pulls straight down, but the blocks are constrained to move along the inclines. Therefore, we must resolve the gravitational force into components parallel to the planes.

For Block A, the force pulling it down the left incline is:
$$F_2 = mg \sin 45^\circ = \frac{mg}{\sqrt{2}}$$

For Block B, the force pulling it down the right incline is:
$$F_1 = 2mg \sin 45^\circ = \sqrt{2}mg$$

Since $F_1 > F_2$, Block B is the dominant force. If there were no friction, the system would accelerate towards the right. The net pulling force driving this tendency is the difference between the two:
$$F_{	ext{net}} = F_1 - F_2 = \sqrt{2}mg - \frac{mg}{\sqrt{2}} = \frac{mg}{\sqrt{2}}$$

### The Friction Arsenal: Normal Forces and Limits

Now, let's look at the defenses. Friction depends on how hard the blocks are pressed against the wedge, which is the normal force ($N$). The normal force balances the perpendicular component of gravity, $mg \cos 	heta$.

For Block A:
$$N_A = mg \cos 45^\circ = \frac{mg}{\sqrt{2}}$$

For Block B:
$$N_B = 2mg \cos 45^\circ = \sqrt{2}mg$$

With the normal forces known, we can calculate the absolute maximum static friction each block can muster before it breaks loose and starts sliding.

Maximum friction for A:
$$(f_{	ext{max}})_A = \mu_A N_A = \frac{2}{3} \left(\frac{mg}{\sqrt{2}}ight) = \frac{\sqrt{2}mg}{3}$$

Maximum friction for B:
$$(f_{	ext{max}})_B = \mu_B N_B = \frac{1}{3} (\sqrt{2}mg) = \frac{\sqrt{2}mg}{3}$$

### The Standstill: Will it Move?

To determine if the system accelerates, we must compare the total available friction with the net pulling force.

The total maximum friction the system can provide is:
$$(f_{	ext{max}})_{	ext{total}} = (f_{	ext{max}})_A + (f_{	ext{max}})_B = \frac{2\sqrt{2}mg}{3} \approx 0.94 mg$$

Our net pulling force is:
$$F_{	ext{net}} = \frac{mg}{\sqrt{2}} \approx 0.71 mg$$

Since $0.71 mg < 0.94 mg$, the net pulling force is simply not strong enough to overcome the combined static friction of both blocks. **The system will not move, and the acceleration of Block A is exactly zero.**

### The Friction Distribution: The Tricky Part

Here is where many students get stuck. The system is at rest, which means the *actual* total friction acting on the blocks must exactly equal the net pulling force ($f_{	ext{req}} = \frac{mg}{\sqrt{2}}$). But how is this friction distributed between Block A and Block B? Do they both use half of their maximum?

No. We must look at the physical cause and effect. Block B is the 'aggressor'. It is the heavier block actively trying to slide down and pull the string. Because Block B is the one initiating the tendency of motion, the surface beneath Block B will be the first to react. Block B's own friction ($f_B$) will activate and try to hold it back.

Only if Block B's friction is completely overwhelmed will it pull the string hard enough to threaten Block A's equilibrium. Let's check if Block B can hold itself.

Block B's maximum friction is $(f_{	ext{max}})_B = \frac{\sqrt{2}mg}{3} \approx 0.47 mg$. But the force it needs to hold back is the net pulling force, which is $\approx 0.71 mg$.

Clearly, Block B's friction is not enough. Therefore, **Block B's friction will completely max out**, acting up the incline to prevent sliding:
$$f_B = \frac{\sqrt{2}mg}{3}$$

### The Final Calculation: Friction on A and Tension

Because Block B's friction maxed out, the remaining unbalanced force is transmitted through the string, trying to drag Block A up the incline. To maintain equilibrium, Block A's friction must step in.

The required friction from Block A is simply the total required friction minus what Block B is already providing:
$$f_A = F_{	ext{net}} - f_B = \frac{mg}{\sqrt{2}} - \frac{\sqrt{2}mg}{3}$$

To subtract these, we find a common denominator:
$$f_A = \frac{3mg - 2mg}{3\sqrt{2}} = \frac{mg}{3\sqrt{2}}$$

**Crucially, because Block A is resisting being pulled UP the incline, its friction $f_A$ must act DOWN the plane.**

Finally, to find the tension ($T$) in the string, we can isolate Block A and apply Newton's First Law. Block A is being pulled up by the tension $T$, and pulled down by both gravity ($mg \sin 45^\circ$) and its own friction ($f_A$).

$$T = mg \sin 45^\circ + f_A$$
$$T = \frac{mg}{\sqrt{2}} + \frac{mg}{3\sqrt{2}}$$
$$T = \frac{3mg + mg}{3\sqrt{2}} = \frac{4mg}{3\sqrt{2}}$$

Rationalizing the denominator gives us the final elegant result:
$$T = \frac{2\sqrt{2}mg}{3}$$

Visualized Solution

$Visualizing the Setup$

$Driving Forces (Gravity)$

$Net Pulling Force$

$Normal Forces$

$Maximum Static Friction$

$Will the System Move?$

$Friction Distribution Logic$

$Actual Friction on Block B$

$Actual Friction on Block A$

$Tension in the String$

The Sigma Insight: Static and Kinetic Friction

The Setup

The Invisible Tug of War

The Friction Arsenal

The Standstill

The Friction Distribution

The Final Calculation

Similar Questions

List-I

List-II

A small block of mass of $0.1 kg$ lies on a fixed inclined plane $P Q$ which makes an angle $θ$ with the horizontal. A horizontal force of $1 N$ acts on the block through its centre of mass as shown in the figure. The block remains stationary if (Take $g = 10 m/s^{2}$ )

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Driving Forces (Gravity)

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Will the System Move?

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

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

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tension in the String