Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure. The c

Q: Why do we assume they move together?

Because the question asks for the maximum force *before* block A slides. Up to this maximum force, static friction is sufficient to keep them moving together with the same acceleration.

Q: Why is the friction from the table calculated using the total mass?

The normal force exerted by the table must support the weight of both block A and block B. Therefore, the normal force is $(m_A + m_B)g$, making the kinetic friction $\mu(m_A + m_B)g$.

Q: What provides the forward acceleration for block A?

The only horizontal force acting on block A is the static friction from block B. As block B is pulled forward, it drags block A along with it via this frictional force.

Q: Why do we use pseudo force in the solution?

Using the non-inertial frame of block B makes it easier to visualize. In block B's frame, block A experiences a backward pseudo force $m_A a$. For A to remain at rest relative to B, this pseudo force must be balanced by the forward static friction.

Q: Does the mass of block A affect the maximum acceleration?

Interestingly, no! The maximum static friction is $\mu m_A g$, and the required force to accelerate it is $m_A a$. The mass $m_A$ cancels out, meaning the maximum acceleration depends only on the coefficient of friction and gravity ($a_{\text{max}} = \mu g$).

Question

Two blocks $A$ and $B$ of masses $m_A = 1	ext{ kg}$ and $m_B = 3	ext{ kg}$ are kept on the table as shown in figure. The coefficient of friction between $A$ and $B$ is $0.2$ and between $B$ and the surface of the table is also $0.2$. The maximum force $F$ that can be applied on $B$ horizontally, so that the block $A$ does not slide over the block $B$ is [Take, $g = 10	ext{ m/s}^2$]

Options:

(A) 12 N

(B) 16 N

(C) 8 N

(D) 40 N

Sigma Percentile · Accepted Answer

Imagine you are standing in a physics lab, looking at two blocks stacked on a table. A force is pulling the bottom block, and you are tasked with finding exactly how hard you can pull before the top block slips off.

This is a classic two-block problem, a favorite in JEE because it tests your ability to shift perspectives—from the macroscopic system to the microscopic interactions between the blocks. Let's break down the physics step by step.

### Analyzing the Setup

We have block $A$ of mass $m_A = 1	ext{ kg}$ resting on block $B$ of mass $m_B = 3	ext{ kg}$. Block $B$ is on a table. A horizontal force $F$ is applied to block $B$.

For block $A$ to stay put on block $B$, they must move together as a single combined system. This means they will share the exact same acceleration, let's call it $a$. As the system tries to move right, the table exerts a kinetic friction force on block $B$ to the left. Since the whole system presses down on the table, the normal force is the total weight.

Let's calculate this friction from the table, $f_1$. The formula is $f_1 = \mu (m_A + m_B) g$. Substituting the given values, we get $f_1 = 0.2 	imes (1 + 3) 	imes 10 = 8	ext{ N}$.

### The Master Equation for the System

Now, applying Newton's second law to the combined system. The net driving force is the applied force $F$ minus the table's friction $f_1$. This net force equals the total mass times the common acceleration.

Mathematically, this is written as $F - f_1 = (m_A + m_B) a$. Substituting our known values, we get $F - 8 = 4a$.

Rearranging this, we find the acceleration $a$ in terms of the unknown force $F$:

$$a = \frac{F - 8}{4}$$

### The Limiting Condition for Block A

Now, let's shift our focus entirely to block $A$. What is pulling block $A$ forward? It is the static friction from block $B$ acting on it to the right. In the frame of block $B$, we can also think of a pseudo force pushing $A$ to the left.

For block $A$ not to slip, the pseudo force trying to slide it backwards must be balanced by the maximum static friction between the blocks. The maximum static friction is $\mu$ times the normal force of $A$, which is $m_A g$.

This gives us the limiting condition: $m_A a \le \mu m_A g$. Notice how the mass of block $A$ cancels out! The maximum acceleration the system can have before $A$ slips is simply $\mu g$.

Calculating this, we get $a_{	ext{max}} = 0.2 	imes 10 = 2	ext{ m/s}^2$.

### Final Calculation

We now have two expressions for the acceleration. We can equate the system's acceleration in terms of $F$ to the maximum allowed acceleration of $2	ext{ m/s}^2$.

$$\frac{F - 8}{4} = 2$$

Multiplying both sides by $4$ gives $F - 8 = 8$. Adding $8$ to both sides, we find that the maximum force $F$ is **$16	ext{ N}$**.

If we apply exactly $16	ext{ N}$, block $A$ is on the verge of slipping. Any force greater than this will cause block $A$ to slide backwards relative to block $B$.

Select Answer:

Visualized Solution

$System Overview$

$Condition for No Slipping$

$Friction from the Table$

$Calculating f_{1}$

$Equation of Motion for System$

$Acceleration of System$

$Forces on Block A$

$Limiting Condition for A$

$Maximum Acceleration$

$Equating Accelerations$

$Solving for F$

$Final Answer$

$The Way Forward$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Master Equation for the System

The Limiting Condition for Block A

Final Calculation

Similar Questions

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A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is

List-I

List-II

Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for No Slipping

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating f1​

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Condition for A

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

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

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

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

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