The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum ho

Q: Why do we treat both blocks as a single system initially?

Since the problem asks for the force to move the blocks *together*, there is no relative motion between them. They share the same acceleration, allowing us to simplify the calculation by treating them as one combined mass.

Q: Why does friction act in the forward direction on the upper block?

Friction always opposes *relative* motion. The lower block is being pulled forward, so relative to the lower block, the upper block tends to slip backward. Therefore, the lower block exerts a forward frictional force on the upper block to pull it along.

Q: Why do we use the limiting value of static friction?

We are asked for the *maximum* force $F$. A larger $F$ means a larger acceleration. The upper block's acceleration is solely provided by static friction. To get the maximum acceleration, we need the maximum possible static friction, which is the limiting friction $\mu N$.

Q: What happens if we apply a force greater than 15 N?

If $F > 15\text{ N}$, the required acceleration for the blocks to move together would exceed $5\text{ m/s}^2$. The static friction cannot provide this much acceleration to the upper block, so the static friction will break, kinetic friction will take over, and the upper block will slip backward relative to the lower block.

Q: Why didn't we include friction from the table in our equations?

The problem explicitly states that "the table is smooth." A smooth surface implies that the coefficient of friction between the table and the lower block is zero, so there is no frictional force from the table to oppose the motion.

Question

The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is .......N. (Take, $g = 10 	ext{ms}^{-2}$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a perfectly smooth table. On this table rests a $2	ext{ kg}$ block, and perched on top of it is a $1	ext{ kg}$ block. We are pulling the bottom block with a horizontal force $F$. The table offers no resistance, but the surfaces between the two blocks are rough, with a coefficient of static friction $\mu = 0.5$.

Our goal is to find the **maximum force** $F$ we can apply such that the two blocks move together as a single unit without slipping against each other.

### The Master Equation

If the blocks are moving together, they share the same acceleration, let's call it $a$. Because they move as one, we can treat them as a single system with a combined mass of $M = m_1 + m_2 = 1	ext{ kg} + 2	ext{ kg} = 3	ext{ kg}$.

According to Newton's Second Law, the total external force equals the total mass times the acceleration:
$$F = (m_1 + m_2)a$$
$$F = 3a \quad \dots (i)$$

To find the maximum force $F$, we first need to figure out the maximum possible acceleration $a$ the system can have before the top block starts slipping.

### The Role of Friction

Let's shift our focus entirely to the top block ($m_1 = 1	ext{ kg}$). We are not applying any direct force to it. So, what is pulling it forward?

It's the **static friction** $f_s$ acting between the two blocks! As the bottom block accelerates forward, it tries to slide out from under the top block. Friction opposes this relative motion, dragging the top block along.

For the top block, Newton's Second Law gives:
$$f_s = m_1 a$$

### Pushing to the Limit

To find the maximum acceleration, we must push the static friction to its absolute limit. The maximum static friction (limiting friction) is given by:
$$f_{s,	ext{max}} = \mu N$$

Here, $N$ is the normal force acting on the top block. Since the top block is in vertical equilibrium, the normal force simply balances its weight:
$$N = m_1 g$$

Substituting this into our friction equation:
$$f_{s,	ext{max}} = \mu m_1 g$$

Now, we equate this maximum friction to the force required to accelerate the top block:
$$\mu m_1 g = m_1 a_{	ext{max}}$$

Notice how the mass of the top block ($m_1$) beautifully cancels out!
$$a_{	ext{max}} = \mu g$$

### Final Calculation

Let's plug in the given values ($\mu = 0.5$ and $g = 10	ext{ m/s}^2$):
$$a_{	ext{max}} = 0.5 	imes 10 = 5	ext{ m/s}^2$$

This is the maximum acceleration the system can sustain. Any higher, and the top block will slip. Finally, we substitute this maximum acceleration back into our master equation to find the maximum force $F$:
$$F_{	ext{max}} = 3 	imes a_{	ext{max}}$$
$$F_{	ext{max}} = 3 	imes 5 = 15	ext{ N}$$

The maximum horizontal force that can be applied is **$15	ext{ N}$**.

Animated Solution for Physics - Laws of Motion: The coefficient of static friction between two blocks is 0.5 and the table is smooth. The maximum horizontal force that can be applied to move the blocks together is .......N. (Take, $g = 10 ms^{- 2}$ )

Enter Numerical Value:

Visualized Solution

$S y s t e m S e t u p$

$C o mm o n A cce l er a t i o n$

$S y s t e m E q u a t i o n$

$F B D o f U pp er B l oc k$

$L imi t in g F r i c t i o n$

$C a l c u l a t in g M a x im u m A cce l er a t i o n$

$V a l u eo f A cce l er a t i o n$

$M a x im u m F or ce$

$T h e W a y F or w a r d$

The Sigma Insight: Static and Kinetic Friction

Analyzing the Setup

The Master Equation

The Role of Friction

Pushing to the Limit

Final Calculation

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What is the maximum value of the force $F$ such that the block shown in the arrangement, does not move? (2003)

Enter Numerical Value:

Visualized Solution

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