A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction b

Q: Why do we take tension acting upwards on the boy?

When the boy pulls the rope downwards, Newton's third law dictates that the rope pulls the boy upwards with an equal and opposite force (Tension).

Q: Why is the tension also acting horizontally on the block?

The problem implies a pulley system where the rope the boy pulls is routed to also pull the block horizontally. The free body diagram provided in the standard solution confirms this setup.

Q: Can we solve this by taking the boy and block as separate systems?

Yes, absolutely! You would just have to account for the normal force and static friction *between* the boy's shoes and the block. However, treating them as a single system elegantly bypasses these internal forces.

Q: Why does the normal reaction not simply equal the total weight?

Because the rope is pulling the boy upwards! This upward tension supports some of the system's weight, meaning the floor doesn't have to push up as hard. Thus, $R = W - T$.

Q: What happens if the boy pulls with a force greater than 30 N?

The horizontal tension will exceed the maximum available static friction. The equilibrium will break, and the block (with the boy on it) will begin to accelerate to the right.

Question

A boy of mass $4	ext{ kg}$ is standing on a piece of wood having mass $5	ext{ kg}$. If the coefficient of friction between the wood and the floor is $0.5$, the maximum force that the boy can exert on the rope, so that the piece of wood does not move from its place is ............... N. (Round off to the nearest integer)
(Take, $g = 10	ext{ ms}^{-2}$)

Sigma Percentile · Accepted Answer

### The Delicate Balance of Friction and Tension

Imagine a boy standing on a heavy wooden block. He grabs a rope and pulls it with all his might. But this isn't just a simple tug-of-war. The rope is routed through a system of pulleys such that as he pulls it downwards, the rope pulls him **upwards**, while simultaneously pulling the wooden block **horizontally** to the right.

The question is: how hard can he pull before the block slips out from under him? To solve this, we need to master the art of the Free Body Diagram.

### Analyzing the Setup

Instead of analyzing the boy and the block separately—which would force us to calculate the messy internal friction between his shoes and the wood—we can treat the boy and the block as a **single, unified system**.

The total mass of our system is $M = 4	ext{ kg} + 5	ext{ kg} = 9	ext{ kg}$.

This means gravity is pulling the entire system downwards with a weight of $W = 90	ext{ N}$.

Now, what are the external forces acting on this system?
1.  **Gravity:** $90	ext{ N}$ acting downwards.
2.  **Normal Reaction ($R$):** The floor pushing upwards.
3.  **Tension ($T$):** The rope pulling the boy upwards.
4.  **Tension ($T$):** The rope pulling the block to the right.
5.  **Static Friction ($f$):** The floor resisting the rightward pull, acting to the left.

### The Master Equations

For the block to remain perfectly stationary, the system must be in static equilibrium. This means the forces must balance perfectly in both the vertical and horizontal directions.

Let's look at the vertical ($y$) axis first. The upward forces must equal the downward forces:

$$R + T = 90$$

Now, let's look at the horizontal ($x$) axis. The rightward pull must be perfectly countered by the leftward friction:

$$T = f$$

### The Friction Trap

Here is where the physics gets beautiful. The boy wants to pull as hard as possible, which means we need the **maximum static friction** to hold the block in place. The law of limiting friction states that $f_{	ext{max}} = \mu R$.

We are given that the coefficient of friction $\mu = 0.5$. Substituting this into our horizontal equation gives:

$$T = 0.5 R$$

Which can be rewritten as:

$$R = 2T$$

Notice the trap! By pulling the rope, the boy is increasing $T$. But because $R = 90 - T$, pulling harder actually **decreases** the normal reaction from the floor. A smaller normal reaction means less friction is available to hold the block in place. He is literally lifting himself into a slip!

### Final Calculation

Now, we simply substitute our relationship for $R$ back into the vertical equilibrium equation:

$$R + T = 90$$

$$(2T) + T = 90$$

$$3T = 90$$

$$T = 30	ext{ N}$$

The maximum force the boy can exert on the rope is exactly **$30	ext{ N}$**. If he pulls even a fraction of a Newton harder, the tension will overcome the dwindling static friction, and the block will slide away.

Enter Numerical Value:

Visualized Solution

$System Definition$

$\sum F_{y} = 0$

$\sum F_{x} = 0$

$f_{max} = μ R$

$Substitution$

$Final Calculation$

$Physical Insight$

The Sigma Insight: Static and Kinetic Friction

The Delicate Balance of Friction and Tension

Analyzing the Setup

The Master Equations

The Friction Trap

Final Calculation

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Visualized Solution

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

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

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

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

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

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

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