One end of massless rope, which passes over a massless and frictionless pulley P is tied to a hook C

Q: Why can't the man climb up the rope?

If the man climbs up, he must accelerate upwards, which means the tension in the rope would be greater than his weight ($600\text{ N}$). Since the rope can only hold $360\text{ N}$, it would break instantly.

Q: Why does accelerating downwards reduce tension?

When you accelerate downwards, part of your weight is used to create that acceleration, leaving less force required from the rope to support you. It's the same feeling as being in an elevator that suddenly goes down—you feel lighter!

Q: What happens if he slides down with a constant velocity?

Constant velocity means zero acceleration. If acceleration is zero, the tension equals his full weight ($600\text{ N}$), and the rope will break. He *must* accelerate to survive.

Q: Is the acceleration $4\text{ m/s}^2$ the maximum or minimum?

It is the *minimum* downward acceleration required. If he accelerates at exactly $4\text{ m/s}^2$, the tension is exactly $360\text{ N}$. If he accelerates faster (e.g., $5\text{ m/s}^2$), the tension drops even further, which is safer for the rope.

Q: Why did the question say 'climb on the rope' if he has to slide down?

It's a classic trick! The word 'climb' often implies going up, but in physics problems, you must follow the math. The math proves that going up or staying still is impossible, so 'climbing' here practically means a controlled slide down.

Question

One end of massless rope, which passes over a massless and frictionless pulley $P$ is tied to a hook $C$ while the other end is free. Maximum tension that the rope can bear is $360	ext{ N}$. With what value of maximum safe acceleration (in $	ext{ms}^{-2}$) can a man of $60	ext{ kg}$ climb on the rope? [Take $g = 10	ext{ m/s}^2$]

Options:

(A) 16

(B) 6

(C) 4

(D) 80

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are the man on the rope. Gravity is relentlessly pulling you down with a force equal to your weight, which is $mg$. To keep you from falling, the rope pulls back up on you with a tension force, $T$.

Let's calculate your weight. With a mass of $60	ext{ kg}$ and acceleration due to gravity $g = 10	ext{ m/s}^2$, your weight is $60 	imes 10 = 600	ext{ N}$.

Here is the terrifying part: the rope has a maximum breaking limit of $T_{\max} = 360	ext{ N}$. If you try to just hang there stationary, the rope needs to supply $600	ext{ N}$ of tension to balance your weight. Since $600	ext{ N} > 360	ext{ N}$, the rope will snap instantly!

### The Impossible Climb

What if you try to climb up? To accelerate upwards, the tension in the rope must be even **greater** than your weight.

If $T - mg = ma$, then $T = mg + ma$. This means the tension would be $600	ext{ N}$ plus some extra force. The rope would break even faster.

So, climbing up is completely out of the question. The only way to survive is to reduce the tension in the rope. How do we do that? By accelerating downwards!

### The Physics of Sliding Down

When you slide down the rope with a downward acceleration $a$, you are allowing gravity to do some of the work.

According to Newton's Second Law, the net downward force is your weight minus the tension. This gives us the master equation: 
$$mg - T = ma$$

Rearranging this to solve for tension, we get:
$$T = mg - ma$$

Notice how the tension $T$ is now **less** than your weight $mg$. The faster you accelerate downwards, the lower the tension becomes. It is exactly like the feeling of becoming lighter when an elevator suddenly drops.

### Final Calculation

To ensure the rope does not break, the tension $T$ must be less than or equal to the maximum allowed tension of $360	ext{ N}$.

Setting up our inequality:
$$mg - ma \le 360$$

Now, we substitute our known values into the raw setup:
$$600 - 60a \le 360$$

Let's carefully solve for your downward acceleration $a$. Moving the terms around:
$$60a \ge 600 - 360$$
$$60a \ge 240$$

Dividing both sides by $60$, we find:
$$a \ge 4	ext{ m/s}^2$$

**The minimum safe downward acceleration is $4	ext{ m/s}^2$.**

Even though the question asks for the safe acceleration to "climb" on the rope, physics dictates that you must slide down. Any acceleration less than $4	ext{ m/s}^2$ will cause the tension to exceed $360	ext{ N}$, and the rope will break.

Select Answer:

Visualized Solution

$Free Body Diagram$

$Checking the Stationary Case$

$Equation for Climbing Down$

$Condition for Safe Sliding$

$Substituting the Values$

$Solving for Acceleration$

$Final Calculation$

$Conclusion$

The Sigma Insight: Newton's Laws of Motion

Analyzing the Setup

The Impossible Climb

The Physics of Sliding Down

Final Calculation

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Select Answer:

Visualized Solution

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

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

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

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

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

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

.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; } Conclusion