A steel block of 10 kg rests on a horizontal floor as shown. When three iron cylinders are placed on

Q: Why do we treat the block and cylinders as one system?

Because they move together with the exact same acceleration. By treating them as a single unit, the internal forces (like the normal force between the cylinders and the block) cancel out, making the calculation much simpler.

Q: Why is the normal reaction not just equal to Mg?

If the system were at rest or moving with a constant velocity, the normal reaction would indeed be equal to $Mg$. However, because the system is accelerating downwards, the floor is effectively 'falling away' from the block, meaning it doesn't have to push up as hard to support it. Hence, $R$ is less than $Mg$.

Q: What is the use of the coefficient of static friction $\mu_s = 0.2$ given in the question?

It is a distractor! In competitive exams, extra information is often provided to test your conceptual clarity. Since there are no horizontal forces and no tendency to slide, friction is zero and $\mu_s$ is irrelevant.

Q: What if the system was accelerating upwards instead?

If the system were accelerating upwards, the net force would be upwards. The equation would become $R - Mg = Ma$, which means $R = M(g+a)$. The normal reaction would be greater than the actual weight.

Q: Does the arrangement of the cylinders (one on top of the other two) matter?

No, the specific arrangement does not matter for finding the normal reaction from the floor, as long as all components are moving together with the same vertical acceleration. Only their total combined mass matters.

Question

A steel block of $10	ext{ kg}$ rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration $0.2	ext{ m/s}^2$. The normal reaction $R$ by the floor, if mass of the iron cylinders are equal and of $20	ext{ kg}$ each, is ...... $	ext{N}$.
[Take, $g = 10	ext{ m/s}^2$ and $\mu_s = 0.2$]

Options:

(A) 716

(B) 686

(C) 714

(D) 684

Sigma Percentile · Accepted Answer

### The Elevator Effect: Finding Normal Reaction in an Accelerating System

Imagine you are standing in an elevator that suddenly starts accelerating downwards. For a brief moment, you feel lighter, right? This physical sensation is exactly what is happening to the steel block and iron cylinders in our problem. Let's break down the physics behind this "apparent weight" phenomenon.

### Analyzing the Setup

We are given a system consisting of a steel block and three iron cylinders resting on a horizontal floor. The entire arrangement is moving downwards with an acceleration of $a = 0.2	ext{ m/s}^2$.

Instead of analyzing the block and each cylinder separately, we can use a powerful physics trick: **treat them as a single combined system**. Since they all move together with the exact same acceleration, the internal normal forces between the cylinders and the block cancel out.

The total mass of our system is simply the sum of all individual masses:
$$M = m_{	ext{block}} + 3 	imes m_{	ext{cyl}}$$
$$M = 10 + 3(20) = 70	ext{ kg}$$

### The Master Equation

Now, let's draw the Free Body Diagram for this $70	ext{ kg}$ super-block. There are only two vertical forces acting on it:
1. The gravitational pull downwards: $W = Mg$
2. The normal reaction from the floor pushing upwards: $R$

Since the system is accelerating downwards, the net force must be directed downwards. According to Newton's Second Law ($F_{	ext{net}} = Ma$), we write our master equation:
$$Mg - R = Ma$$

### Final Calculation

We have all the pieces of the puzzle. Let's substitute the known values ($M = 70	ext{ kg}$, $g = 10	ext{ m/s}^2$, $a = 0.2	ext{ m/s}^2$) into our equation:
$$70(10) - R = 70(0.2)$$
$$700 - R = 14$$

Rearranging to solve for the normal reaction $R$:
$$R = 700 - 14$$
$$R = 686	ext{ N}$$

**The Red Herring:** Did you notice the problem provided the coefficient of static friction, $\mu_s = 0.2$? This is a classic distractor! Because there are no horizontal forces applied and no tendency for the blocks to slide horizontally, friction never comes into play. Always trust your Free Body Diagram to tell you exactly which variables matter.

Select Answer:

Visualized Solution

$M_{total} = m_{block} + 3 m_{cyl}$

$F_{net} = M a$

$70 (10) - R = 70 (0.2)$

$R = 686 N$

$Final Answer$

$The Way Forward$

The Sigma Insight: Newton's Laws of Motion

The Elevator Effect

Analyzing the Setup

The Master Equation

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; } Mtotal​=mblock​+3mcyl​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 70(10)−R=70(0.2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R=686 N

.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