Newton's laws applications — elevator, pulley, spring scale, connected bodies

Question

How do we apply Newton’s laws to common scenarios — a person in an elevator, spring scale readings, and connected body systems?

Solution — Step by Step

A person of mass $m$ stands on a weighing scale in an elevator. The scale reads the normal force $N$ (apparent weight), not the actual weight $mg$ .

Elevator accelerating upward (or decelerating while going down):

N - mg = ma \implies N = m(g + a)

Scale reads more than actual weight — you feel heavier.

Elevator accelerating downward (or decelerating while going up):

mg - N = ma \implies N = m(g - a)

Scale reads less — you feel lighter.

Free fall ( $a = g$ ):

N = m(g - g) = 0

Weightlessness — the scale reads zero.

A spring scale reads the tension in it. Two key scenarios:

Scenario 1: A spring balance hangs from the ceiling with a mass $m$ attached below. Reading = $mg$ .

Scenario 2: Two masses $m_1$ and $m_2$ pull from opposite ends of a spring balance (like a tug of war). Reading = tension in the spring.

A spring scale reads the force exerted on ONE end, not the sum of forces on both ends. If 5 kg hangs on each side, the reading is $5g$ (not $10g$ ). Think of it as: the spring must exert 5g on each side to keep them stationary, so the tension throughout is 5g.

Three blocks $m_1$ , $m_2$ , $m_3$ connected by strings on a smooth surface, pulled by force $F$ on $m_1$ :

System acceleration:

a = \frac{F}{m_1 + m_2 + m_3}

Tension between $m_1$ and $m_2$ :

T_1 = (m_2 + m_3) \cdot a

Tension between $m_2$ and $m_3$ :

T_2 = m_3 \cdot a

The trick: for tension in any string, consider the mass being pulled by that string (everything on the other side from the applied force).

System approach: Treat all connected bodies as one system to find acceleration. Internal forces (tensions) cancel out.

Individual FBD approach: Draw FBD for each body separately to find internal forces (tensions).

Use the system approach first (to get $a$ ), then individual FBDs (to get tensions). This two-step method works for almost every connected body problem.

flowchart TD
    A["Newton's Laws Application"] --> B{"Scenario type?"}
    B -->|"Elevator"| C["N = m times g plus-or-minus a"]
    B -->|"Spring scale"| D["Reading = tension in spring"]
    B -->|"Connected bodies"| E["Step 1: System approach for a"]
    E --> F["Step 2: Individual FBD for tensions"]
    C --> G{"Direction of acceleration?"}
    G -->|"Upward"| H["Apparent weight increases"]
    G -->|"Downward"| I["Apparent weight decreases"]
    G -->|"Free fall"| J["Weightlessness: N = 0"]

Why This Works

Newton’s second law ( $\vec{F}_{net} = m\vec{a}$ ) applies to every object individually. When objects are connected, they share the same acceleration (assuming inextensible connections), so the system can be treated as one composite object for finding that acceleration. The individual FBDs then reveal the internal forces that enforce this shared motion.

Alternative Method

For elevator problems with multiple stages (accelerating, constant velocity, decelerating), draw an acceleration-time graph first. Then calculate the apparent weight for each phase. This visual approach prevents sign errors and is especially useful for multi-part problems in NEET.

Common Mistake

In elevator problems, students confuse “elevator moving up” with “elevator accelerating up.” An elevator moving upward at constant velocity has zero acceleration — the apparent weight equals actual weight ( $N = mg$ ). The change in apparent weight happens only during acceleration or deceleration, not during constant velocity motion. NEET 2022 and CBSE boards test this distinction regularly.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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