A man in a lift — apparent weight when lift accelerates up at 2 m/s²

Question

A man of mass 60 kg stands in a lift. Find his apparent weight when the lift accelerates upward at 2 m/s². (Take $g = 10$ m/s²)

Solution — Step by Step

Apparent weight is not the gravitational force on the man — it’s the normal force the lift floor exerts on him. This is what a weighing machine inside the lift would read. When the lift is stationary, apparent weight = actual weight = $mg$ . When the lift accelerates, apparent weight changes.

Two forces act on the man:

Weight $W = mg$ — downward (gravitational pull)
Normal force $N$ — upward (floor pushing up on him)

The man accelerates upward at $a = 2$ m/s² along with the lift.

Take upward as positive. The net force equals mass × acceleration:

N - mg = ma

The normal force must exceed gravity to produce upward acceleration — this makes intuitive sense. You feel “heavier” when a lift shoots upward.

N = m(g + a)

N = 60 \times (10 + 2)

N = 60 \times 12 = 720 \text{ N}

The apparent weight of the man = 720 N

His actual weight = $60 \times 10 = 600$ N

So his apparent weight is 120 N more than his actual weight — he feels 20% heavier. If he were standing on a weighing machine, it would show 72 kg instead of 60 kg.

Why This Works

Newton’s 2nd law applies in the ground (inertial) frame. In this frame, the man is genuinely accelerating upward, so the net force must be upward. This means the upward normal force must exceed the downward gravitational force — hence $N > mg$ .

The formula $N = m(g + a)$ for upward acceleration and $N = m(g - a)$ for downward acceleration are the two key results. They follow directly from $F_{net} = ma$ applied in the vertical direction.

Quick memory trick: up = add, down = subtract. When the lift goes up (accelerating), add $a$ to $g$ . When the lift goes down (accelerating), subtract $a$ from $g$ . This trick applies whether you’re asked for acceleration or apparent weight.

Alternative Method — Pseudo Force Approach (Non-Inertial Frame)

In the reference frame of the lift (non-inertial frame accelerating at $a$ upward), we introduce a pseudo force $= ma$ downward on the man.

Now the man is in equilibrium in this frame:

N = mg + ma = m(g + a) = 60(10 + 2) = 720 \text{ N}

Same answer. For board exams, the Newton’s 2nd law approach (Step 3) is preferred — pseudo forces are a JEE concept.

Common Mistake

Students often write $N = mg - ma$ when the lift goes up, confusing the sign. Remember: when the lift accelerates upward, the net force is upward, so $N > mg$ , giving $N = m(g+a)$ . The subtraction formula $N = m(g-a)$ applies when the lift accelerates downward (like when it starts moving down, or is in free fall). If $a = g$ in a falling lift, $N = 0$ — the person experiences weightlessness.

Question

Solution — Step by Step

Why This Works

Alternative Method — Pseudo Force Approach (Non-Inertial Frame)

Common Mistake

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