Tension in Strings

Introduction

One of the most common and fascinating applications of Newton’s Laws of Motion in mechanics is the elevator problem. It deals with understanding how the tension in a string or the reading on a weighing machine (apparent weight) changes when an object is inside an elevator that is moving in different ways.

The concept is simple but highly important: when an object is hanging by a string or resting on a scale inside a moving elevator, the forces acting on it depend not just on gravity but also on the elevator’s acceleration. This problem is a classic in physics, and is frequently asked in competitive exams like JEE, NEET, SSC, and in general mechanics courses.

At the heart of this problem lies the concept of tension in strings. Let’s dive deep into it step by step.

Understanding Tension

Tension is the pulling force transmitted through a string, rope, cable, or chain when it is pulled tight by forces acting at both ends.

It always acts along the length of the string.
It is always directed away from the object it is attached to.
It adjusts itself depending on the situation, so long as the string is massless and inextensible (ideal string).

Example: If you hang a block from the ceiling using a rope, the rope is under tension. That tension is what balances the block’s weight.

Newton’s Second Law – The Foundation

To solve elevator problems, we rely on Newton’s second law of motion: F=maF = m aF=ma

Where:

FFF = Net force acting on the body
mmm = Mass of the body
aaa = Acceleration of the body

Also, gravitational force (weight) is always acting downward: W=mgW = m gW=mg

Where:

WWW = Weight of the body
ggg = Acceleration due to gravity (9.8 m/s29.8 \, m/s^29.8m/s2)

Forces in Elevator Problems

When an object of mass mmm is in an elevator, the following forces act on it:

Weight (mg) – always downward.
Tension (T) – upward force from the string (or Normal reaction if placed on a scale).
Net Force (ma) – depends on the elevator’s motion.

The Free Body Diagram (FBD) for the object will always have:

Downward arrow = mgmgmg
Upward arrow = TTT
Net Force = difference of the two

Different Cases

Case 1: Elevator at Rest or Moving with Constant Velocity

If the elevator is stationary or moving with constant velocity, acceleration a=0a = 0a=0. T=mgT = mgT=mg

👉 Tension equals the actual weight.

Case 2: Elevator Accelerating Upward

If the elevator moves upward with acceleration aaa: T−mg=maT – mg = maT−mg=ma T=m(g+a)T = m(g + a)T=m(g+a)

👉 Tension is greater than weight.

This explains why you feel heavier in an elevator going upward.

Case 3: Elevator Accelerating Downward

If the elevator moves downward with acceleration aaa: mg−T=mamg – T = mamg−T=ma T=m(g−a)T = m(g – a)T=m(g−a)

👉 Tension is less than weight.

That’s why you feel lighter in a descending elevator.

Case 4: Elevator in Free Fall

If the cable snaps and the elevator falls freely: a=ga = ga=g. T=m(g−g)=0T = m(g – g) = 0T=m(g−g)=0

👉 Tension is zero, i.e., the object becomes weightless.
This is the concept of apparent weightlessness.

Case 5: Elevator Moving Upward but Decelerating

If the elevator was moving upward but slowing down, the acceleration is downward.
So tension decreases: T=m(g−a)T = m(g – a)T=m(g−a)

Case 6: Elevator Moving Downward but Decelerating

If the elevator was moving downward but slowing down, the acceleration is upward.
So tension increases: T=m(g+a)T = m(g + a)T=m(g+a)

Apparent Weight

The concept of apparent weight is central here. Wapparent=T=m(g±a)W_{apparent} = T = m(g \pm a)Wapparent=T=m(g±a)

When elevator accelerates upward → apparent weight > real weight.
When elevator accelerates downward → apparent weight < real weight.
In free fall → apparent weight = 0.

👉 That’s why you feel lighter or heavier in a moving elevator.

Numerical Examples

Example 1:

A 10 kg block is hanging in an elevator. Take g=10 m/s2g = 10 \, m/s^2g=10m/s2.

Elevator at rest:

T=10×10=100 NT = 10 \times 10 = 100 \, NT=10×10=100N

Elevator moving upward with a=2 m/s2a = 2 \, m/s^2a=2m/s2:

T=10(10+2)=120 NT = 10(10 + 2) = 120 \, NT=10(10+2)=120N

Elevator moving downward with a=2 m/s2a = 2 \, m/s^2a=2m/s2:

T=10(10−2)=80 NT = 10(10 – 2) = 80 \, NT=10(10−2)=80N

Elevator in free fall:

T=0T = 0T=0

Example 2: Apparent Weight on Scale

Suppose a person of mass 70 kg stands on a weighing machine inside an elevator.

At rest:

Wapparent=70×9.8=686 NW_{apparent} = 70 \times 9.8 = 686 \, NWapparent=70×9.8=686N

Elevator accelerating upward at 2 m/s22 \, m/s^22m/s2:

Wapparent=70(9.8+2)=826 NW_{apparent} = 70 (9.8 + 2) = 826 \, NWapparent=70(9.8+2)=826N

👉 The person feels heavier.

Elevator accelerating downward at 3 m/s23 \, m/s^23m/s2:

Wapparent=70(9.8−3)=476 NW_{apparent} = 70 (9.8 – 3) = 476 \, NWapparent=70(9.8−3)=476N

👉 The person feels lighter.

In free fall:

Wapparent=0W_{apparent} = 0Wapparent=0

👉 The person experiences weightlessness.

Real-Life Applications

Elevators: The changing readings in a weighing scale inside an elevator.
Astronauts in space: They experience weightlessness similar to free fall.
Theme park rides: Roller coasters and drop towers create apparent heaviness or lightness.
Rocket launches: Astronauts feel several times their body weight because acceleration is very high.

Key Takeaways

Tension (or apparent weight) depends on the elevator’s acceleration.
Upward acceleration → tension increases.
Downward acceleration → tension decreases.
Free fall → tension becomes zero.
Apparent weight = m(g±a)m(g \pm a)m(g±a).