String Tension Equation- Physics Problem Solutions

What Is String Tension and Why You Need the Right Equation

String tension is the pulling force transmitted through a string, rope, cable, or any flexible connector. It's measured in Newtons (N). Every time you see a hanging weight, a pulley system, or a rope pulling something, string tension is doing the work.

The basic string tension formula depends on what you're solving for. Most problems fall into three categories:

• Static equilibrium — object isn't moving
• Accelerating systems — objects are speeding up or slowing down
• Conical pendulums and circular motion — tension has angular components

Most students mess up because they apply the wrong equation to the wrong situation. Let's fix that.

The Core String Tension Equation

For a mass hanging from a string at rest, the tension equals the weight:

T = mg

Where:

• T = tension in Newtons
• m = mass in kilograms
• g = gravitational acceleration (9.8 m/s² on Earth)

That's the simple case. Things get messy when the mass accelerates.

Problem Type 1: Accelerating Vertical Motion

When a mass hangs from a string and moves up or down with acceleration, use Newton's second law combined with tension.

For upward acceleration:

T − mg = ma

T = m(g + a)

For downward acceleration:

mg − T = ma

T = m(g − a)

Example: Upward Accelerating Elevator

A 50 kg person stands in an elevator accelerating upward at 3 m/s². Find the tension in the cable.

Solution:

T = m(g + a)

T = 50(9.8 + 3)

T = 50 × 12.8

T = 640 N

The cable feels a heavier load than just the person's weight because the acceleration adds extra force.

Example: Downward Accelerating Elevator

Same 50 kg person, but now the elevator accelerates downward at 3 m/s².

Solution:

T = m(g − a)

T = 50(9.8 − 3)

T = 50 × 6.8

T = 340 N

The tension drops because gravity and acceleration fight each other.

Problem Type 2: Angled Strings (Tension with Components)

When strings pull at angles, you must break tension into horizontal and vertical components. This is where most students give up.

For a string at angle θ from vertical:

• Vertical component: T_y = T cos(θ)
• Horizontal component: T_x = T sin(θ)

Example: Suspended Sign

A 12 kg sign hangs from two strings at 45° angles from the ceiling. Find the tension in each string.

Solution:

First, total weight = mg = 12 × 9.8 = 117.6 N downward

Two strings share the load. Since the setup is symmetric:

2(T cos 45°) = 117.6

2(T × 0.707) = 117.6

1.414 T = 117.6

T = 83.2 N in each string

Problem Type 3: Pulley Systems

Pulleys change direction but don't reduce force (unless it's a movable pulley). The tension stays the same throughout an ideal rope.

Ideal Pulley (massless, frictionless)

For a massless pulley with a 10 kg mass on one side and a 6 kg mass on the other:

System acceleration:

a = g(m₁ − m₂) / (m₁ + m₂)

a = 9.8(10 − 6) / (10 + 6)

a = 9.8 × 4 / 16

a = 2.45 m/s²

Tension in the rope:

T = 2m₁m₂g / (m₁ + m₂)

T = 2(10)(6)(9.8) / 16

T = 73.5 N

The heavier mass accelerates down. The lighter mass accelerates up.

Problem Type 4: Conical Pendulum

A conical pendulum has a mass swinging in a horizontal circle while attached to a string. The string makes an angle θ with vertical.

Horizontal component of tension provides centripetal force:

T sin(θ) = m × v² / r

Vertical component balances weight:

T cos(θ) = mg

Combining these:

tan(θ) = v² / (rg)

Example: Conical Pendulum

A 2 kg mass swings on a 1.5 m string, making a 30° angle with vertical. Find the tension and speed.

Tension:

T = mg / cos(θ)

T = (2 × 9.8) / cos(30°)

T = 19.6 / 0.866

T = 22.6 N

Radius:

r = L sin(θ) = 1.5 × sin(30°) = 1.5 × 0.5 = 0.75 m

Speed:

v = √(rg tan(θ))

v = √(0.75 × 9.8 × 0.577)

v = 2.06 m/s

Quick Reference: String Tension Equations

How to Solve Any String Tension Problem

Follow these steps. Every time. No exceptions.

Step 1: Draw a Free Body Diagram

Sketch the mass and all forces acting on it. Label tension as an arrow pulling away from the mass.

Step 2: Identify Force Directions

Weight always points down (mg). Tension always pulls away from the mass along the string.

Step 3: Choose Your Coordinate System

Align one axis with motion direction. For angled problems, break tension into components along these axes.

Step 4: Apply Newton's Second Law

ΣF = ma for each axis separately.

Step 5: Solve for Tension

Plug in known values. Isolate T. Calculate.

Common Mistakes That Kill Your Answers

• Forgetting acceleration direction — tension is larger when acceleration is upward, smaller when downward
• Using the wrong angle — cos(θ) is for vertical components when θ is from vertical; sin(θ) is from horizontal
• Assuming pulleys reduce force — they only change direction, not magnitude (in ideal systems)
• Mixing up mass and weight — convert mass to weight by multiplying by g
• Ignoring multiple strings — if two strings hold a mass, tension is split unless stated otherwise

When Tension Equals Zero

In free fall, a string has zero tension. The mass and string fall together with acceleration g. The string isn't pulling on anything because everything accelerates at the same rate.

This matters in elevator problems: if the cable snaps, the elevator free-falls and the normal force (not tension) becomes zero. The tension in any connecting rope would also be zero.

Real-World Application: Crane Lifting

A crane lifts a 500 kg steel beam at constant velocity. Then it accelerates upward at 1.5 m/s².

Constant velocity (equilibrium):

T = mg = 500 × 9.8 = 4,900 N

Accelerating upward:

T = m(g + a) = 500(9.8 + 1.5) = 500 × 11.3 = 5,650 N

The crane cable must be rated for at least 5,650 N to handle the acceleration phase safely.

Final Point

String tension problems follow the same logic every time: identify forces, apply Newton's laws, solve for what you need. The equations change based on geometry and motion, but the method stays fixed. Practice the free body diagram. Get that right, and the math takes care of itself.