Dynamic Tension and Stress
Dynamic tensile stress equals the internal tension obtained from Newton's laws divided by cross-sectional area, with centripetal and tangential contributions evaluated at the stated position.
Why this shows up in the exam
Rotating masses on wires · Pulley-connected blocks · Vertical-circle safety calculations
Learn the idea
In accelerating or circular systems, determine tension dynamically before dividing by area. A wire does not know the attached mass; it responds to the actual instantaneous tension, which may include acceleration or centripetal-force requirements.
🧠 Memory hook: Dynamics first, stress second.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- sigma = T/A — stress after the actual internal tension is known
- T - mg cos(theta) = mv²/r — representative radial balance with sign and angle defined for the chosen geometry
- T = mu m g for Atwood-type acceleration only after solving dynamics — tension must follow the system equations rather than defaulting to mg
How to approach it
- 1Draw a free-body diagram at the stated instant
- 2Solve Newton's law for the internal tension
- 3Divide by area and compare with the allowed stress
Common slip-ups that cost marks
- •Setting tension equal to mg in an accelerating system
- •Using speed without its location in a vertical circle
- •Equating breaking stress directly to force
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.
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