Question

EXAMPLE [6] Hooks' law After a fall, a 95 kg rock climber finds himself dangling from the end of a rope that had been 15 m long and 9.6 mm in diameter but has stretched by 2.8 cm. For the rope, calculate. (i) the strain, (ii) the stress and (ii) the modulus of elasticity.​

Answer 1

Answer:

Explanation:

If  L(=1500cm) is the unstretched length of the rope and ΔL=28.cm  is the amount it stretches, then the strain is :

                              ΔL/L=(28cm)(1500cm)=1.9×10−3

(b) The stress is given by F/A where F is the stretching force applied to one end of the rope and A is the cross-sectional area of the rope. Here F is the force of gravity on the rock climber. If m is the mass of the rock climber then F=mg. If r is the radius of the rope then  A=πr2. Thus the stress is:

                         FA​ = mg / πr2​

                              = (95kg)(9.8m/s2) / π(4.8×103m)2​

                              = 1.3 × 107 N/m2

(c) Young’s modulus is the stress divided by the strain:

                                             E = (1.3 \times 10^7N/m^2

) / (1.9 \times 10^{–3}) = 6.9 \times 10^9 N/m^2

Hope this will help you