Young's Modulus and Material Strength Calculations

44. (a) The Young’s modulus is given by

(b) Since the linear range of the curve extends to about 2.9 × 10⁸ N/m², this is approximately the yield strength for the material.

46. Since the force is (stress × area) and the displacement is (strain × length), we can write the work integral (eq. 7-32) as

  W =

  = A (differential strain)L  = AL (differential strain)

which means the work is  (wire-area) × (wire-length) × (graph-area-under-curve).  Since the area of a triangle (see the graph in the problem statement) is  (base)(height)  then we determine the work done to be

            W = (2.00 x 10^-6 m²)(0.800 m)(1.0 × 10^-3)(7.0 × 10⁷ N/m²) = 0.0560 J .

48. 46. Since the force is (stress × area) and the displacement is (strain × length), we can write the work integral (eq. 7-32) as

  W =

  = A (differential strain)L  = AL (differential strain)

which means the work is  (thread cross-sectional area) × (thread length) × (graph-area-under-curve). The area under the curve is 

(a) The kinetic energy that would put the thread on the verge of breaking is simply equal to W:

(b) The kinetic energy of the fruit fly of mass 6.00 mg and speed 1.70 m/s is

(c) Since

, the fruit fly will not be able to break the thread.

(d) The kinetic energy of a bumble bee of mass 0.388 g and speed 0.420 m/s is

(e) On the other hand, since

, the bumble bee will be able to break the thread.

49. The flat roof (as seen from the air) has area A = 150 m × 5.8 m = 870 m². The volume of material directly above the tunnel (which is at depth d = 60 m) is therefore

V = A × d = (870 m²) × (60m) = 52200 m³.

Since the density is r = 2.8 g/cm³ = 2800 kg/m³, we find the mass of material supported by the steel columns to be m = rV = 1.46 × 10⁸ m³.

(a) The weight of the material supported by the columns is mg = 1.4 × 10⁹ N.

(b) The number of columns needed is

50. On the verge of breaking, the length of the thread is

,

where 

is the original length, and

, as given in the problem. The free-body diagram of the system is shown on the right. The condition for equilibrium is

where m is the mass of the insect and 

. Since the volume of the thread remains constant is it is being stretched, we have

, or

. The vertical distance

 is

.

Thus, the mass of the insect is

or 0.421 g.