13.4. Young's modulus for iron is 1.9 x 10' Pa. When an iron wire 1.0 m long with a cross-sectional area of 4.0 mm supports a 100 kg load, the wire stretches by
(

0.0129 m

Explanation:

ΔL = FL / (EA)

where ΔL is the deflection,

F is the force,

L is the initial length,

E is Young's modulus,

and A is the cross sectional area.

F = mg = 100 kg × 9.8 m/s² = 9800 N

A = 4.0 mm² × (1 m / 1000 mm)² = 4×10⁻⁶ m²

ΔL = (9800 N) (1.0 m) / ((1.9×10¹¹ Pa) (4×10⁻⁶ m²))

ΔL = 0.0129 m

given data:

determine the horse power (h.p) = ?

force applied to move the box(f) = 10 pounds,

box lifted from a distance         (d) = 3 feet

time taken to lift                         (t) = 10 sec,

the horse power is defined as amount of work required to move a load. the horse power is equal to 550 foot-pound per second

                  1 h.p   = 550 foot-pound/second

it can also be defined as the the load that need to be move a body for certain distance in a certain amount of time.

mathematically,

                      h.p = (force × distance)/ time

                            = (10 × 3)/ 10

                            = 30/10

                            = 3 foot -pound/sec

hence, the horse power is needed to move the box is 3 foot-pound/second

important note: 1 h.p = 550 foot-pound/second = 746 watts