Explanation:
1. Circumference of the handle (where the effort is applied) is pi times diameter.
C = πd
C = 1.5π inches
C ≈ 4.71 inches
2. Mechanical advantage is the ratio of distance in over distance out.
MA = din / dout
MA = 1.5 / 0.25
MA = 6
3. Mechanical advantage is the ratio of force out over force in.
MA = Fout / Fin
6 = Fout / 10 lb
Fout = 60 lb
The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².
<h3>What is normal stress?</h3>
If the direction of deformation force is perpendicular to the cross-sectional area of the body, the stress is called normal stress. Changes in wire length and body volume will be normal.
σ = P/A
Where, σ = Normal stress
P = Pressure
A = Area
1 Kg = 9.81 N
800 kg = 7848 N
Since the rod is half bronze and half steel
800 kg = 7848/2
= 3924 N
Pₙ = Fₙ = 3924 N [n = Bronze]
Pₓ = 3924 N [x = steel]
Given,
σₙ = 90MPa
σₓ = 120MPa
Aₙ = ?
Aₓ = ?
Aₙ = Pₙ/σₙ
Aₙ = 3924/90
Aₙ = 43.6 mm²
Aₓ = Pₓ/σₓ
Aₓ = 3924/120
Aₓ = 32.7 mm²
To know more about normal stress, visit:
brainly.com/question/28012990
#SPJ9
Answer:
Determine the added thrust required during water scooping, as a function of aircraft speed, for a reasonable range of speeds.= 132.26∪
Explanation:
check attached files for explanation
Answer is: $637.28; just did the math but i really don’t want to type it all out.
