Answer:
P=740 KPa
Δ=7.4 mm
Explanation:
Given that
Diameter of plunger,d=30 mm
Diameter of sleeve ,D=32 mm
Length .L=50 mm
E= 5 MPa
n=0.45
As we know that
Lateral strain



We know that




So the axial pressure


P=740 KPa
The movement in the sleeve


Δ=7.4 mm
Answer:
F = 0.535 N
Explanation:
Let's use the concepts of energy, at the highest and lowest point of the trajectory
Higher
Em₀ = U = mg y
Lower
= K = ½ m v²
Emo =
mg y = ½ m v2
v = √ 2gy
y = L - L cos θ
v = √ (2g L (1-cos θ))
Now let's use Newton's second law n at the lowest point where the acceleration is centripetal
F = ma
a = v² / r
In turning radius is the cable length r = L
F = m 2g (1-cos θ)
Let's calculate
F = 2 1.25 9.8 (1 - cos 12)
F = 0.535 N
<span>65W * 8h * 3600s/h = 1.9e6 J = 447 Cal </span>