Answer:
a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,
b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm
Explanation:
Given that:
γ= 9.5 kN/m³ = 9500N/m3
b = 6 inches = 0.1524 m
t = 0.0013 mm
d = 2 inches = 0.0508 m
n = 1750 rpm
L = 9 ft = 2.7432 m
Ks = 1.25
g = 9.81 m/s²
a)
b)
dip =
Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
= U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
= K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)