A banked circular highway is designed for traffic moving
1 answer:
Answer:
0.063
Explanation:
velocity of the car, v = 40 km/h = 11.11 m/s
radius, r = 200 m
Let the coefficient of friction is μ.
The coefficient of friction relates to the velocity on banked road is given by
where, v is the velocity, r be the radius of the curve road and μ is coefficient of friction.
By substituting the values, we get
μ = 0.063
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The value of normal force as the slider passes point B is
- 6 mg
The value of h when the normal force is zero
- 3R/2
The normal force is calculated using the work energy principle which is applied as below
K₁ + U₁ = K₂
k represents kinetic energy
U represents potential energy
the subscripts 1,2 , and 3 = a, b, and c
for 1 to 2
K₁ + W₁ = K₂
0 + mg(h + R) = 0.5mv²₂
g(h + R) = 0.5v²₂
v²₂ = 2g(1.5R + R)
v²₂ = 2g(2.5R)
v²₂ = 5gR
Using summation of forces at B
Normal force, N = ma + mg
N = m(a + g)
N = m(v²₂/R + g)
N = m(5gR/R + g)
N = 6mg
for 1 to 3
K₁ + W₁ = K₃ + W₃
0 + mgh = 0.5mv²₃ + mgR
gh = 0.5v²₃ + gR
0.5v²₃ = gh - gR
v²₃ = 2g(h - R)
at C
for normal force to be zero
ma = mg
v²₃/R = g
v²₃ = gR
and v²₃ = 2g(h - R)
gR = 2gh - 2gR
gR + 2gR = 2gh
3gR = 2gh
3R/2 = h
Learn more about normal force at:
#SPJ1
Answer:
Magnitude of y-component=7 units
Direction= angle of vector A with x-axis=
Explanation:
Explained solution is in the picture attached
