Answer:
SKID
Explanation:
In general, airplane tracks are flat, they do not have cant, consequently the friction force is what keeps the bicycle in the circle.
Let's use Newton's second law, let's set a reference frame with the horizontal x-axis and the vertical y-axis.
Y axis y
N- W = 0
N = W
X axis (radial)
fr = m a
the acceleration in the curve is centripetal
a = 
the friction force has the expression
fr = μ N
we substitute
μ mg = m v²/r
v = 
we calculate
v = 
v = 1,715 m / s
to compare with the cyclist's speed let's reduce to the SI system
v₀ = 18 km / h (1000 m / 1 km) (1 h / 3600 s) = 5 m / s
We can see that the speed that the cyclist is carrying is greater than the speed that the curve can take, therefore the cyclist will SKID
<span>B) 0.6 N
I suspect you have a minor error in your question. Claiming a coefficient of static friction of 0.30N is nonsensical. Putting the Newton there is incorrect. The figure of 0.25 for the coefficient of kinetic friction looks OK. So with that correction in mind, let's solve the problem.
The coefficient of static friction is the multiplier to apply to the normal force in order to start the object moving. And the coefficient of kinetic friction (which is usually smaller than the coefficient of static friction) is the multiplied to the normal force in order to keep the object moving. You've been given a normal force of 2N, so you need to multiply the coefficient of static friction by that in order to get the amount of force it takes to start the shoe moving. So:
0.30 * 2N = 0.6N
And if you look at your options, you'll see that option "B" matches exactly.</span>
Step 1 : Get your supply list together
Step 2 : Pick what model you want to do
Step 3 : Ask for a partner
Step 4 : Complete the model and take your time.
Step 5 : Read the directions carefully
Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa
<u>Explanation:</u>
Given -
Stress Direction, A = [1 0 0 ]
Slip plane = [ 1 1 1]
Normal to slip plane, B = [ 1 1 1 ]
Critical stress, Sc = 2.92 MPa
Let the direction of slip on = [ 1 1 0 ]
Let Ф be the angle between A and B
cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3
cos Ф = 1/√3
σ = Sc / cosФ cosλ
For slip along [ 1 1 0 ]
cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1
cos λ = 1/√2
Therefore,
σ = 2.92 / 1/√3 1/√2
σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa
Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa
