The answer & explanation for this question is given in the attachment below.
The sports car has a weight of 4900 lblb and center of gravity at GG. If it starts from rest it causes the rear wheels to slip a
s it accelerates. The coefficients of static and kinetic friction at the road are μsμs = 0.5 and μkμkmu_k = 0.35, respectively. Neglect the mass of the wheelDetermine how long it takes for it to reach a speed of 10 ft/s.What are the normal reactions at each of the rear wheels on the road?What are the normal reactions at each of the front wheels on the road?
1 answer:
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Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
