Answer:
The answer is μ=1.2
Explanation:
Step one
F=μN
F =friction force
μ =coefficient of friction
N=normal force
Step two
Given
F=12,000N
N= weight of car 10,000N
Step three
Substituting our values in the formula we have
12,000=μ*10,000
μ=12000/10000
μ=1.2
coefficient of friction. A measure of the amount of resistance that a surface exerts on or substances moving over it, equal to the ratio between the maximal frictional force that the surface exerts and the force pushing the object toward the surface.
Answer:
minimum coefficient of static friction, between the tires and road surface = 1.2
Explanation:
Analysis of this problem ;
Friction keeps the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. Now, we know that the maximum static friction (at which the tires roll but do not slip) is μsN, where μs is the static coefficient of friction and N is the normal force. Thus, the normal force equals the car’s weight on level ground, so that N=mg.
Thus the frictional force in this situation is given by;
Ff = μsN = μsmg
Now, from the question, weight = 10,000N while frictional force(Ff) = 12,000N
So, mg = weight;thus;
12000 = μs(10,000)
μs = 12000/10000 = 1.2
