Friction Forces and max braking force

[DeadFishFactory]

1. The coefficient of static friction between the tires of a car and a dry road is .62. The mass of the car is 1500 kg. What is the maximum braking force is obtainable:
a. horizontal to the road.
b. on an 8.6 degree downgrade?



2. F=mA f(s)=M(s)N N=mg F=f(s)-mgsin(x)



3.
a. f(s)=.62(9.8)(1500)sin90=9114 N
b. f(s)=.62(9.8)(1500)sin8.6=2198 N

My problem is that "b" isn't correct, I checked the back of the book I was using and they got 9000 N. Can anyone show me how to do this problem? I know I suck at physics. Also, if possible, can anyone explain to me how to find the tension of a string? Thanks!

 

[Dick]

If horizontal is 90 degrees, then 8.6 degrees downgrade must be 81.4 degrees. Most people would probably call horizontal 0 degrees and use a cos instead of sin. The tension on a string is "how much it pulls", sorry but that question is really vague.

 

[m2003]

First, let's clarify the problem. The maximum braking force refers to the maximum force that can be applied by the brakes to stop the car. In this case, we are assuming that the brakes are applied at the tires and the car is not skidding on the road.

To find the maximum braking force, we need to use the formula F=μN, where μ is the coefficient of static friction and N is the normal force. The normal force is the force exerted by the road on the tires and is equal to the weight of the car, which is given by mg.

a. For a horizontal road, the normal force is equal to the weight of the car, so N=mg=1500 kg x 9.8 m/s^2 = 14700 N. Therefore, the maximum braking force is F=μN=0.62 x 14700 N = 9114 N.

b. For an 8.6 degree downgrade, the normal force is equal to the component of the weight of the car perpendicular to the road, which is given by mgcosθ, where θ is the angle of the incline. In this case, θ=8.6 degrees, so the normal force is N=mgcos8.6=1500 kg x 9.8 m/s^2 x cos8.6 = 14796 N. Therefore, the maximum braking force is F=μN=0.62 x 14796 N = 9171 N.

The reason why the answer in the book is different is because they may have used a different value for the coefficient of static friction or rounded their answers differently.

To find the tension of a string, we need to use the formula T=mg+ma, where T is the tension, m is the mass of the object, g is the acceleration due to gravity, and a is the acceleration of the object. This formula is based on Newton's Second Law, which states that the net force on an object is equal to its mass times its acceleration.

In order to use this formula, we need to know the mass of the object and the acceleration of the object. The acceleration of the object can be calculated using the formula a=F/m, where F is the net force acting on the object and m is the mass of the object.

For example, if we have a mass of 5 kg hanging from a string and the string is accelerating upwards at