A 984 kg car is driving along a boardwalk. The car's center of mass is 33 cm in front of the rear wheels, and the distance between the front and rear wheels is 2 m. What total force does the car's front tires exert on the boardwalk? Give your answer in Newtons.
Set the total moment about a line through the back tires' ground contact points equal to zero.
984*g*0.33 - F*2.0 = 0
Solve for F
To find the total force exerted by the car's front tires on the boardwalk, we need to consider the center of mass and the weight distribution of the car.
First, let's find the weight of the car. We can use the formula:
Weight = mass × gravitational acceleration
The mass of the car is given as 984 kg. The gravitational acceleration can be taken as approximately 9.8 m/s^2. Therefore, the weight of the car is:
Weight = 984 kg × 9.8 m/s^2 = 9643.2 N
Now, we need to consider the weight distribution of the car. The car's center of mass is located 33 cm (or 0.33 m) in front of the rear wheels. Since the car's weight acts downward, it can be assumed to act at its center of mass.
Next, we need to determine the weight distribution between the front and rear wheels. The distance between the front and rear wheels is given as 2 m.
The proportion of weight on the front tires can be calculated using the lever rule:
Weight on front tires = (distance from center of mass to front wheels / wheelbase) × total weight
Distance from center of mass to front wheels = 2 m - 0.33 m = 1.67 m
Wheelbase = 2 m
Weight on front tires = (1.67 m / 2 m) × 9643.2 N = 8372.27 N
Therefore, the total force exerted by the car's front tires on the boardwalk is 8372.27 Newtons.