A car (m = 1870 kg) is parked on a road that rises 12.1 ° above the horizontal. What are the magnitudes of (a) the normal force and (b) the static frictional force that the ground exerts on the tires?
Well, well, well. Looks like this car is taking on an uphill challenge! Let's see what's going on here, shall we?
(a) The normal force is the force that the ground exerts on an object to support its weight. In this case, since the car is on an upward slope, the normal force is gonna be a bit different than usual. To find it, you'll need to break down the weight of the car into its components. So, we have the weight of the car acting vertically downward, and we need to find its component perpendicular to the slope. Using a bit of trigonometry, we can find that the normal force is given by:
Normal Force = Weight * cos(θ)
Where θ is the angle of the incline. In this case, the weight is the mass of the car multiplied by the acceleration due to gravity (9.8 m/s²). Plug in the numbers, do the math, and voila! You'll find the magnitude of the normal force acting on the car.
(b) Ah, the static frictional force. This force comes into play when an object is at rest or is in motion with a constant velocity. It acts parallel to the surface and opposes the impending motion. The magnitude of the static frictional force can be found using the formula:
Static Frictional Force = µs * Normal Force
Here, µs is the coefficient of static friction, which varies depending on the materials in contact. It's kind of like a love/hate relationship between the tires and the road. The tires love the road enough to maintain static equilibrium, but they hate it if it tries to make them move. So, multiply the coefficient of static friction by the normal force, and you'll have the magnitude of the static frictional force.
Hope that clears things up a bit! Keep your tires on the ground and your sense of humor intact!
To find the magnitudes of the normal force and the static frictional force, we can first draw a free-body diagram for the car on the inclined road:
|\
| \
F_f | \
↓-----|--\-----
\ | \
\ | \
\ | \
\| \
↓-------\
m * g
Here, F_f represents the static frictional force, and m * g represents the weight of the car (where m is the mass of the car and g is the acceleration due to gravity).
(a) The normal force (N) exerted by the ground on the car is perpendicular to the incline. It can be calculated using the following relationship:
N = m * g * cos(θ)
where θ is the angle of inclination (12.1° in this case).
So, N = 1870 kg * 9.8 m/s^2 * cos(12.1°).
Using a calculator, the normal force N is approximately 18226 N.
(b) The static frictional force (F_f) acts parallel to the incline and prevents the car from sliding down. Its magnitude can be found using the following equation:
F_f ≤ μ_s * N
where μ_s is the coefficient of static friction between the tires and the road surface.
Since the car is parked, the static frictional force is at its maximum and is equal to μ_s * N.
Assuming a coefficient of static friction of μ_s = 0.7, we can calculate:
F_f = 0.7 * 18226 N.
Using a calculator, the static frictional force F_f is approximately 12758 N.
To find the magnitudes of the normal force and the static frictional force, we need to decompose the gravitational force acting on the car into its components perpendicular and parallel to the inclined plane.
First, we calculate the perpendicular component of the gravitational force (mg) acting on the car. The perpendicular component is given by:
F_perpendicular = mg * cos(θ)
Where:
m = mass of the car = 1870 kg
g = acceleration due to gravity = 9.8 m/s²
θ = angle of inclination = 12.1°
Substituting the given values into the equation, we get:
F_perpendicular = (1870 kg) * (9.8 m/s²) * cos(12.1°)
Now, let's calculate the parallel component of the gravitational force acting on the car. The parallel component is given by:
F_parallel = mg * sin(θ)
Substituting the values:
F_parallel = (1870 kg) * (9.8 m/s²) * sin(12.1°)
(a) The normal force (N) exerted by the ground on the tires is equal in magnitude and opposite in direction to the perpendicular component of the gravitational force. So, the normal force is:
N = F_perpendicular
(b) The static frictional force (f_s) is what prevents the car from sliding down the inclined plane. Its magnitude can vary between zero and a maximum value given by:
f_s ≤ μ_s * N
Where:
μ_s = coefficient of static friction
Since the car is parked and not sliding, the static frictional force will have its maximum value, which occurs when f_s = μ_s * N.
So, the static frictional force is:
f_s = μ_s * N
To get the actual value of the static frictional force, you need to know the coefficient of static friction (μ_s) between the tires and the road surface. Once you have that value, you can substitute it into the equation to find the static frictional force.
Please provide the coefficient of static friction to complete the calculations.
M*g = 1870 * 9.8 = 18,326 N.=Wt. of car.
Fp = 18,326*sin12.1 = 3841 N.
a. Fn = 18,326*Cos12.1=17,919 N.= Normal.
b. Fp-Fs = M*a.
3841-Fs = 1870*0 = 0.
Fs = 3841 N. = Force of static friction.