Question

Braking Load A Chevrolet Silverado with a gross weight of 4500 pounds is parked on a street with a \(10^{\circ}\) grade. Find the magnitude of the force required to keep the Silverado from rolling down the hill. What is the magnitude of the force perpendicular to the hill?

Short answer

The force required to keep the Silverado from rolling down is approximately 781 pounds, while the force perpendicular to the hill is approximately 4432 pounds.

Step-by-step solution

- Identify the forces involved

List all the forces acting on the Chevrolet Silverado. The main forces are the gravitational force, the force parallel to the hill (parallel component of gravity), and the force perpendicular to the hill (normal component of gravity).

- Determine gravitational force

The gravitational force acting on the Silverado is equivalent to its weight, which is given as 4500 pounds.

- Calculate the parallel component of the gravitational force

Use the formula for the parallel component: \[ F_{\text{parallel}} = F_{\text{gravity}} \times \text{sin}(\theta) \]Substitute the given values: \[ F_{\text{parallel}} = 4500 \times \text{sin}(10^\text{°}) \]Use a calculator to find \[ F_{\text{parallel}} = 4500 \times 0.1736 \]Which gives: \[ F_{\text{parallel}} \approx 781 \text{ pounds} \]

- Calculate the perpendicular component of the gravitational force

Use the formula for the perpendicular component: \[ F_{\text{perpendicular}} = F_{\text{gravity}} \times \text{cos}(\theta) \]Substitute the given values: \[ F_{\text{perpendicular}} = 4500 \times \text{cos}(10^\text{°}) \]Use a calculator to find \[ F_{\text{perpendicular}} = 4500 \times 0.9848 \]Which gives: \[ F_{\text{perpendicular}} \approx 4432 \text{ pounds} \]

Key concepts

gravitational force

Gravitational force is one of the fundamental forces in physics. It is the force that attracts any object with mass toward the center of the Earth.
Essentially, gravitational force is what gives objects weight.
In our example with the Chevrolet Silverado, its weight of 4500 pounds is due to gravitational force.
Here we need to use a simple equation to find this:
\( F_{gravity} = mass \times gravity \.\)
For simplicity, the weight is already given directly here and is equivalent to gravitational force.
The force that keeps our truck firmly grounded is this gravitational pull acting straight downward.

force components

Force components break down a complex force into simpler parts. When a vehicle is on an incline, gravitational force is split into two distinct components:
- Parallel component: acts along the slope of the hill and causes the vehicle to potentially slide down.
- Perpendicular component: acts perpendicular to the slope and presses the vehicle against the hill.
The mathematical expressions for these components depend on trigonometric functions like sine and cosine.
For example, the parallel component is found by multiplying the gravitational force by the sine of the angle of the incline:
\[ F_{\text{parallel}} = F_{\text{gravity}} \times \text{sin}(\theta) \]
Meanwhile, the perpendicular component is identified using the cosine of the angle:
\[ F_{\text{perpendicular}} = F_{\text{gravity}} \times \text{cos}(\theta) \]
In this exercise, we calculated the parallel component as approximately 781 pounds and the perpendicular component as 4432 pounds.
This breakdown helps us understand how the forces play out and keep the vehicle in place.

trigonometric functions in physics

Trigonometric functions like sine, cosine, and tangent are key tools in physics. They help resolve forces, velocities, and even projectile motions into components.
Sine (sin) and cosine (cos) are particularly useful when dealing with forces on an incline:
- The sine function relates to the opposite side over hypotenuse in a right-angled triangle.
- The cosine function relates to the adjacent side over hypotenuse.
In our problem, the driveway angle is given as \(10^{\circ}\). We use sine and cosine to find force components.
With the equation\[ F_{\text{parallel}} = 4500 \times \text{sin}(10^{°}) \]
we get the parallel force, and with\[ F_{\text{perpendicular}} = 4500 \times \text{cos}(10^{°}) \]
we get the perpendicular force.
These functions help in calculating precise magnitudes of both the force pulling the truck down the hill and the force pressing it against the surface.
Understanding trigonometric functions is vital for solving numerous physics problems involving angles and forces.