A 1300 kg El Camino is parked on a 20 degree hill.(20 degrees above x axis)The driver of the El Camino returns, starts the car, and accelerates up the hill at 2 m/s^2. Find Sigma-Fx and Sigma-Fy. What is the normal Force acting on the car, and what is the force being supplied by the engine pushing the car up the hill?
You do not say what directions are x and y
gravity force down = m g = 1300 * 9.81
component of gravity force normal to road = m g cos 20
that is balanced by force up from road normal to road, no acceleration normal to road
Component of gravity force down slope
= m g sin 20
force up slope from engine (actually from road friction on tires)
= F
total force up road parallel to road
= F - m g sin 20
force = mass * acceleration
F - m g sin 20 = m a = 1300* 2 = 2600
so
F = 1300*9.81 *sin 20 + 2600
To find the values of Sigma-Fx and Sigma-Fy, we need to analyze the forces acting on the El Camino. Here's how you can calculate them step by step:
1. Resolve the force of gravity into its components:
Gravity can be split into two components: one parallel to the incline (downhill), and the other perpendicular to the incline (normal force). The parallel component is given by:
F_parallel = m * g * sin(theta)
where m is the mass of the El Camino (1300 kg), g is the acceleration due to gravity (9.8 m/s^2), and theta is the angle of the incline (20 degrees).
2. Calculate the normal force:
The normal force (F_normal) is the force exerted by the surface on the car perpendicular to the incline. It can be found by:
F_normal = m * g * cos(theta)
3. Determine the force being supplied by the engine:
The force provided by the engine (F_engine) is the net force acting in the direction of acceleration. Since the car is moving up the hill, F_engine is equal to:
F_engine = m * a + F_parallel
where a is the acceleration of the car (2 m/s^2) and F_parallel is the parallel component of the force of gravity.
4. Find Sigma-Fx and Sigma-Fy:
Sigma-Fx is the net force acting in the x-direction, parallel to the incline. In this case, the only force acting in the x-direction is the force of gravity's parallel component:
Sigma-Fx = F_parallel
Sigma-Fy is the net force acting in the y-direction, perpendicular to the incline. In this case, the forces acting in the y-direction are the normal force and the perpendicular component of gravity:
Sigma-Fy = F_normal - m * g * cos(theta)
5. Calculate the normal force and force supplied by the engine:
Using the values obtained in steps 1-4, substitute them into the respective formulas to find the desired values.
So, the normal force acting on the car is F_normal = m * g * cos(theta), and the force being supplied by the engine pushing the car up the hill is F_engine = m * a + F_parallel.