Question
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?
Answers
Damon
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
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