A 1030 kg car travelling on a road that runs straight up a hill reaches the rounded crest at 10.4 m/s. If the hill at that point has a radius of curvature (in a vertical plane) of 59.0 m, what is the effective weight of the car at the instant it is horizontal at the very peak?
To find the effective weight of the car at the instant it is horizontal at the peak, we need to consider both the gravitational force and the centripetal force acting on the car.
The centripetal force is provided by the vertical component of the normal force, which can be calculated using the radius of curvature and the car's velocity. The centripetal force is given by:
Fc = m * v^2 / r
Where:
Fc is the centripetal force
m is the mass of the car (1030 kg)
v is the velocity of the car (10.4 m/s)
r is the radius of curvature (59.0 m)
Let's calculate the centripetal force:
Fc = (1030 kg) * (10.4 m/s)^2 / (59.0 m)
Fc ≈ 1835.56 N
Now, the effective weight of the car at the peak is the difference between the gravitational force (mg) and the centripetal force (Fc). The gravitational force can be calculated as:
Fg = m * g
Where:
Fg is the gravitational force
m is the mass of the car (1030 kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Let's calculate the gravitational force:
Fg = (1030 kg) * (9.8 m/s^2)
Fg = 10094 N
Finally, we can calculate the effective weight of the car at the peak:
Effective Weight = Fg - Fc
Effective Weight = 10094 N - 1835.56 N
Effective Weight ≈ 8258.44 N
Therefore, the effective weight of the car at the instant it is horizontal at the very peak is approximately 8258.44 N.