Question

A car manufacturer claims that you can drive its new vehicle across a hill with a 46° slope before the vehicle starts to tip. If the vehicle is 2.1 m wide, how high is its center of gravity? Round to three decimal places and express the answer in terms of SI units.

Answer 1

To determine the height of the car's center of gravity, we can use the concept of torque.

Step-by-step explanation:

To determine the height of the car's center of gravity, we can use the concept of torque. The car will start to tip over when the torque due to its weight is equal to the torque due to the slope. The torque due to the weight can be calculated by taking the product of the weight and the perpendicular distance from the center of gravity to the slope. The torque due to the slope can be calculated by multiplying the weight of the car by the sine of the slope angle and the distance between the wheels.

Let's assume the height of the center of gravity is h. The perpendicular distance from the center of gravity to the slope is h*sin(46°). The torque due to the weight is (weight of the car)*(h*sin(46°)). The torque due to the slope is (weight of the car)*(g*sin(46°))*(width of the car)/2, where g is the acceleration due to gravity.

Setting the two torques equal to each other and solving for h, we have:
(weight of the car)*(h*sin(46°)) = (weight of the car)*(g*sin(46°))*(width of the car)/2
Cancelling out the weight of the car and sin(46°), we get:
h = (g*(width of the car))/2