An automobile with a mass of 1310 kg has 3.15 m between thefront and rear axles.

Question

An automobile with a mass of 1310 kg has 3.15 m between thefront and rear axles. Its center of gravity is located 1.78 mbehind the front axle. The automobile is on levelground. (a) Determine the magnitude of the force from theground on each front wheel (assuming equal forces on the frontwheels). (b) Determine the magnitude of the force from theground on each rear wheel (assuming equal forces on the rearwheels). I dont know where to go with this. An automobile with a mass of 1310 kg has 3.15 m between thefront and rear axles. Its center of gravity is located 1.78 mbehind the front axle. The automobile is on levelground. (a) Determine the magnitude of the force from theground on each front wheel (assuming equal forces on the frontwheels). (b) Determine the magnitude of the force from theground on each rear wheel (assuming equal forces on the rearwheels). I dont know where to go with this.

Explanation / Answer

The car's mass isM = 1310 kg   The horizontal distancebetween the car's axles is L = 3.15 m   The horizontal distancegoing from the axle in the back all the way to the center ofmass l =(3.151.78) m      =1.37m   F1 isthe force exerted on each one of the front wheels F2 is the force exerted on each back to thewheel.

a) The torques about the rear axle, andcalculate
                 = FL
                F1 = (Mgl)/2L                     = (1310 kg)(9.80 m/s2)(1.37 m)/2(3.15m)                     = 2791.75 N b) Equilibrium of forces leads to thisequation,               2F1 + 2F2 =Mg,
                    F2 = Mg-2F1 /2
     and weget F2 =3628 N

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