A $1500 - k g$ automobile has a wheel base (the distance between the axles) of $3.00 m$ . The automobile's center of mass is on the centerline at a point $1.20 m$ behind the front axle. Find the force exerted by the ground on each wheel.

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Solution

Given,
mass of automobile is $1500 k g$
Distance between the axles is $3 m$
centre of mass from from front axle is $1.2 m$
At equilibrium
$F_{T o t a l} = 0 τ_{t o t a l} = 0$
$F_{T o t a l} = F_{1} + F_{2} - m g = 0 F_{1} + F_{2} = m g$
Substitute all value in above equation
$F_{1} + F_{2} = 1500 \times 9.8 F_{1} + F_{2} = 14700$
$τ_{t o t a l} = 0 = F_{1} \times (3 - 1.2) - F_{2} \times 1.2 0 = 1.8 F_{1} - {1.2 F}_{2} F_{1} = \frac{1.2}{1.8} F_{2} F_{1} = \frac{2}{3} F_{2}$
Now,
$\frac{2}{3} F_{2} + F_{2} = 14700 F_{2} = 8820 N$
And
$F_{1} = \frac{2}{3} \times 8820 F_{1} = 5880 N$

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