A car weighs 1800 kg. The distance between its front and back axles is 1.8 m. Its centre of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel.
Solution and Explanation
Mass of the car, m = 1800 kg
Distance between the front and back axles, d = 1.8 m
Distance between the C.G. (centre of gravity) and the back axle = 1.05 m
The various forces acting on the car are shown in the following figure.
Rf and Rb are the forces exerted by the level ground on the front and back wheels respectively. At translational equilibrium :
Rf + Rb = mg =
1800 × 9.8
= 17640 N ...(i)
For rotational equilibrium, on taking the torque about the C.G., we have :
Rf (1.05) = Rb (1.8 - 1.05)
Rf × 1.05 = Rb × 0.75
\(\frac{R_f }{ R_b}\)b =\(\frac{ 0.75 }{ 1.05} =\frac{ 5 }{ 7}\)
\(\frac{R_b }{ R_f }\)= \(\frac{7 }{ 5}\)
\(R_b = 1.4 \,R_f...(ii)\)
Solving equations (i) and (ii), we get : 1.4 Rf + Rf = 17640
Rf =\(\frac{ 17640 }{ 2.4}\)= 7350 N
∴ Rb = 17640 - 7350 = 10290 N
Therefore, the force exerted on each front wheel =\(\frac{ 7350 }{ 2 }\)= 3675 N, and
The force exerted on each back wheel = \(\frac{10290}{ 2}\) = 5145 N
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].