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
Find the value of m if \(M = 10\) \(kg\). All the surfaces are rough.
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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.
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
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].