a car of mass 1200 kg is towed up a ghat having an angle of inclination of 30 . if the coefficient of rolling friction between the tyres and the road is 0.004 ,find the force required to tow up the car at uniform speed ? pls experts answer it fast its urgent ....

Dear Student,

Please find below the solution to the asked query:

Given information is,
Mass of the car $M = 1200 kg$; Angle of inclination $\theta = {30}^0$; coefficient of rolling friction ${\mu }_r = 0.004$.

The situation is shown in the below diagram.


So, in order to pull it up with constant velocity, we have to apply a force F as,
$$\begin{gathered} F = Mg \sin \theta + F_{fri} = Mg \sin \theta + \mu Mg \cos \theta = \frac{Mg}{2}\left(1+\sqrt{3}\right) \\ \Rightarrow F = \frac{1200\times 10\times \left(2.732\right)}{2} = 6000\times 2.732 \\ \Rightarrow F= 16.39 kN \end{gathered}$$



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