Question

The 1500-kg truck reaches a speed of 50 km/h from rest in a distance of 60 m up the 10% incline with constant acceleration. Calculate the normal force under each pair of wheels and the effective coefficient of friction between the tires and the road during this motion.​

Answer 1

Given :

The mass of the truck = m = 1500 kg

The distance cover by truck = d = 60 meters

The final speed = $V_f$ = 50 km/h

                                 = 13.89 m/s

To Find :

a ) Normal force between each pairs of wheels

b ) The effective coefficient of friction between the tires and the road during this motion

Solution :

From figure

∵ Initially truck is at rest , So, $V_i$ = 0

So,   $V_f$²  -  $V_i$² = 2 a s                              ( from linear acceleration )

where d = distance cover

           a = acceleration

i.e  ( 13.89 m/s )²  - 0 = 2 × a × 60 meters

or,  192.9 = 120 a

∴           a = $\dfrac{192.9}{120}$

i.e   acceleration = a = 1.607  m/s²

Again

From figure

Ф = 5.71°

Now, From equilibrium force

Force along y-axis

 $N_1 + N_2$ = 1500 × 9.81 × cosФ

               =  1500 × 9.81 × cos 5.71°

              = 14641.9 Newton            ..........1

Force along x-axis

 F - W sinФ = ma

or,  F = 150 × 1.607 × 9.81 + 1500  × sin 5.71°

i.e  F = 3880 Newton

Again

Moment at Center of Gravity = 0

i.e  1.5 $N_1$ + 3880 × 0.6 -  1.5 $N_2$ = 0

Or,  1.5 $N_1$  -  1.5 $N_2$ = 2328

i.e      $N_1$  -   $N_2$ = 1552                  ................2

Solving eq 1 and eq 2

  (   $N_1$  -   $N_2$ ) + ( $N_1 + N_2$ ) = 14642 + 1552

Or, 2 $N_1$ = 16194

∴       $N_1$ = $\dfrac{16194}{2}$

i.e      $N_1$  = 8097 Newton

Put the value of  $N_1$ in eq 2

$N_1$  -   $N_2$ = 1552

or,  $N_2$ = 8097 - 1552 = 6545 Newton

Again

 ∵    $N_1$  $> N_2$

So,  F = $\mu _k N_1$

i.e  $\mu _k$ = $\dfrac{F}{N_1}$

Or,    $\mu _k$ = $\dfrac{3880}{8097}$

Or,    $\mu _k$ = 0.47

Hence,

a )The normal force under each pair of wheels are 8097 Newton and 6545 Newton

b) The effective coefficient of friction between the tires and the road during this motion is 0.47   Answer