Question

A vehicle of mass 700 kg accelerates uniformly from rest to a velocity of 60 kmh−1 in 10 s whilst ascending a 150 gradient. The frictional resistance to motion is 0.5 kN. Making use of D’Alembert’s principle, determine:

i) the tractive effort between the wheels and the road surface

ii) the work done in ascending the slope

iii) the average power developed by the engine.

Answer 1

The picture shows forces acting on the vehicle.

There are: the gravitational force $\vec Q=m\vec g,$ the reaction of road’s surface $\vec R$ and frictional force $\vec F,$ working against the vehicle’s velocity \vec \text{v}.

The vehicle is moved uphill by the tractive force $\vec T$ which does the real work.

The force $\vec Q$ can be split into 2 compounds: $\vec Q_p,$ perpendicular to the road surface and $\vec Q_t$ parallel to the road.

Write the Newton’s second law of dynamics for the vehicle as follows:

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m\vec a=\vec T+\vec Q_t+\vec F \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$

The angle between $\vec Q$ and $\vec Q_p$ equals $\alpha$ (the same as the slope of the road). Therefore:

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m a=T-mg\sin \alpha-F \ \ \ \ \ \ \ \ \ \ (2)$$

The acceleration $a$ may be calculated as $a=\text{v}/t,$ where $\text{v}$ is the given velocity, and $t$ is the acceleration time .

$$\ \ \ \ \ \ \ \ \ \ \ \ \ T=m(\dfrac{\text{v}}{t})+mg\sin \alpha+F \ \ \ \ \ \ \ \ \ \ (3)$$

$m=700kg$

$g=9.81m/s^2$

$\text{v}=60km/h=\dfrac{50}{3}m/s$

$t=10s$

$F=0.5kN=500N$

$\tan\alpha=0.15$

$\sin\alpha\approx0.14834$

$$T=700kg(\dfrac{\dfrac{50}{3}m/s}{10s})+700kg(9.81m/s^2)(0.14834)$$

$$+500N =2685\ N=2.685\ kN$$

The work $W$ done in ascending the slope equals $T\times s,$ where $s$ is the slope length. 

$$s=\dfrac{at^2}{2}=\dfrac{(\dfrac{v}{t})t^2}{2}=\dfrac{vt}{2}$$

In a uniformly accelerating movement the distance $s$ can be calculated by multiplying time by the average velocity.

The amount of work $W$ equals:

$$W=Ts=T(\dfrac{vt}{2})$$

$$W=2685\ N(\dfrac{50}{3}m/s)(\dfrac{10s}{2})$$

$$=223776\ J=223.776\ kJ$$

The average power $P$ equals $W$ divided by time $t,$ therefore:

$$P=\dfrac{W}{t}$$

$$P=\dfrac{223776\ J}{10s}=22377.6\ W\approx22.4\ kW$$