# Math Help - truck problem

1. ## truck problem

A heavy truck of weight 400kN is shown in Fig. 4, parked on an extremely steep hill of slope theta with its hand
brake applied to the rear brakes only. Determine the reactions between the road and the truck’s wheels in
terms of theta. Find the steepest value of theta that the truck can manage without sliding if the coefficient of friction
between the tyres and the slippery road is m = 0.8. Would it be better if the truck were parked facing downhill?

Could someone explain how to solve this problem? Cheers.

2. Hello Haris
Originally Posted by Haris

A heavy truck of weight 400kN is shown in Fig. 4, parked on an extremely steep hill of slope theta with its hand
brake applied to the rear brakes only. Determine the reactions between the road and the truck’s wheels in
terms of theta. Find the steepest value of theta that the truck can manage without sliding if the coefficient of friction
between the tyres and the slippery road is m = 0.8. Would it be better if the truck were parked facing downhill?

Could someone explain how to solve this problem? Cheers.
Resolve horizontally, resolve vertically and take moments (for example, about the point of contact of the rear wheel with the road).

Don't forget to include the weight of the truck - apart from that, you have all the forces you need.

Then say that if the rear wheel is on the point of slipping, the friction force $= \mu \times$ normal contact force.

Provided you can deal OK with the sines and cosines of the angle $\theta$, it's very straightforward.

If you can't get it to work, post the equations that you've written down, and we'll take it from there.

3. I really am stuck. First of all how would you resolve the forces? Cheers.

4. Hello Haris

Call the points of contact with the road of the rear wheel and front wheel R and F respectively, and the respective normal contact forces $N_R$ and $N_F$. Call the friction force at R, $F_R$. All measured in kN.

Resolve vertically: $N_R\cos\theta+N_F\cos\theta+F_R\sin\theta = 400$ (1)

Resolve horizontally: $N_R\sin\theta+N_F\sin\theta=F_R\cos\theta$ (2)

Take moments about R: $400(3-1.3\tan\theta)\cos\theta=7N_F$ (3)

Equation (3) gives you $N_F$ directly. Substitute this into (1) and (2) and solve for the other two forces.

P.S.

It gives simpler equations if you resolve at right angles to the plane:

$N_R+N_F=400\cos\theta$ (1)

... and up the plane:

$F_R=400\sin\theta$ (2)