Some Examples of Static Equilibrium — Question 8

Question 8

An uphill parked vehicle transfers normal load toward its downhill axle

A vehicle of weight $W$ is static on a slope $\theta$. Along the slope, the downhill axle is at $x=0$, the uphill axle at $x=L=2\,\mathrm m$, and the center is at $x=1\,\mathrm m$, height $h=.50\,\mathrm m$ normal to the slope. Tire friction acts along the slope.

1. Find both normal reactions as functions of $\theta$.

2. Find the uphill-contact loss angle and the total friction coefficient needed at $20^\circ$, assuming the braking force can be appropriately shared.

Question 8 - Solution

An uphill parked vehicle transfers normal load toward its downhill axle

(a) Derivation and calculation

Use $x$ uphill and $y$ normal to the slope. About the downhill axle, $$LN_u-Wx_G\cos\theta+Wh_G\sin\theta=0.$$ With $L=2$, $x_G=1$, $h_G=.5$ m, $$\boxed{N_u=W(.5\cos\theta-.25\sin\theta)},$$ $$\boxed{N_d=W(.5\cos\theta+.25\sin\theta)}.$$

(b) Result and interpretation

The uphill axle loses contact at $$N_u=0\Rightarrow\tan\theta=2\Rightarrow\boxed{\theta_{\mathrm{tip}}=63.4^\circ}.$$ For sliding equilibrium with shareable friction, $$f_d+f_u=W\sin\theta\le\mu_s(N_d+N_u)=\mu_sW\cos\theta.$$ At $20^\circ$, $$\boxed{\mu_s\ge\tan20^\circ=.364}.$$ The friction and contact inequalities must both hold.

Physical check

The two normal reactions sum to $W\cos\theta$, and the downhill load share increases with slope.