根据《公路路基设计规范》（JTG D30-2015）第7.2.2条第5款，$T_{1} = F_{s}W_{1}\sin \alpha_{1} - W_{1}\cos \alpha_{1}\tan \varphi_{1} - c_{1}L_{1} = 1.2\times 2640\times \sin 20^{\circ} - 2640\times \cos 20^{\circ}\tan 10^{\circ} - 15\times 24 = 286.1kN$ $\varphi_{2} = \cos (20^{\circ} - 12^{\circ}) - \sin (20^{\circ} - 12^{\circ})\tan 10^{\circ} = 0.966$ $T_{2} = F_{s}W_{2}\sin \alpha_{2} + \varphi_{2}T_{1} - W_{2}\cos \alpha_{2}\tan \varphi_{2} - c_{2}L_{2}$ $= 1.2\times 6200\sin 12^{\circ} + 0.966\times 286.1 - 6200\cos 12^{\circ}\tan 10^{\circ} - 15\times 23$ $= 408.9kN$ $\varphi_3 = \cos (12^\circ -5^\circ) - \sin (12^\circ -5^\circ)\tan 10^\circ = 0.971$ $T_{3} = F_{s}W_{3}\sin \alpha_{3} + \varphi_{3}T_{2} - W_{3}\cos \alpha_{3}\tan \varphi_{3} - c_{3}L_{3}$ $= 1.2\times 3750\sin 5^{\circ} + 0.971\times 408.9 - 3750\cos 5^{\circ}\tan 10^{\circ} - 15\times 29 = -304.5 < 0$ 当 $\mathrm{T_3 < 0}$ 时，取 $\mathrm{T_3 = 0}$ ，所以在正常工况下第三块 $(\mathrm{W}_3)$ 产生的剩余下滑力取0。

参考答案: B