Question 3 of 5: Combined Toppling (Block A) and Sliding (Block B) — Verification-Case Statics

Question 3: Combined Toppling (Block A) and Sliding (Block B) — Verification-Case Statics (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Plane inclination $\psi=30^\circ$; friction angle on every surface $\phi=35^\circ$; block A: width (along plane) $b_A=0.75$ m, height (normal to plane) $h_A=3$ m; block B is twice as heavy as A ($W_B=2W_A$), thickness (normal to plane) $t$ unknown; A and B share a common face, in contact only over B's height $t$ (from the plane up to the top of B).

Find. The thickness $t$ of block B, and a demonstration that block A has no tendency to slip at C.

Approach. Take moments about C for block A (weight vs. the interblock thrust $P$ from B, applied at height $t$) to get one equation in $P$ and $t$; take force equilibrium along the plane for block B at its limiting (sliding) friction to get $P$ in terms of $W_B$; combine to solve for $t$, then check the force balance at C for block A against the available friction.

1. Isolated-block check — confirm A alone would topple. A rectangular block standing normal to a plane inclined at $\psi$, with width $b$ and height $h$, topples about its downslope toe when $\tan\psi > b/h$. $$\tan(30^\circ)=0.577\quad\text{vs.}\quad b_A/h_A = 0.75/3.0=0.250$$ Since $0.577>0.250$, block A would topple even in isolation — it needs a stabilizing thrust from B to remain in the stated limiting equilibrium, consistent with the problem.
2. Moment equilibrium of A about C. Work in plane-local axes ($x'$ along the plane, downslope positive; $y'$ normal to the plane, outward positive), origin at C. A's centroid sits at $(x_A',y_A')=(-b_A/2,\,h_A/2)=(-0.375,\,1.5)$ m (upslope of C, half-height up). Gravity resolves to $(W_A\sin\psi,\,-W_A\cos\psi)$ in these axes. The interblock thrust $P$ from B acts on A at the top of the shared face, position $(0,t)$, purely normal to that face (along $-x'$, i.e. pushing A back upslope — the standard frictionless-release-joint idealization for a thin vertical(ish) contact). $$M_{weight}=x_A'(-W_A\cos\psi)-y_A'(W_A\sin\psi) = -0.375(0.866)W_A - 1.5(0.5)W_A = -0.425\,W_A$$ $$M_P = t\cdot P \quad(\text{CCW, stabilizing})$$ Limiting equilibrium ($\sum M_C=0$): $\ -0.425\,W_A + tP = 0\ \Rightarrow\ P = \dfrac{0.425\,W_A}{t}$
3. Force equilibrium of B (limiting sliding). B receives the reaction $+P$ from A (Newton's third law, downslope push) in addition to its own weight $W_B=2W_A$; at the point of sliding, base friction is fully mobilized at $\phi=35^\circ$. $$N_B=W_B\cos\psi,\qquad W_B\sin\psi+P-N_B\tan\phi=0$$ $$P = W_B(\cos\psi\tan\phi-\sin\psi)=2W_A(0.866\times0.700-0.500)=2W_A(0.106)=0.213\,W_A$$
4. Solve for $t$. Equating the two expressions for $P$: $$0.213\,W_A=\frac{0.425\,W_A}{t}\ \Rightarrow\ t=\frac{0.425}{0.213}=\boxed{2.00\ \text{m}}$$
5. No-slip check for A at corner C. With $P=0.213\,W_A$ known, resolve force equilibrium of A along and normal to the plane: $$N_A=W_A\cos\psi=0.866\,W_A,\qquad f_A = P-W_A\sin\psi = 0.213\,W_A-0.500\,W_A=-0.287\,W_A$$ so the friction actually mobilized at C has magnitude $|f_A|=0.287\,W_A$ (acting upslope, resisting the net downslope pull). Available friction at C is $N_A\tan\phi=0.866(0.700)\,W_A=\boxed{0.606\,W_A}$. $$\text{FS against slip at C} = \frac{0.606\,W_A}{0.287\,W_A}\approx\boxed{2.1}\ \gg1$$ Mobilized friction (0.287$W_A$) is well below the available friction (0.606$W_A$), so block A has no tendency to slip at C — its limiting mode is toppling, exactly as stated in the problem.
   Check: the A–B contact is idealized as smooth (normal force only, no shear on that face) and applied at the top of B's height $t$ — the standard simplifying assumption for this class of two-block toppling/sliding verification problem, and the one that reproduces the clean $t=2.00$ m result.