The zero-couple equations in the two coupled horizontal blocks are
$$
L\Omega_y=-\frac3{13}U_x,
\qquad
L\Omega_x=-\frac3{13}U_y,
$$
and $\Omega_z=0$. The force equations then become
$$
k_x=5U_x+L\Omega_y=\frac{62}{13}U_x,
\qquad
k_y=\frac{62}{13}U_y,
\qquad
k_z=5U_z.
$$
Therefore
$$
\boxed{\alpha=\frac{13}{62},\qquad\gamma=\frac3{62}},
$$
and
$$
\boxed{
\mathbf U=\left(\frac{13}{62}k_x,
\frac{13}{62}k_y,\frac15k_z\right),
\qquad
L\boldsymbol\Omega
=-\frac3{62}(k_y,k_x,0)}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-329/2/d/solution