Source: High school physics (Chinese)
Problem
A uniform wooden cube of mass $m$ and side length $l$ is placed on a plane inclined at an angle $\theta$.
- If the block remains at rest, find the magnitude of the normal force and the position of its line of action.
- An upward pushing force $F$, parallel to the incline and passing through the cube's center, is applied. The cube remains at rest. The coefficient of static friction is $\mu$. What are the new magnitude and line of action of the normal force? Are there constraints on the magnitude of $F$?
[Q1] $N = mg \cos\theta$. The line of action is at a distance $x = \frac{l}{2} \tan\theta$ from the center of the base, towards the lower edge. [Q2] $N' = mg \cos\theta$. The line of action is at a distance $x' = \frac{l}{2} (\tan\theta - \frac{F}{mg \cos\theta})$ from the center of the base. Yes, the magnitude of $F$ is constrained by both no-slip and no-tip conditions.
Let the coordinate system be aligned with the incline (x-axis parallel, y-axis perpendicular). The center of mass (CM) is at height $l/2$ from the base. Let $x$ be the distance of the normal force's line of action from the CM along the incline.
[Q1] For static equilibrium, the net force and net torque are zero. Force balance perpendicular to the incline:
$$\sum F_y = N - mg \cos\theta = 0 \implies N = mg \cos\theta$$Force balance parallel to the incline:
$$\sum F_x = mg \sin\theta - f_s = 0 \implies f_s = mg \sin\theta$$Torque balance about the CM:
$$\sum \tau_{CM} = N x - f_s \frac{l}{2} = 0$$ $$(mg \cos\theta) x = (mg \sin\theta) \frac{l}{2} \implies x = \frac{l}{2} \tan\theta$$The line of action is at a distance $x$ from the center, towards the lower edge.
[Q2] With the applied force $F$ passing through the CM: Force balance perpendicular to the incline is unchanged:
$$N' = mg \cos\theta$$Force balance parallel to the incline:
$$\sum F_x = F + f_s' - mg \sin\theta = 0 \implies f_s' = mg \sin\theta - F$$Torque balance about the CM, with new normal force position $x'$:
$$\sum \tau_{CM} = N' x' - f_s' \frac{l}{2} = 0$$ $$(mg \cos\theta) x' = (mg \sin\theta - F) \frac{l}{2} \implies x' = \frac{l}{2} \left( \frac{mg \sin\theta - F}{mg \cos\theta} \right) = \frac{l}{2} \left( \tan\theta - \frac{F}{mg \cos\theta} \right)$$Yes, there are constraints on $F$. The cube must not slip or tip. No-slip condition: $|f_s'| \le \mu N' \implies |mg \sin\theta - F| \le \mu mg \cos\theta$. This gives $mg(\sin\theta - \mu\cos\theta) \le F \le mg(\sin\theta + \mu\cos\theta)$. No-tip condition: The normal force must act within the base, so $|x'| \le l/2$. This gives $| \tan\theta - \frac{F}{mg \cos\theta} | \le 1$, which simplifies to $mg(\sin\theta - \cos\theta) \le F \le mg(\sin\theta + \cos\theta)$. Both sets of conditions must be satisfied for the cube to remain at rest.