Question

A uniform solid cube of mass m and side length l is placed on smooth horizontal surface in figure. A force (F) is applied on block in vertical direction.

• A. When $F=\frac{Mg}{4}$, the cube will remain stationary but the effective normal will shift at a distance $\frac{l}{3}$ from point A.
• B. $F=\frac{3mg}{5}$, the cube will start rotating and slip at point A but never loose constant.
• C. $F=\frac{4mg}{3}$, the cube will lift up.
• D. $F=-\frac{mg}{3}$, the cube will remain stationary but the effective normal reaction will shift a distance $\frac{5l}{8}$ from A.

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

To analyze the motion of the uniform solid cube, we consider the forces and torques acting on it. Let 'm' be the mass of the cube and 'l' be its side length. The force 'F' is applied vertically at the top right edge. Point A is the bottom left edge. The surface is smooth, so there is no friction.

Forces involved:

1. Weight (mg): Acts vertically downwards through the center of mass, at a horizontal distance of l/2 from point A.
2. Applied Force (F): Acts vertically upwards at the top right edge, at a horizontal distance of l from point A.
3. Normal Reaction (N): Acts vertically upwards from the surface at a distance 'x' from point A.

Conditions for Motion:

1. Vertical Equilibrium (No Lifting): For the cube to remain in contact with the surface, the net vertical force must be zero, and the normal reaction N must be non-negative. $\Sigma F_y = N + F - mg = 0$ $N = mg - F$ The cube lifts off the surface if $N \le 0$, which means $mg - F \le 0 \implies F \ge mg$. If $F > mg$, the cube lifts up.

2. Rotational Equilibrium (No Tipping): The cube tends to tip about point A. For rotational equilibrium, the net torque about any point must be zero. Let's take torques about point A.

• Torque due to F (clockwise): $\tau_F = F \times l$
• Torque due to mg (counter-clockwise): $\tau_{mg} = mg \times (l/2)$
• Torque due to N (counter-clockwise): $\tau_N = N \times x$ (where x is the distance of N from A)

For static equilibrium, the sum of torques about any point is zero. A convenient point to take torques is the point of application of the normal force, (x, 0). $\Sigma \tau_x = F(l-x) - mg(l/2 - x) = 0$ $Fl - Fx - \frac{mgl}{2} + mgx = 0$ $x(mg - F) = \frac{mgl}{2} - Fl$ Since $N = mg - F$, we have: $xN = \frac{mgl}{2} - Fl$ $x = \frac{\frac{mgl}{2} - Fl}{N} = \frac{\frac{mgl}{2} - Fl}{mg - F}$

The cube will start to tip (rotate) about point A when the normal force effectively shifts to point A (i.e., $x=0$). Setting $x=0$ in the torque equation: $F \times l - mg \times (l/2) = 0$ $F = \frac{mg}{2}$ If $F > mg/2$, the cube will start rotating (tipping) about point A. If $F < mg/2$, the cube remains stationary and the normal force shifts.

Analysis of Options:

(A) When $F=\frac{Mg}{4}$, the cube will remain stationary but the effective normal will shift at a distance $\frac{l}{3}$ from point A.

• Check for lifting: $F = Mg/4 < Mg$. So, it will not lift. $N = Mg - Mg/4 = 3Mg/4 > 0$.
• Check for tipping: $F = Mg/4 < Mg/2$. So, it will not tip. It remains stationary.
• Calculate the position 'x' of the normal force: $x = \frac{\frac{Mgl}{2} - (\frac{Mg}{4})l}{Mg - \frac{Mg}{4}} = \frac{\frac{Mgl}{4}}{\frac{3Mg}{4}} = \frac{l}{3}$ This statement is consistent with our calculations. So, (A) is correct.

(B) $F=\frac{3mg}{5}$, the cube will start rotating and slip at point A but never loose constant.

• Check for lifting: $F = 3mg/5 < mg$. So, it will not lift. $N = mg - 3mg/5 = 2mg/5 > 0$.
• Check for tipping: $F = 3mg/5 = 0.6mg$. Since $0.6mg > 0.5mg$ (which is $mg/2$), the cube will start rotating (tip) about point A.
• "slip at point A": On a smooth surface, there's no friction for horizontal slipping. However, in this context, "slip at point A" is often used to describe the onset of rotation about point A, implying that A is the sole contact point.
• "never loose constant" (likely typo for "never lose contact"): When the cube tips about A, it remains in contact with the surface at point A. Considering the common phrasing in such problems, this statement is consistent. So, (B) is correct.

(C) $F=\frac{4mg}{3}$, the cube will lift up.

• Check for lifting: $F = 4mg/3 = 1.33mg$. Since $1.33mg > mg$, the cube will lift up from the surface (N becomes negative). This statement is consistent with our calculations. So, (C) is correct.

(D) $F=-\frac{mg}{3}$, the cube will remain stationary but the effective normal reaction will shift a distance $\frac{5l}{8}$ from A. The diagram shows F acting upwards. If F is interpreted as a component, then $F = -mg/3$ means the force is directed downwards with a magnitude of $mg/3$. Let's assume F is a downward force of magnitude $mg/3$.

• Vertical Equilibrium: The normal force N must balance the weight and the downward applied force. $N - mg - F = 0 \implies N = mg + F = mg + mg/3 = 4mg/3$. Since $N > 0$, the cube remains in contact and stationary vertically.
• Rotational Equilibrium: Taking torques about point A.
  • Torque due to F (downward, at l from A): $F \times l$ (counter-clockwise)
  • Torque due to mg (downward, at l/2 from A): $mg \times (l/2)$ (counter-clockwise)
  • Torque due to N (upward, at x from A): $N \times x$ (clockwise) For equilibrium: $N \times x - F \times l - mg \times (l/2) = 0$ $x = \frac{F \times l + mg \times (l/2)}{N}$ Substitute $F = mg/3$ and $N = 4mg/3$: $x = \frac{(\frac{mg}{3})l + mg(\frac{l}{2})}{\frac{4mg}{3}} = \frac{\frac{mgl}{3} + \frac{mgl}{2}}{\frac{4mg}{3}} = \frac{\frac{2mgl + 3mgl}{6}}{\frac{4mg}{3}}$ $x = \frac{\frac{5mgl}{6}}{\frac{4mg}{3}} = \frac{5mgl}{6} \times \frac{3}{4mg} = \frac{5l}{8}$ Since $x = 5l/8 < l$ (the width of the base), the normal force acts within the base, so the cube remains stationary. This statement is consistent under the interpretation that $F=-mg/3$ means a downward force of magnitude $mg/3$. So, (D) is correct.

All options (A), (B), (C), and (D) are correct based on the typical interpretations of such problems.

The final answer is $\boxed{A, B, C, D}$

Explanation of the solution:

1. Identify Forces and Conditions: The forces are weight (mg), applied force (F), and normal reaction (N). Conditions for motion are lifting (N=0 when F=mg) and tipping (normal force shifts to edge A when F=mg/2).
2. Derive Normal Force Position: For static equilibrium, the position 'x' of the normal force from point A is given by $x = \frac{\frac{mgl}{2} - Fl}{mg - F}$.
3. Evaluate Option (A): For $F=Mg/4$, calculate $x$. $x = l/3$. Since $F < Mg/2$ and $F < Mg$, the cube remains stationary. So (A) is correct.
4. Evaluate Option (B): For $F=3mg/5$, compare with tipping and lifting conditions. $F > Mg/2$ but $F < Mg$, so the cube tips (rotates) about A without lifting. "Slip at A" is interpreted as tipping about A. "Never lose contact" at A is true for tipping. So (B) is correct.
5. Evaluate Option (C): For $F=4mg/3$, compare with lifting condition. $F > Mg$, so the cube lifts up. So (C) is correct.
6. Evaluate Option (D): For $F=-mg/3$, interpret as a downward force of magnitude $mg/3$. Calculate N and then x. $N = 4mg/3$. $x = 5l/8$. Since $N > 0$ and $x < l$, the cube remains stationary. So (D) is correct.