Question

A uniform solid cube of mass m and side length $l$ is placed on smooth horizontal surface as shown 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 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)

A, B, C

Answer 2

The problem describes a uniform solid cube of mass $m$ and side length $l$ placed on a smooth horizontal surface. A vertical force $F$ is applied to the block. The figure shows the force $F$ applied at the top right edge and point A as the bottom right corner.

However, based on standard physics problems of this type and the provided similar question, it's highly probable that the intended setup is for the force $F$ to be applied at the top left edge, and point A to be the bottom right corner. This setup creates a torque that tends to tip the cube. If $F$ were applied directly above A, it would create no torque about A, and the analysis would be different. We will proceed with the interpretation that $F$ is applied at the top left edge and A is the bottom right corner, consistent with the similar question.

Forces and Torques:

1. Weight (mg): Acts vertically downwards through the center of mass (CM) of the cube. The CM is at a horizontal distance of $l/2$ from both the left and right edges.
2. Applied Force (F): Acts vertically upwards at the top left edge.
3. Normal Reaction (N): Acts vertically upwards from the horizontal surface. Its point of application (let's say at distance $x$ from point A) shifts to maintain equilibrium or cause rotation.

Equilibrium Conditions:

I. Vertical Translational Equilibrium: For the cube to remain in contact with the surface and not lift up, the net vertical force must be zero or positive (if accelerating upwards, but here we consider static conditions or just starting motion). $N + F = mg$ $N = mg - F$ For the cube to remain in contact, $N \ge 0$, which implies $mg - F \ge 0 \implies F \le mg$. If $F > mg$, the cube will lift up.

II. Rotational Equilibrium (about point A, the bottom right corner): The cube will tend to rotate (tip) about point A if the net torque about A is non-zero. Let's consider torques about point A:

• Torque due to F ($\tau_F$): The force $F$ is applied at the top left edge. The perpendicular distance from A (bottom right corner) to the line of action of $F$ is $l$. This torque tends to rotate the cube anti-clockwise. $\tau_F = F \times l$ (anti-clockwise)
• Torque due to Weight ($\tau_{mg}$): The weight $mg$ acts at the center of mass, which is at a horizontal distance of $l/2$ from A. This torque tends to rotate the cube clockwise. $\tau_{mg} = mg \times (l/2)$ (clockwise)
• Torque due to Normal Reaction ($\tau_N$): Let the normal reaction $N$ act at a distance $x$ from point A. This torque tends to rotate the cube clockwise. $\tau_N = N \times x$ (clockwise)

For rotational equilibrium, the sum of torques about A must be zero: $\sum \tau_A = 0$ $\tau_F - \tau_{mg} - \tau_N = 0$ (taking anti-clockwise as positive) $F \times l - mg \times (l/2) - N \times x = 0$ $N \times x = F \times l - mg \times (l/2)$

Substitute $N = mg - F$: $(mg - F)x = F \times l - mg \times (l/2)$ $x = \frac{Fl - mg(l/2)}{mg - F}$

Conditions for Tipping: The cube will remain stationary without tipping as long as the effective normal reaction $N$ acts within the base of the cube, i.e., $0 \le x \le l$. Tipping occurs when $x$ moves to the edge, i.e., $x \le 0$. Setting $x=0$: $Fl - mg(l/2) = 0 \implies Fl = mg(l/2) \implies F = mg/2$. If $F > mg/2$, then $Fl - mg(l/2) > 0$, so $x > 0$. This means the net torque due to F and mg tends to rotate the cube anti-clockwise. This would mean the normal reaction shifts to a point to the left of A. If $x$ calculated is positive, it means N acts between A and the center of the base. Let's re-evaluate the torque balance. If $F$ is at top left, and A is bottom right. $F$ causes an anti-clockwise torque $Fl$. $mg$ causes a clockwise torque $mg(l/2)$. For rotational equilibrium, the normal force $N$ must provide a clockwise torque $Nx$. So, $Fl = mg(l/2) + Nx$. $Nx = Fl - mg(l/2)$. $x = \frac{Fl - mg(l/2)}{N} = \frac{Fl - mg(l/2)}{mg - F}$. The cube tips when $N$ shifts to A, meaning $x=0$. This happens when $Fl - mg(l/2) = 0 \implies F = mg/2$. If $F > mg/2$, then $Fl > mg(l/2)$, so $x > 0$. The normal reaction shifts to the left of A. This means the cube rotates about A. If $F < mg/2$, then $Fl < mg(l/2)$, so $x < 0$. This implies the normal reaction would need to be to the right of A, which is impossible. This means the cube will not tip about A. Instead, the torque due to weight is dominant, and the normal reaction shifts towards the center of the base (or towards the left edge).

Let's re-evaluate the similar question's method. In the similar question, $F$ is at top-left, A is bottom-right. $\tau_1 = F \times a$ (anticlockwise). $\tau_2 = Mg \times (a/2)$ (clockwise). For no motion, $\tau_1 = \tau_2$ is not the condition. The condition for tipping is when the net torque about A becomes positive (anti-clockwise) or the normal force shifts to the edge. The similar question states: "The cube will rotate only, when $\tau_1 > \tau_2$" i.e., $F \times a > Mg \times (a/2) \implies F > Mg/2$. This means if $F > Mg/2$, the cube will start rotating about A. If $F < Mg/2$, the cube will not rotate. The normal reaction will shift such that the torque due to N balances the net torque. If $F < Mg/2$, the torque due to weight ($Mg(a/2)$ clockwise) is greater than the torque due to F ($Fa$ anticlockwise). The net torque is clockwise. To balance this, the normal reaction must shift away from A (to the left of A). Let's measure $x$ from the center of the base. Or, let's measure $x$ from the left edge (point B). Then the normal reaction $N$ acts at $x_B$ from B. Torque about B (left bottom corner): $\tau_F = 0$ (F is applied above B). $\tau_{mg} = mg \times (l/2)$ (clockwise). $\tau_N = N \times x_B$ (anti-clockwise). $\tau_A = F \times l$ (clockwise) So, $N \times x_B + F \times l = mg \times (l/2)$. $N x_B = mg(l/2) - Fl$. $x_B = \frac{mg(l/2) - Fl}{mg - F}$. The cube tips about A (right edge) when $x_B \ge l$. This means $N$ shifts to A. The cube tips about B (left edge) when $x_B \le 0$. This means $N$ shifts to B.

Let's go back to the similar question's logic for $x$. $x$ is distance from A. $N \times x = Mg(a/2) - Fa$. (This equation implies that $x$ is measured from A towards the left, and $N$ provides a clockwise torque about A). If $F = Mg/4$, $x = \frac{Mg(a/2) - (Mg/4)a}{Mg - Mg/4} = \frac{Mga/4}{3Mg/4} = a/3$. This $x=a/3$ is a positive value, meaning the normal reaction acts at $a/3$ from A towards the left. This is within the base. This means the similar question uses $x$ as the distance from A towards the center of the base.

Let's use this convention for the current problem: $x$ is the distance of the normal reaction from A, measured towards the center of the base (left). So, $N \times x = \tau_{mg} - \tau_F = mg(l/2) - Fl$. $x = \frac{mg(l/2) - Fl}{mg - F}$.

Now, let's evaluate each option using this formula and the conditions derived:

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

• $F = mg/4$.
• Check for lifting: $F = mg/4 < mg$, so it won't lift.
• Check for tipping: $F = mg/4 < mg/2$. Since $F < mg/2$, the torque due to weight is greater than the torque due to $F$. The cube will not tip about A. It will remain stationary.
• Calculate $x$: $N = mg - F = mg - mg/4 = 3mg/4$. $x = \frac{mg(l/2) - (mg/4)l}{3mg/4} = \frac{mgl/4}{3mg/4} = l/3$. This means the normal reaction acts at $l/3$ from A (towards the left). This is within the base ($0 \le l/3 \le l$). Therefore, option (A) is correct.

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

• $F = 3mg/5 = 0.6 mg$.
• Check for lifting: $F = 0.6 mg < mg$, so it won't lift. $N = mg - 3mg/5 = 2mg/5 > 0$, so it won't lose contact.
• Check for tipping: $F = 0.6 mg > mg/2 = 0.5 mg$. Since $F > mg/2$, the torque due to $F$ ($Fl$) is greater than the torque due to weight ($mg(l/2)$). This will cause the cube to rotate about point A (the bottom right corner). "Slip at point A" can be interpreted as "pivot at point A". Therefore, option (B) is correct.

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

• $F = 4mg/3 \approx 1.33 mg$.
• Check for lifting: $F = 4mg/3 > mg$. As $F > mg$, the normal reaction $N = mg - F$ would be negative. This means the cube will lift up from the surface. Therefore, option (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.

• $F = mg/3$.
• Check for lifting: $F = mg/3 < mg$, so it won't lift.
• Check for tipping: $F = mg/3 < mg/2$. Since $F < mg/2$, the cube will remain stationary (no tipping).
• Calculate $x$: $N = mg - F = mg - mg/3 = 2mg/3$. $x = \frac{mg(l/2) - (mg/3)l}{2mg/3} = \frac{mgl/2 - mgl/3}{2mg/3} = \frac{mgl/6}{2mg/3} = \frac{l}{6} \times \frac{3}{2} = \frac{l}{4}$. The option states $x = 5l/8$. Since $l/4 = 2l/8 \ne 5l/8$, this option is incorrect. Therefore, option (D) is incorrect.

Based on the analysis, options (A), (B), and (C) are correct.

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

Explanation (Minimal): Assume force $F$ is applied at the top left edge and point A is the bottom right corner.

1. Vertical Equilibrium: $N = mg - F$. Cube lifts if $F > mg$.
2. Rotational Equilibrium (about A): $N \times x = Fl - mg(l/2)$, where $x$ is the distance of normal reaction from A (towards left). Cube tips if $F > mg/2$.

• Option A ($F = mg/4$):

  • $F < mg$ (no lift).
  • $F < mg/2$ (no tip, stationary).
  • $N = mg - mg/4 = 3mg/4$.
  • $x = \frac{(mg/4)l - mg(l/2)}{3mg/4} = \frac{-mgl/4}{3mg/4}$ (error in sign convention of my formula, it should be $mg(l/2) - Fl$ for positive $x$ towards left from A, as per similar question). Let's use the similar question's equation for $x$: $x = \frac{Mg(a/2) - Fa}{Mg - F}$. This implies $x$ is measured from A towards the left. $x = \frac{mg(l/2) - (mg/4)l}{mg - mg/4} = \frac{mgl/4}{3mg/4} = l/3$. Correct.

• Option B ($F = 3mg/5$):

  • $F < mg$ (no lift).
  • $F = 0.6mg > mg/2$ (tips about A).
  • Correct.

• Option C ($F = 4mg/3$):

  • $F > mg$ (lifts up).
  • Correct.

• Option D ($F = mg/3$):

  • $F < mg$ (no lift).
  • $F < mg/2$ (no tip, stationary).
  • $x = \frac{mg(l/2) - (mg/3)l}{mg - mg/3} = \frac{mgl/6}{2mg/3} = l/4$.
  • Option states $5l/8$. Incorrect.