Question

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is $\mu = 0.4$, the maximum possible value of $100 \times \frac{b}{a}$ for a box not to topple before moving is _______

Answer 1

75

Answer 2

To solve this problem, we need to consider two conditions for the box:

1. Condition for sliding (moving): The box starts to slide when the applied force $F$ overcomes the maximum static friction force.
2. Condition for toppling: The box starts to topple when the torque due to the applied force about the pivot edge exceeds the torque due to its weight about the same edge.

Let $M$ be the mass of the cubical box and $g$ be the acceleration due to gravity. The side of the cube is $a$. For a uniform cubical box, its center of mass (CM) is at a height of $a/2$ from the base. The force $F$ is applied at a point $b$ above its center of mass. Therefore, the height of the point of application of force $F$ from the base is $h = \frac{a}{2} + b$.

1. Condition for sliding: When the box is about to slide, the applied force $F$ is equal to the maximum static friction force $f_{s,max}$. The normal force $N$ acting on the box is equal to its weight $Mg$. $N = Mg$ The maximum static friction force is $f_{s,max} = \mu N = \mu Mg$. So, the minimum force required to slide the box is: $F_{slide} = \mu Mg$

2. Condition for toppling: The box will topple about the edge closest to the applied force. Let's consider torques about this edge (pivot point P). The weight $Mg$ acts downwards through the CM, which is at a horizontal distance of $a/2$ from the pivot edge. This creates a restoring torque. $\tau_{Mg} = Mg \times \frac{a}{2}$ (counter-clockwise, resisting toppling)

The applied force $F$ acts horizontally at a height $h = \frac{a}{2} + b$ from the base. This creates a toppling torque. $\tau_F = F \times h = F \left(\frac{a}{2} + b\right)$ (clockwise, causing toppling)

For the box to be on the verge of toppling, the toppling torque must be equal to the restoring torque: $F_{topple} \left(\frac{a}{2} + b\right) = Mg \frac{a}{2}$ So, the force required to topple the box is: $F_{topple} = \frac{Mg \frac{a}{2}}{\frac{a}{2} + b}$

3. Condition for not toppling before moving: For the box not to topple before moving, it must slide first. This means the force required to slide must be less than or equal to the force required to topple: $F_{slide} \le F_{topple}$ Substitute the expressions for $F_{slide}$ and $F_{topple}$: $\mu Mg \le \frac{Mg \frac{a}{2}}{\frac{a}{2} + b}$

Cancel $Mg$ from both sides: $\mu \le \frac{\frac{a}{2}}{\frac{a}{2} + b}$

Rearrange the inequality to find the maximum value of $b/a$: $\mu \left(\frac{a}{2} + b\right) \le \frac{a}{2}$ $\frac{\mu a}{2} + \mu b \le \frac{a}{2}$ $\mu b \le \frac{a}{2} - \frac{\mu a}{2}$ $\mu b \le \frac{a}{2} (1 - \mu)$ $\mu b \le \frac{a}{2} (1 - \mu)$ $\frac{b}{a} \le \frac{1 - \mu}{2\mu}$

Given the coefficient of friction $\mu = 0.4$. Substitute $\mu = 0.4$ into the inequality: $\frac{b}{a} \le \frac{1 - 0.4}{2 \times 0.4}$ $\frac{b}{a} \le \frac{0.6}{0.8}$ $\frac{b}{a} \le \frac{6}{8}$ $\frac{b}{a} \le \frac{3}{4}$ $\frac{b}{a} \le 0.75$

The maximum possible value of $\frac{b}{a}$ is $0.75$. The question asks for the maximum possible value of $100 \times \frac{b}{a}$. $100 \times \frac{b}{a} = 100 \times 0.75 = 75$.

The final answer is $\boxed{75}$.