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Current Question (ID: 10158)
Question:
$\text{A uniform cube of mass } m \text{ and side } a \text{ is placed on a frictionless horizontal surface. A vertical force } F \text{ is applied to the edge as shown in the figure. Match the following (most appropriate choice).}$ $\text{List- I}$ $\text{(a) } mg/4 < F < mg/2$ $\text{(b) } F > mg/2$ $\text{(c) } F > mg$ $\text{(d) } F = mg/4$ $\text{List- II}$ $\text{(i) cube will move up.}$ $\text{(ii) cube will not exhibit motion.}$ $\text{(iii) cube will begin to rotate and slip at A.}$ $\text{(iv) normal reaction effectively at } a/3 \text{ from A, no motion.}$
Options:
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1. $\text{a - (i), b - (iv), c - (ii), d - (iii)}$
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2. $\text{a - (ii), b - (iii), c - (i), d - (iv)}$
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3. $\text{a - (iii), b - (i), c - (ii), d - (iv)}$
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4. $\text{a - (i), b - (ii), c - (iv), d - (iii)}$
Solution:
\text{Hint: Recall the concept of toppling.}
\text{Step 1: Find torque due to } F \text{ and } mg \text{ about point A.}
\text{Consider the given diagram, the moment of the force } F \text{ about point } A\text{:}
\tau_1 = r \times F \text{ (anti-clockwise)}
\text{The moment of weight } mg \text{ of the cube about point } A\text{:}
\tau_2 = mg \times \frac{a}{2} \text{ (clockwise)}
\text{Step 2: Equate the torques for no motion and find the value of } F\text{.}
\text{Cube will not exhibit motion if } \tau_1 = \tau_2
\text{In this case, both the torques will cancel the effect of each other:}
F \times a = mg \times \frac{a}{2} \implies F = \frac{mg}{2}
\text{Step 3: Find the value of } F \text{ for which the cube will rotate.}
\text{Cube will rotate only when } \tau_1 > \tau_2
F \times a > mg \times \frac{a}{2} \implies F > \frac{mg}{2}
\text{Step 4: If the normal reaction is acting at } \frac{a}{3} \text{ from point A and for no motion find } F \text{ and interpret.}
\text{Let the normal reaction be acting at } \frac{a}{3} \text{ from point } A\text{, then:}
mg \times \frac{a}{3} = F \times a \text{ or } F = \frac{mg}{3} \text{ (For no motion)}
\text{When } F = \frac{mg}{4}\text{, which is less than } \frac{mg}{3}\text{:}
\left(F < \frac{mg}{3}\right)
\text{there will be no motion.}
\text{Hence, option (2) is the correct answer.}
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