A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point directly above the centre of the face, at a height $$3a/4$$ above the base. If m $$=$$ $$6$$ kg then find the minimum value of F in N for which the cube begins to tip about an edge.

Question

A uniform cube of side a and mass m rests on a rough horizontal table. A horizontal force F is applied normal to one of the faces at a point directly above the centre of the face, at a height $$3a/4$$ above the base. If m $$=$$ $$6$$ kg then find the minimum value of F in N for which the cube begins to tip about an edge.

Infinity Learn · Accepted Answer

In the limiting case normal reaction will pass through A. The cube will tip about A if torque of F exceeds the torque of mg.

Hence, $F (\frac{3 a}{4}) > mg (\frac{a}{2}) or F > \frac{2}{3} mg = \frac{2}{3} \times 6 \times 10 = 40 N$

Therefore, minimum value of F is 40 N.

Rotational motion

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In the limiting case normal reaction will pass through A. The cube will tip about A if torque of F exceeds the torque of mg.
Hence, $F (\frac{3 a}{4}) > mg (\frac{a}{2}) or F > \frac{2}{3} mg = \frac{2}{3} \times 6 \times 10 = 40 N$
Therefore, minimum value of F is 40 N.

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Rotational motion

Remember concepts with our Masterclasses.

In the limiting case normal reaction will pass through A. The cube will tip about A if torque of F exceeds the torque of mg.Hence, F3a4>mga2 or F>23mg=23×6×10=40 NTherefore, minimum value of F is 40 N.

Ready to Test Your Skills?

free study material

In the limiting case normal reaction will pass through A. The cube will tip about A if torque of F exceeds the torque of mg.
Hence, $F (\frac{3 a}{4}) > mg (\frac{a}{2}) or F > \frac{2}{3} mg = \frac{2}{3} \times 6 \times 10 = 40 N$
Therefore, minimum value of F is 40 N.