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Mobile App Chapter 6: Q81GP (page 138)
A cubic crate of side\(s = 2.0\;{\rm{m}}\)is top-heavy: its CG is 18 cm above its true center. How steep an incline can the crate rest on without tipping over? [Hint: The normal force would act at the lowest corner].
Short Answer
The steep created by an inclined cube is \(40.2^\circ \).
Step by step solution
Given data
The side of the cube is\(s = 2\;{\rm{m}}\).
The center of gravity is \(CG = 18\;{\rm{cm}}\).
Understanding the free body diagram of the cubic crate
In this problem, first, draw a free body diagram of the cubic crate on the verge of tipping. In this case, if a vertical line projected downward from the center of gravity falls outside the base of support, the cube crate will topple.
Free body diagram of the cube and calculation of the inclined angle
The free body diagram is as follows:
The relation to find the inclined angle can be written as:
\(\tan \theta = \frac{{\left( {\frac{s}{2}} \right)}}{{\left( {\frac{s}{2} + CG} \right)}}\)
On plugging the values in the above relation, you get:
\(\begin{array}{c}\tan \theta = \left[ {\frac{{\left( {\frac{{2\;{\rm{m}}}}{2}} \right)}}{{\left( {\left( {\frac{{2\;{\rm{m}}}}{2}} \right) + 18\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}}} \right]\\\tan \theta = 0.847\\\theta = 40.2^\circ \end{array}\)
Thus, \(\theta = 40.2^\circ \) is the required angle.
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