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Chapter 12: Problem 39

A uniform rectangular block is twice as long as it is wide. Letting \(\theta\) be the angle that the long dimension makes with the horizontal, determine the angular positions of any static equilibria, and comment on their stability.

Short Answer

Expert verified
The possible angular positions for static equilibrium are \( \theta = 0, \pi, 2\pi \). Of these, only \( \theta = \pi \) is stable, as any marginal displacement from this position will not cause the block to topple over, whereas for \( \theta = 0, 2\pi \), any minor deviation will cause the block to topple over, hence they are unstable.

Step by step solution

01

Expressing height as a function of \( \theta \)

Assuming the thickness of the block to be negligible compared to its length and width, the vertical height of the block from the contact point with the surface can be shown by a trigonometric equation, due to symmetry. This is because the height will vary depending on the angle the block makes with the surface. Hence, the height, \( h \), can be given as function of angle \( \theta \), \( h = l \sin (\theta) \), where \( l \) is the length of the block.
02

Setting up and solving the equilibrium condition

To find the static equilibria, it is necessary to set the net torque to zero. The net torque about the contact point is the gravitational torque, \( \tau = Mgh \sin(\theta) \), where \( M \) is the mass of the block, \( g \) is the gravitational acceleration, and \( h \) is the height of the centre of mass from the contact point. Equating to zero, we get the bicubic equation, \( \sin^3(\theta) = 0 \). Solving this equation gives the possible values for \( \theta = 0, \pi, 2\pi \).
03

Analysing the stability of the equilibria

To determine the stability of the equilibria, we need to consider small perturbations about these equilibrium points. If a small perturbation leads the system to return to the equilibrium point, then the equilibrium is said to be stable, otherwise unstable. In this case, when \( \theta = 0, 2\pi \), any perturbation will cause the block to topple over and hence, are unstable equilibria. Whereas \( \theta = \pi \) represents the block being flat on the longer side, and any small perturbations will not cause the block to topple, thus representing a stable equilibrium.

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