Why does resolving forces in one direction give a completely different answer to resolving the opposite way?

Question

I can solve parts i), ii) and am able to show that $R=0$ for part iii). In this question $g$ is the acceleration of free fall taken to be $9.8$

Using Newtons 2nd law [$F=ma$] for the last part I resolved forces perpendicular to $PQ$ in the direction of $S$ to obtain $\displaystyle\ S- \frac{4mg}{5} = \frac{ma}{2}\alpha$ (1)

Where $\alpha=$ Tangential angular acceleration.

To obtain an expression for $\alpha$ I took moments about $X$ using Newtons 2nd law in angular form [$C=I\alpha$], where $C$ is applied couple or torque, and $I$ is the moment of inertia for the lamina.

This yielded: $\displaystyle\frac{2mga}{5} = \frac{4ma^2\alpha}{3}$

Solving gives $\displaystyle\alpha=\frac{3g}{10a}$ (which the mark-scheme agrees with).

Substitution of $\displaystyle\alpha=\frac{3g}{10a}$ in (1) yields $\displaystyle\ S=\frac{19mg}{20}$ and according to the mark-scheme this is incorrect.

The correct answer according to the mark-scheme is $\displaystyle\ S=\frac{13mg}{20}$ obtained from $\displaystyle\ \frac{4mg}{5} -S = \frac{ma}{2}\alpha$. This is what the mark-scheme says:

So the real question is: Why must $S$ and $\displaystyle\frac{ma}{2}\alpha$ be in opposite directions?

Thank you kindly,

with best regards.

Answer 1

The two forces acting on the lamina in the direction parallel to $S$ are $S$ itself and $\dfrac45 mg,$ the component in that direction of the gravitational force on the lamina. That component of gravitational force is opposite $S$. So you are correct that the force in the direction of $S$ is $$S - \frac45 mg.$$ Note that this is the force in an upward direction. (Not straight up, but still upward.) If that force were positive, the center of mass of the lamina would be accelerated upward.

But the pivot is inert. The only force it exerts on the lamina is the equal and opposite reaction necessary to keep the pivot point of the lamina from falling. It can't stop the center of mass of the lamina from falling, which is why the rotation of the lamina is accelerating at the point in time shown in the illustration. The center of mass is accelerating in a downward direction at that time.

So the net force on the center of mass parallel to $S$ has to be downward, that is, it must be opposite the direction of $S$, not the same direction. In other words, the net force in the direction of $S$ is negative. The force acting in the direction of positive $\alpha$ is the positive-valued force that is opposite in direction to $S$, that is, $$\frac45 mg - S.$$

Answer 2

i)$$I=m\frac{((2a)^2+(3a)^2)}{12}+m\left(\frac{3a}2-a\right)^2=\frac43ma^2$$ ii)$$\frac12I\omega_i^2+mga(1-\cos\theta)=\frac12I\omega_f^2\\ \implies w_f^2=\frac{6g}{5a}$$ iii)$$R=mg\cos\theta-m\omega_f^2\left(\frac{3a}2-a\right)=0$$ $$\alpha=\frac{\tau}I=\frac{mg\sin\theta\times\left(\frac{3a}2-a\right)}{\frac43ma^2}=\frac{3g}{10a}$$ $$mg\sin\theta-S=m\left(\frac{3a}2-a\right)\alpha\\ \implies S=\frac{13}{20}mg$$

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Explanation:

Take directions with signs:

Towards $mg\sin\theta$ perpendicular to PQ positive and opposite negative, then using your same expression: $$\displaystyle\ (-S)+\left(+\frac{4mg}{5}\right) = \left(+\frac{ma}{2}\alpha\right)$$ Actually this is a remainder of some vector form: $$\vec S+\frac{4mg}{5}\hat d =\left(\frac{ma}2\right)\hat d$$ where $\hat d$ is in direction perpendicular to PQ along $mg\sin\theta$

Conclusively and easily $S$ and $\frac {ma}2\alpha$ are in opposite directions.