Fuzzy dice

Draw a free-body diagram. Let $F$ be the force put on the pedals. Let $T$ be the tension in the string. Then, using Newton's Second Law in the $x$ and $y$ directions, respectively, we have: $\begin{aligned} T \sin \theta &=& F, \\ T \cos \theta &=& mg. \end{aligned}$ Then, $F = mg \tan \theta \implies a = \dfrac {F}{m} = g \tan \theta$ .

Plugging in $g = \text {9.8 } \text {m/s}^2$ and $\theta = 15^\circ$ , we get $a = \boxed {2.626 \, \text{m/s}^2}.$

First, assume your frame of reference to be the car, then a pseudo force will be acting on the dice.

Since, the dice will be in equilibrium at $\theta = 15^\circ$ . So, the force of gravity (acting on dice) will be balanced by the tension's vertical component and the pseudo force will be balanced by the tension's horizontal component, therefore -

Let the tension in string be $T$

$T sin \theta = ma$

$T cos \theta = mg$

Dividing the two equations we get $a = g \times tan \theta = 9.8 \times tan 15^\circ = 2.626$

We can draw a triangle of forces for the dice that looks something like this I\, where the acute angle $\theta$ of $15^{o}$ is at the top, the bottom left is a $90^{o}$ angle and the bottom right is $75^{o}$ . From this we can work out, using trigonometry, that the magnitude of the adjacent is x $\cos 15$ and the opposite is x $\sin 15$ , where x is an unknown constant. We know the only force acting downwards is gravity with a magnitude 9.8 $ms^{-2}$ , and so

9.8 = x $\cos 15$

x = 9.8 / $\cos 15$

And since the only force acting horizontally is the acceleration, we can work out

a = x $\sin 15$

a = 9.8 / $\cos 15$ multiplied by $\sin 15$

a = 2.63 $ms^{-2}$

Let the horizontal acceleration be $a$

Since the angle to the horizontal is $90-15=75^\circ$ we can know $a tan 75^\circ$ is the vertical component of the acceleration of the string is exerting on dice.

Since there is no net vertical acceleration

$a tan 75^\circ = g$

$a=\frac{g}{ tan 75^\circ}= \boxed{2.626}$

Initially when the dice are hanging down, the only force acting on them is g Later when the vehicle moves, another force (lets say, x ) starts acting perpendicular to the vertical. And the dice are now making a 15° angle with the vertical, i.e. the resultant force is making a 15° angle to the vertical. Now as the vectors g and x are already perpendicular and the resultant is making an angle of 15°, we can solve by using vector addition and trigonometry $\tan 15$ = $\frac{x}{g}$

Consider the case when the dice make an angle of $15^{\circ}$ with respect to the vertical. In that case, since the dice are in equilibrium, the forces acting on the dice must cancel each other. $$ Let the tension in the sting be $T$ . Now the vertical component of $T$ must cancel out the downward force on the dice $mg$ So we get $T \cos 15^{\circ} = mg$ and similarly, $T \sin 15^{\circ}$ must cancel the additional force that is acting on the dice, let it be $F$ . So $T \sin 15^{\circ} = F$ .

$$ Dividing the above two equations, we get $\tan 15^{\circ} = \frac{F}{g}$ $$ Therefore, $F = \tan 15^{\circ} * g \implies \boxed{ F \approx 2.626}$

We can resolve Tension $T$ in the string in two components $Tsin\theta$ and $Tcos\theta$ .

$Tcos\theta = mg$ (For equilibrium in y-direction) and $Tsin\theta = ma$ (For equilibrium in x-direction).Solving , $a=gtan\theta = \boxed{2.62}$