Sliding down a sliding incline - 2

Figure 1. is a snapshot in time of the initial setup.

There are two frames of reference $X_1,Y_1$ attached to the incline, and $X_2,Y_2$ fixed to the ground. $X_1,Y_1$ is non-inertial and the force balance with respect to the block must include inertial terms to account for this. The Free Body Diagram of the incline is shown in Figure 2.

In the absence of friction we only need consider the equation of motion in $X_2$ . The equation of motion is given by:

$\displaystyle \sum{F_{x_2}} = M \ddot{x_2} = N\sin\theta$ ...Eq1

Now turn to Figure 3., the Free Body Diagram of the Block.

The form of Newtons Law with non-inertial frames is given by $m\vec{\ddot{r}} = \vec{F} - m\vec{A}$

$\displaystyle \sum{F_{x_1}} = m\ddot{x_1} = mg\sin\theta + m\ddot{x_2}cos\theta$ ...Eq2

The last term on the right is the negative of the projection $\vec{a_2} \rightarrow \vec{a_1}$ . If you imagine we are riding on the block we are going to feel an additional inertial force pulling us down the incline. The same way when you accelerate forward in a car you are pushed back into the seat and visa versa.

Moving on to the force balance in $Y_1$ . There is no acceleration in the $Y_1$ direction. The force balance on the block in this direction given by:

$\displaystyle \sum{F_{y_1}} = 0 = N - mg\cos\theta + m\ddot{x_2}sin\theta$ ...Eq3.

Now we have these three equations of motion, that we can solve to find $\ddot{x_2}$ and $\ddot{x_1}$ .

Solving for $N$ in Eq3.

$\displaystyle N = mg\cos\theta - m\ddot{x_2}sin\theta$

Substitute $N$ into Eq1. and solve for $\ddot{x_2}$ and simplify.

$\displaystyle \ddot{x_2} = \frac{mg \sin\theta\cos\theta}{M+m\sin^2\theta}$

Then take that result and substitute into Eq2. and simplify using a common denominator.

$\displaystyle \ddot{x_1} = \frac{(M+m) g \sin\theta}{M+m\sin^2 \theta}$

Next, we need to figure out how long the force from the block is applied to the incline. All accelerations are constant, so we can just use kinematic relationships with the initial velocity of the block is zero, and the distance traveled down the incline $l$

$\displaystyle l = \frac{1}{2}\ddot{x_1}t^2 = \frac{1}{2}\left( \frac{(M+m) g \sin\theta}{M+m\sin^2 \theta}\right)t^2$

or

$\displaystyle t^2 = \frac{2l(M+m\sin^2\theta)}{(M+m) g \sin\theta}$

Similarly, $\Delta x_2$ is found by substituting $t^2$ , and simplifying. letting $l = \frac{b}{cos\theta}$ where $b$ is the length of the base.

$\displaystyle \Delta x_2 = \frac{1}{2}\ddot{x_2}t^2 = \frac{1}{2}\left(\frac{mg \sin\theta\cos\theta}{M+m\sin^2\theta}\right)\left(\frac{2l(M+m\sin^2\theta)}{(M+m) g \sin\theta}\right) = \left(\frac{m}{M+m}\right)b$

Finally, plugging in values $\Delta x_2 = \SI{20}{\cm}$