The system is as follows –
Friction exists only between the 2 blocks.
I am trying to find out the accelerations of $m_1$ and $m_2$.
Let $a_2$ be acceleration of $m_2$, and $a_x$ and $a_y$ be the accelerations of $m_1$ in the respective directions.
Let $R$ be the normal reaction between the 2 blocks, and $N$ be the normal reaction between $m_2$ and floor.
Balancing components across the axes, I get the following equations –
$$N = m_2g + R\cos\theta \tag{1}$$
$$m_2a_2 = R\sin\theta \tag{2}$$
$$a_x = R(\sin\theta + \mu_s\cos\theta) \tag{3}$$
$$a_y = R(\cos\theta + \mu_s\sin\theta) – m_1g \tag{4}$$
I don’t think $(1)$ is necessary, since friction is not involved between the blocks and the ground.
Leaving that aside, I have 3 equations in 4 variables: $a_x, a_y, a_2, R$.
Is there are any way I could perhaps get a 4th equation so that the system of equations could be solved?
I can get $|a_1|$ in terms of $R$ from the expressions for $a_x$ and $a_y$, but I don’t think that would help.
Best Answer
If you consider the two blocks as a system, it can be observed that there is no external force acting on the system in the horizontal direction, so the center of mass will have no acceleration in the horizontal direction. This give:
$m_1a_x=m_2a_2...(4)$
There you have your fourth equation.
A better method to solve this problem would be to observe $m_1$ from the frame of $m_2$. We will have to apply a pseudo force on $m_1$ equal to $m_1a_2$ in magnitude. In this frame $m_1$ is constrained to move along the incline only, so you can consider it to have an acceleration $a_1$ along this incline instead of assuming $a_x$ and $a_y$. This reduces no. of variables will get.