# A block on a wedge

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.

• What does Newton's third law tell you about the friction? What does it tell you about $a_{x}$ and $a_{2}$? Are your equations quite right? – Jerry Schirmer Jun 23 '13 at 4:36
• I'm not able to write out a full answer right now, but could you use center of mass to equate (with a proportionality factor based on their relative masses) the x-axis accelerations of the two masses? – AlexQueue Jun 23 '13 at 4:37

$m_1a_x=m_2a_2...(4)$
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.