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.
No comments:
Post a Comment