Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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

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:

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.

share|cite|improve this answer

You can find the fourth equation by using constraint relations. Try to find a relation between the displacements of the wedge and the block and then differentiate it twice wrt time. there you get a relation between the accelerations. here it will be X of block = sine component of acceleration of wedge.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.