# Two blocks and a frictionless pulley problem

Block B ($m_{B}$=0.36 kg) is connected to a lightweight rope that passes over a lightweight, low-friction pulley.The other end of the rope is connected to Block A ($m_{A}$=0.72 kg), which is on a low-friction surface inclined at an angle θ above the horizontal. The pulley is attached to the top of the inclined plane.

If the two blocks remain at rest when released, what is the incline angle? If the mass of Block B is increased, does Block A slide up the incline, down the incline, or remain at rest? What if the mass of Block A is increased instead?

If the masses of Blocks A and B remain fixed at the values given in the problem statement above, for what range of incline angles will Object A move up the incline? Down the incline?

So far, I've drawn the free body diagrams and go the following equations based on Newton's Second Law:

For Block B: $$T-m_{B}g=m(-a)$$

For Block A: $$F_{x}=T-m_{A}\sin\theta g=m_{A}a$$ and $$N-m_{A}g \cos\theta=0$$

To me it seems like the same question is being asked twice but I'm not sure how to approach it. Can I get some help?

• Your equations are right (apart you should write $m_B$ instead of $m$ in the first one). What's the problem then? Jun 19 '16 at 19:41
• I'm having trouble finding ways to answer: 1)If the two blocks remain at rest when released, what is the incline angle? If the mass of Block B is increased, does Block A slide up the incline, down the incline, or remain at rest? What if the mass of Block A is increased instead? If the masses of Blocks A and B remain fixed at the values given in the problem statement above, for what range of incline angles will Object A move up the incline? Down the incline? Jun 19 '16 at 19:57

Subtracting the first equation from the second yields: $$m_Bg - m_Ag \sin\theta = (m_A + m_B)a$$ Note that $m_A = 2m_B$, so we have: $$m_Bg - 2m_Bg \sin\theta = 3m_Ba$$ Now if the blocks remain at rest when released, then $a = 0$, so: \begin{align*} m_Bg - 2m_Bg \sin\theta &= 0 \\ m_Bg &= 2m_Bg \sin\theta \\ \frac{1}{2} &= \sin\theta \\ \theta &= 30^\circ \end{align*}
• Now, would it be fair to say that if $m_Asin30$ is less than $m_B$ then a moves positively which means that B will weigh downwards, and $m_Asin30$ is greater than $m_B$ then a moves negatively meaning B will move upwards and A will move down the slope? Is that how to interpret all this? Jun 20 '16 at 3:05