Two bodies of mass $m_1=50 \text{kg}$ and $m_2 = 10 \text{kg}$ are connected with a light rope through a pulley (no friction between the rope and the pulley). $m_2$ moves on surface with angle $37^{\circ}$ with a friction coefficient $\mu_k=0.5$. Force $\vec F$ acts on $m_1$. The system was moving with a constant velocity for some time until the force stopped and then the system started to accelerate. What's the acceleration of the system?

Free body diagram

I assumed that the system accelerates to the left, therefore the equation for the $m_1$ would be: $ \begin{cases} N_1=m_1 g\\ T-N_1 \mu=m_1a \end{cases} $

Therefore $T=m_1(a+g \mu)$

For $m_2$ the equation would be: $ \begin{cases} N_2=m_2 g \cos 37^{\circ}\\ m_2 g \sin 37^{\circ} - T - N_2 \mu=m_2a \end{cases} $

Therefore $m_2 g \sin 37^{\circ} -T- m_2 g \cos 37^{\circ} \mu = m_2a $

The solution would be:

$100 \cdot 0.2024 - 50a -250=10a \\ a \approx -3.829 \frac{\text{m}}{\text{s}^2}$

Which is wrong according to the answers. The answer should be $\approx -5.83\frac{\text{m}}{\text{s}^2}$ which you can get pretty close if instead of $10a$ in the last step it was $-10a$. So I'm pretty sure that my mistake here is with the signs (of the acceleration), but I can't figure out why I'm wrong.


closed as off-topic by AccidentalFourierTransform, ZeroTheHero, Yashas, Qmechanic May 2 '17 at 4:57

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  • $\begingroup$ Could you take the numbers out and replace them with variables (or at least numbers with units)? It's really hard to follow what you're doing when all I see are numbers with no way of determining their significance. $\endgroup$ – Dan Mar 30 '13 at 23:01

Answer: The problem says that the system was moving right with constant velocity. So, all the friction forces will work towards left.

Hence, the equations will become $$T+N_1 \mu = m_1 a$$


$$m_2 g \sin 37°- T + m_2 g \cos 37° = m_2a$$

Solving these will give you the right answer!

Elementary, my dear Watson!


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