Second Law of Dynamics and Pulleys

Question

The system in the figure is initially at rest . If the system starts to move , evaluate the acceleration of the block $A$. Mass of the block $A$ is $m_A = 60\;\mathrm{kg}$ ; mass of blocks $B$ and $C$ are same and $m_B=m_C=20\;\mathrm{kg} $. (There's no friction between the bodies and the mass of the pulleys is negligible).

Answer: $a_{A} = 2\;\mathrm{m/s^2}$

First problem: my solution isn't working.

I've been trying to solve this problem in the non inertial frame of reference, i.e., in the frame of reference of the block $A$, using Einstein's Equivalence Principle, where inertial forces are acting. Inertial forces acts on $A$, $B$ and $C$, and $A$ is at relative rest in it's own frame of reference. In the frame of reference of $A$, $B$ is moving to the left while $C$ move to the left and moves down simultaneously. I've written the equations using the Newton's Second Law for the three bodies, but my answer differ from the answer of the book.

Second problem: I dont understand the solution's manual.

The book solve the problem in the inertial frame of reference of Earth. He writes Newton's Second Law for the three bodies (no problem here) and then say that we can consider the block $A$ as a "giant moving pulley", so we have $\vec{a_{A}} = \frac{\vec{a_{B}}+\vec{a_{C}}}{2}$, and then, combining the Second Law with this vector equation, we have $a_{A} = 2\;\mathrm{m/s^2}$ I understand the vector equation for moving pulleys, but I don't understand why we can take the block $A$ as a "giant moving pulley". Can someone explain this please?

Answer 1

I think it is easier to consider to the string length. String length is constant, so we have:

$$x_C+2x_A-x_B=\textrm{constant}$$ $$\Longrightarrow\;a_A=\large{\frac{a_B-a_C}2}\;\tag 1$$ On the other hand, we know: $$m_Cg-T=m_Ca_C\;\tag 2$$ $$T=m_Ba_B\;\tag 3$$ $$-2T=m_Aa_A\;\tag 4$$ (You can obtain last three equations by drawing the free body diagram for $A$, $B$ and $C$)

Now, we have four equations and four unknowns so that we can solve them.

Answer 2

The system can be reduced to one with only one pulley:

Using Newton's second law you can get three equations for each of the masses but as is often the case in such problems you need a fourth as there are four unknowns.
The fourth equation you find by looking at the geometry of the system assuming that the string is inextensible.

A way of getting that fourth equation is as follows:
Assume that mass $m_B$ is stationary and move mass $m_C$ a distance $x_C$ to the right.
Because the length of the string is constant mass $m_A$ moves a distance $\frac{x_C}{2}$ to the right.
Now repeat with mass $m_C$ fixed and mass $m_B$ moving a distance $x_C$ to the right.
In this case mass $m_A$ moves a distance $\frac{x_B}{2}$ to the right.

Now using superposition with both mass $m_B$ and $m_C$ moving $x_B$ and $x_C$ to the right means that mass $m_A$ moves a distance $\frac{x_B + x_C}{2}$ to the right $( =x_A)$.

Now differential with respect to time twice to get $a_A =\dfrac{a_B + a_C}{2}$.

Answer 3

Although Block B rests on Block A, the motion of Block B does not affect the motion of Block A, since there is no friction between them. The only forces on Block A are two tension forces in the string.

So you can imagine that Block B rests on the smooth table, instead of on Block A. Both Blocks A and B then slide on the smooth table. Block A is attached to the 2 small pulleys, and can be lumped with them into a single pulley which slides to the right, as in Farcher's diagram.

A word of caution : unlike the pulleys Block A does not rotate, so the "giant pulley" idea could not be used if the rotational KE of the pulleys had to be taken into account.

If you don't like the giant pulley idea (I don't), draw separate free-body diagrams for each of Blocks A, B, C. The equations of motion are :
$m_A a_A = 2T$
$m_B a_B = -T$
$m_C a_C = m_C g - T$.
As lucas shows, the accelerations are related by the fixed length of the string.