Three equal weights A, B and C of mass 2kg each are hanging on an ideal string passing over an ideal pulley.

Question

The tension in the string between B and C is?

Let’s consider B and C as one unit of mass 4kg. The tension on both sides of the string will be T. Also assume that the the acceleration is upwards for A and downwards for B and C. Let g=10

The the equations shall be

For A $$T-20=2a$$ And for B and C $$40-T=4a$$

Solving these we get the a=30. You may have realized by now that I have done something wrong, but I haven’t yet. The right answer is 13N. How should I solve it?

EDIT:

The tension on the string for A will be T, but the downward force due to gravity will be 2(10)=20N

So net force is $F=T-20=(2)(a)$

Answer 1

Let's denote $T_{AB}$ to be the tension in the rope $AB$ and $T=T_{BC}$.

Then from kinematics, we know that accelerations of bodies are $a_A=a$, $a_B=a_C=-a$ (axis goes upwards). Newton's law for 3 bodies:

$$ ma = T_{AB} - mg,\qquad (A)\\ m(-a) = T_{AB}-T_{BC}-mg,\qquad(B)\\ m(-a) = T_{BC}-mg.\qquad(C) $$

We have 3 equations and 3 unknowns ($a$, $T_{AB}$, $T_{BC}$). We substract $(A)$ from $(B)$: $$ -ma-ma = -T_{BC},\qquad ma = \frac12 T_{BC}.\qquad (*) $$

Then we add $(C)$ and $(*)$: $$ ma - ma = \frac12T_{BC} + T_{BC}-mg, \qquad T_{BC}=\frac23mg. $$

If we let $g=10~\mathrm{m^2/s}$, then $T_{BC}=\frac23\times2\,\mathrm{kg}\times10~\mathrm{m^2/s}=13.3~\mathrm N$.

Answer 2

If you add the two equations you get $$20=6a\\ a=\frac {20}6\\ T=\frac 23\cdot 20=\frac {40}3N$$ This is not $13$, but close.