Why are the two acceleration considered as a1 and a2 here? What is the reason for this even though the tension of both the strings are same.
T is equal to 5gTcos 53 for left side and 5gTcos37 for the right side
Why are the two acceleration considered as a1 and a2 here? What is the reason for this even though the tension of both the strings are same.
T is equal to 5gTcos 53 for left side and 5gTcos37 for the right side
As posed, you have three unknowns - $a1$, $a2$, and $a3$. The tension on the cable is everywhere uniform, but the accelerations are different because the static force components affecting the tension are different.
$F=ma$ at each mass. So for mass $m1$, the tension is
$m1*(g*cos53 +a1)$. Likewise
$m2*(g*cos37 + a2)$ and
$1/2*m3*(g*cos90 +a3)$.
These are all equal. And the length of the line doesn't change.
You can assume that you start from a stationary condition (all velocities are initially zero) - that doesn't change the accelerations. You can choose any initial length that like - that doesn't change the accelerations either.
So $m1(g*cos53 + a1) = 1/2 m3(g*cos90 + a3) = m2(g*cos37 +a2) $ from equal tension.
And $a1 = 2*a3 - a2$ from constant length.
$ 5 kg (0.600g + a1) = 3 kg (1g + a3) = 5 kg ( 0.799g + a2)$
$0.01 kg + 5 kg * a1/g = 3 kg * a3/g = 0.993 kg + 5 kg * a2/g$
$0.0033 + 1.666 a1/g = a3/g = 0.331 + 1.666 a2/g$
$0.0033 + 1.666 (2a3 - a2)/g = a3/g$
$0.0033 - 1.666 a2/g = -2.332 a3/g$
$-0.0014 + 0.7144 a2/g = a3/g $
$-0.0014 + .7144 a2/g = 0.331 + 1.666 a2/g$
$-0.3324 = .9516 a2/g$
$a2 = -0.349 g$
$a3 = -0.2509 g$
$a1 = -0.1529 g$