Really Need YOUR Help!!!?

the system is not in equilibrium and moves towards the side of the m1block. let there be tension t in the string and let the blocks move at acceleration a.

then for m1

m1g - t =m1 a ...................(1)

for m2

t -m2gsin36 = m2a ...............(2)

adding (1) &(2) and solving we get

a=4.6131866m/s sqr. towards direction of motion i.e.in direction m1 moving down which is also the acceleration of m2 .i.e 5kg block

subtiuting in any one of the eqns we get

t=51.86 N towards the opp. direction of motion

Forces on m1: m1*g and T, and according to 2nd law of motion: m1*a = m1*g - T, so T = m1*(g - a)

Forces on m2: m2*g downward, Fn - normal force perpendicular to incline, T along the incline. Resultant force is along the incline and according to 2nd law of motion equal to m2*a. Looking at components of forces parallel to incline we can write: m2*a = T - m2*g*sin(A), so T = m2*a + m2*g*sin(A) = m2*(a + g*sin(A)).

Now equate 2 expressions for T and get

m1*(g - a) = m2*(a + g*sin(A)).

m1*g - m2*g*sin(A) = m2*a + m1*a

a = (m1*g - m2*g*sin(A))/(m1 + m2)

To find T substitute a into T = m1*(g - a) and get

T = m1((m2*g + m2*g*sin(A))/(m1 + m2)) = m1*m2/(m1 + m2) *g(1 + sin(A))

Okay, I don't understand any of that 0-o