Conceptually, you know that the blocks will have the SAME acceleration and the SAME tension.
This is easier to see with a Free-Body Diagram, but I’ll try to explain it as best as I can. You need to first pick a direction that the blocks are moving. For this part, it doesn’t matter which way you designate the motion, it is just for keeping our positive a negative labels consistent. I am going to state that the hanging block is moving DOWN and the other block is moving UP THE INCLINE.
Now, just solve the following equation for both blocks:
ΣF = m(a)
Where:
ΣF = sum of forces acting along the direction of motion
m = mass
a = acceleration
BLOCK 1
ΣF1 = m1(a)
Where:
ΣF1 = T – W1sin30
m1 = mass of block 1 = 3.70kg
a = acceleration = unknown
T = tension = unknown
W1 = weight of block 1 = m1(g) = 37 N
T – W1(sin28) = m1(a)
T – (37 N)(sin 28) = (3.70kg)a
T – 17.37N = (3.70kg)(a)
BLOCK 2
ΣF2 = m2(a)
Where:
ΣF2 = W2 – T
m2 = mass of block 2 = 1.86kg
a = acceleration = unknown
T = tension = unknown
W2 = weight of block 2 = m2(g) = 18.6N
W2 – T = m2(a)
18.6N – T = (18.6kg)(a)
Now, solve for T in the block 1 equation and plug that into the equation for block 2
T = 17.37N + (3.70kg)(a)
18.6N – (17.37N + (3.70kg)(a)) = (1.86Kg)(a)
Now solve for a
