Please help me out here. thanks
in the attachment you'll find a your slightly modified diagram.
You are dealing with a parallelogram of forces. Take the left half of the parallelogram. Since AD || BC you only have to calculate the side AD = BC. Use Sine rule:
AD/70 N = sin(45°)/sin(30°)
AD = 140*1/2*sqrt(2) N = 70*sqrt(2) N ≈ 98.994949.. N ≈ 99 N
This is a Newton's 2nd Law problem.
Let the tension on the left be T1 = 70 N and the tension on the right be T2. Let the mass be m. I will use a coordinate system with +x to the right and +y upward.
Now take Newton's 2nd Law in each coordinate direction acting on the point where the 3 strings meet:
(Sum)F_x = -70*cos(45) + T2*cos(60) = 0 <-- Since this point has no mass (and is not accelerating either.)
(Sum)F_y = 70*sin(45) + T2*sin(60) - mg = 0
From the x equation we know that:
T2 = 70*[cos(45)/cos(60)] N = 98.9949 N
As a bonus:
Inserting this into the second equation we get:
70*sin(45) + (98.9949)*sin(60) - mg = 0
m = 1/g*[70*sin(45) + (98.9949)*sin(60)] kg =