A 20-kg ornamental star is suspended by two chains fastened to horizontal beams at different heights with angles as shown. Determine the tension in each chain. (the angles are 40 and 60)
|dw:1590366465327:dw| referenced from the original diagram
let's call the tension force on the left rope T1 and the tension force on the right T2 reduce the tension forces into their horizontal and vertical components. horizontal: the left rope has horizontal component T1*cos(40°) and the right rope has horizontal component T2cos(60°) vertical: the left rope has vertical component T1*sin(40°) and the right rope has vertical component T2sin(60°) |dw:1590366572573:dw| |dw:1590366718824:dw|
now, since the ball is stationary, that means forces sum to 0 in both the horizontal and vertical directions horizontal: T1cos(40) = T2cos(60) vertical: considering gravity, T1sin(40) + T2sin(60) = mg where m is the mass of the ball (20kg) and g is the gravitational constant essentially you have a system of equations with two unknowns. You can solve the horizontal equation for either T1 or T2 by dividing both sides by either cos(40) or cos(60) as appropriate. then plug this into the vertical equation to reduce the vertical equation into one variable. solve for that tension, then plug back into the horizontal equation to solve for the other tension.