Question

Two cords support a chandelier in the manner shown in (Figure 1), the upper cord makes an angle of 45° with the ceiling

Problem 9.07 static Equilibrium

Answer 1

Vasº pfly fix (45° F eff fix fix bu

Apply static equilibrium conditions, the net force along and
у
directions are equal to zero. Mathematically given as,

ΣF = 0

And

ΣF, = 0

If we look at the point where the cords come together, you can express the equilibrium of that point as follows: (The vector sum of F₁, F₂, and F₃ - where is the weight of the chandelier)

In the x direction:
F_1x + F_2x + F_3x = 0
-F₁cos(45^o)+ F_2x + 0 = 0

So we know that F_2x = F₂ (It has no y component) = F₁cos(45^o)

In the y direction
F_1y + F_2y + F_3y = 0

Now, F₁ is always going to be larger than F₂ and F₃ which are both equal to the weight(mg), since sin(45^o) is less than one,
and F₁ = (mg)/sin(45^o)

Now let's solve the problem, making F₁ just equal to 1390 N, as this would be the maximum it could be:

In the x direction:
F_1x + F_2x + F_3x = 0
-(1390 N)cos(45^o)+ F_2x + 0 = 0
-982.9 N + F_2x + 0 = 0
So, F_2x = 982.9 N

In the y direction,
F_1y + F_2y + F_3y = 0
(1390 N)sin(45^o)+ 0 - mg = 0
982.9 - mg = 0

So, the weight (mg (max))= 982.9 N = 980 N _Ans.