Question

Determine the maximum weight of the flowerpot that can besupported without exceeding a cable tension of 50 lb in eithercable AB or AC.

Answer 1

Answer:

Maximum weight = 76.64 lb

Explanation:

In the second figure attached it can be seen the free body diagram  of the problem.

The equation of equilibrium respect x-axis is

$\sum F_x = 0$

$F_{AC} \times sin 30 - F_{AB} \times \frac{3}{5} = 0$

$F_{AC} = 1.2 \times F_{AB}$

So, cable segmet AC supports bigger tension than segment AB. If $F_{AC} = 50 lb$ then

$F_{AB} = \frac{F_{AC}}{1.2}$

$F_{AB} = 41.67 lb$

The equation of equilibrium respect y-axis is

$\sum F_y = 0$

$F_{AB} \times \frac{4}{5} + F_{AC} \times cos 30 - W = 0$

$41.67 lb \times \frac{4}{5} + 50 lb \times cos 30 - W = 0$

$W = 76.64 lb$