Question

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom rises from pin A to B, at angle phi from horizontal. A cable goes up from the load to B and then left to a pulley.
Determine the greatest weight of the crate that can be supported without causing the cable to fail if ϕ=30∘. Neglect the size of the winch.

Answer 1

Explanation:

Let us assume that forces acting at point B are as follows.

        $\sum F_{x} = 0$

        $T + F_{AB} Sin 60$ = 0 ...... (1)

       $\sum F_{y}$ = 0

       $F_{AB} Cos 60 + W$ = 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

               $\sigma_{allow} = \frac{T}{A}$

       T = $(20 \times 1000) \frac{\pi}{4} \times (0.5)^{2}$

          = 3925 kip

From equation (1),   $F_{AB}Sin (60^{o})$  = -3925

               $F_{AB} \times -0.304 8$ = -3925

             $F_{AB}$ = 12877.29 kip

From equation (2),    -12877.29 (Cos 60) + W = 0

         $-12877.29 kip \times \frac{1}{2} + W$ = 0

                           W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.