Question

Knowing the P is 1110 N, determine the tension in the rope if the frame is in equilibrium. Express your answer in kN to the nearest 10 N.

Answer 1

Explanation:

It is given that the value of P is 1110 N. And, for pin connection we have only two connections which are $A_{x}$ and $A_{y}$. Let T be the tension is rope.

So,  $\sum F_{y} = 0$ and $A_{y}$ - 1110 = 0

              $A_{y} = 1110 N$

     $\sum F_{x}$ = 0

And,    $T - A_{x}$ = 0

                  T = $A_{x}$

Also, $\sum M_{A}$ = 0

                1110(0.75 + 0.75 + 0.75) - T(0.5 + 0.1) = 0

                    2497.5 - 0.6T = 0

                         T = 4162.5 N

                             = 4.16 kN

Therefore, we can conclude that the tension in the rope if the frame is in equilibrium is 4.16 kN.