Question

consider the figgure below and the following information

support E is fixed

Cable CD is in tension

Pin B is fixed in member AC

Determine the magnitude of the support reaction , in N at pin A

Consider the figure below and the following information: Support E is fixed. Cable CD is in tension. Pin B is fixed in member AC. Determine the magnitude of the support reaction, in N, at pin A. D 60 N 100 mm 120 N B 160 mm 140 mm A E 180 mm

Answer 1

Solution:

Given that:

• Draw FBD of member FGB and calculate pin force at B, using equlibrium equation:

In case GON Asing 120N L H B a 160mm c=140mm Gi May K b=180mm

$\varepsilon M_{G}=0\, \, \textrm{(Net moment of all forces about point G=0)}$

$=120\times a-N\sin \theta \times c-N\cos \theta \times b=0\, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\left\{\begin{matrix} \tan \theta =\frac{140}{180}\\ \sin \theta =0.61 \\ \cos \theta =0.79 \end{matrix}\right.$

$=120\times 160-0.61\,N\times 140-0.79\,N\times 180=0$

$\Rightarrow N=84.36\,N\, \, \,\rightarrow \textrm{Pin force at B}$

• Draw FBD of member ABC and calculate pin force at A.

$N\cos \theta =84.36 \times 0.79=66.64\,N$

$N\sin \theta =84.36 \times 0.61=51.46\,N$

Now equilibrium equations:

It e=240mm 51.46 B d=1401 Ax 'Ay

$\varepsilon M_{A}=0$

$51.46\times d=T\times \rho =0$

$\Rightarrow 51.46\times 140-T\times 240=0$

$T=30\,N\, \, \rightarrow \textrm{Tension in cable}$

$\varepsilon F_{x}=0$

$A_{X}+51.46-30=0$

$A_{X}=-21.46\,N\, \, \, \rightarrow \textrm{Support reaction at A}$