Question

3: Determine the horizontal and vertical components of reaction at the pin A. (20pts) B A w 30°/ m m I

Use 5 m, 10 m, and 50 N/m

Answer 1

Solution - Drawing the free body diagram of the member AB we get -

W B N/m ) 60 А 30° А. 20/3 m 5 m Av m

Here value of W is -

$W = \frac{1}{2}(50\: \frac{N}{m})(10 \: m) = 250 \: N$

Now applying the equilibrium condition.

Taking sum of moment about point A equal to zero we get -

$(R_{B})(5) - (250)(\frac{20}{3}) = 0$

Now taking sum of forces in horizontal direction equal to zero we get -

$A_{H} = 166.667 \: (\rightarrow)$

Now taking sum of forces in vertical direction equal to zero we get -

$A_{V} = 38.675 \: (\uparrow)$

Thus horizontal reaction at the pin A is _{$A_{H} = 166.667 \: (\rightarrow)$}

And vertical reaction at the pin A is _{$A_{V} = 38.675 \: (\uparrow)$}