howework: f5-1:determine the horizontal and vertical componengts of ewaction at the supports.neglect the…

Explanation:

Step1: Resolve the 2500 - N force

The horizontal component of the 2500 - N force is $F_{x}=2500\times\frac{4}{5}=2000$ N and the vertical component is $F_{y}=2500\times\frac{3}{5}=1500$ N.

Step2: Sum of moments about point A

Let the vertical reaction at $B$ be $V_{B}$ and the horizontal reaction at $A$ be $H_{A}$ and vertical reaction at $A$ be $V_{A}$. $\sum M_{A}=0$: $V_{B}\times(1.5 + 1.5)-1500\times1.5-900 = 0$. $3V_{B}-2250 - 900=0$. $3V_{B}=3150$. $V_{B}=1050$ N.

Step3: Sum of horizontal forces

$\sum F_{x}=0$: $H_{A}-2000 = 0$, so $H_{A}=2000$ N.

Step4: Sum of vertical forces

$\sum F_{y}=0$: $V_{A}+V_{B}-1500 = 0$. $V_{A}+1050 - 1500=0$. $V_{A}=450$ N.

Answer:

Horizontal reaction at $A$: $H_{A}=2000$ N, Vertical reaction at $A$: $V_{A}=450$ N, Vertical reaction at $B$: $V_{B}=1050$ N.