A 17.8-kg bucket of water is sitting on the end of a 3.7-kg, 3.00-m long board. The board is attachedto the wall at the left end and a cable is supporting the board in the middle. A. Determine the magnitude of the vertical component of the wall’s force on the board in newtons. B. What direction is the vertical component of the wall’s force in the board. c. The angle between the cable and the board is 40 degrees. Determine the magnitude of the tension in the cable in newtons.

[tex]\begin{gathered} F_y\cdot\left(-\frac{d}{2}\right)+\left(-Mg\right)\cdot\left(\frac{d}{2}\right)=0 \\ \\ \text{ We replacing:} \\ \\ F_y\cdot\left(-\frac{3}{2}\right)+\left(-17.8\cdot9.8\right)\cdot\left(\frac{3}{2}\right)=0 \\ \\ F_y=-\frac{2}{3}\cdot\left(17.8\cdot9.8\right)\cdot\left(\frac{3}{2}\right) \\ \\ F_y=-174.44\text{ N} \\ \\ |F_y|=174.44\text{ N} \end{gathered}[/tex]

B.

The negative sign shows that the force is downward. Therefore, the correct answer is downward

C.

We can determine the value of the tension, by means of the following balance at point A, thus

[tex]\begin{gathered} T\cdot\sin\theta\cdot\left(\frac{L}{2}\right)+\left(-Mg\right)\cdot\left(L\right)+mg\left(\frac{L}{2}\right)=0\: \\ \\ \text{ We replacing:} \\ \\ T\cdot\sin\:40°\:\cdot\left(\frac{3}{2}\right)+\left(-17.8\cdot9.8\right)\cdot\left(3\right)+3.7\cdot9.8\cdot\left(\frac{3}{2}\right)=0\: \\ \\ T=\frac{\left(17.8\cdot9.8\right)\cdot\left(3\right)-3.7\cdot9.8\left(\frac{3}{2}\right)}{\sin\:40°\:\cdot\left(\frac{3}{2}\right)} \\ \\ T=\frac{523.32-54.39}{0.96418}=\frac{468.93}{0.96418} \\ \\ T=486.35\text{ N} \end{gathered}[/tex]