91Ó°ÊÓ

Theconcreteblock of mass$225kg$

Strut of mass$45.0kg$

$$\begin{gathered} \mathrm{ϕ}=30.0° \\ \mathrm{θ}=45.0° \end{gathered}$$

In the free body diagram of the given situation, we notice the acting torques on the body both horizontally and vertically.Here, the acting torque results from the applied force along the length of the cable hanging from the strut and along the weight of the block to balance it at the strut.The vertical force acting on the body about the hinge is due to the tension resulting from the hanging to the strut, also the downward pull due to the weight of the block, and the downward pull due to the weight of the strut that acts at the center of the strut. Then using Newton's laws of motion, we balance the acting torques in an equation separately for both the horizontal and the vertical directions. Then, accordingly, calculate the required values as per the problem

We note that the angle between the cable and the strut is

$\begin{array}{rcl}\mathrm{Î\pm }& =& \mathrm{θ}-\mathrm{Ï•}\\ & =& 45°-30°\\ & =& 15°\\ & & \end{array}$

The angle between the strut and any vertical force(like the weights in the problem) is

$\begin{array}{rcl}\mathrm{β}& =& 90°-45°\\ & =& 45°\\ & & \end{array}$

Denoting$M=225kgandm=45.0kg$and $\mathrm{â„“}$as the length of the boom, we compute torques about the hinge and find

$\begin{array}{rcl}T& =& \frac{Mg\mathrm{â„“}sin\mathrm{β}+mg\frac{\mathrm{â„“}}{2}sin\mathrm{β}}{\mathrm{â„“}sin\mathrm{Î\pm }}\\ & =& \frac{Mgsin\mathrm{β}+mgsin\frac{\mathrm{β}}{2}}{sin\mathrm{Î\pm }}\\ & & \end{array}$

Theunknown lengthcancels out and we obtain$T=6.63×{10}^{3}N$

The tension in the cable is.$6.63×{10}^{3}N$

(b)

Since the cable is at$30°$from horizontal, then horizontal equilibrium of forces requires that the horizontal hinge force be

$\begin{array}{rcl}Fx& =& Tcos30°\\ & =& 5.74×{10}^{3}N\\ & & \end{array}$

Horizontal component of the force is$5.74×{10}^{3}N$

(c)

Vertical equilibrium of forces gives the vertical hinge force component

$\begin{array}{rcl}Fy& =& Mg+mg+Tsin30°\\ & =& 5.96×{10}^{3}N\\ & & \end{array}$

Vertical component of the force is$5.96×{10}^{3}N$