Chapter 12, Problem 17P

To determine

To Find:The tension in the backstay and the normal force that the deck exerts on the mast.

Answer to Problem 17P

The tension in thee backstay is $692\text{ N}$ .

The normal force exerted by the deck on the mast is $2.54\text{ kN}$ .

Explanation of Solution

Given data:

Free body diagram for the given situation is as follows:

Formula used:

Apply the condition of translational and rotational equilibrium.

  $\begin{array}{l}{\displaystyle \sum F=0}\\ {\displaystyle \sum \tau =0}\end{array}$

Calculation:

Apply the condition of rotational equilibrium to the mast. The net torque due to tension by the wires is equal to zero and is represented as:

  ${\displaystyle \sum \tau =0}$

Torque due to the left side of the pole is given as

  $d_B{T}_F\sin \theta =0$

Torque due to right side of the pole is given as

  $d_B{T}_F\sin {45}^{\circ }=0$

Since the torque due to right side of the wire is in clock wise direction, it is negative.

Thus, the torque of horizontal components according to the condition of rotational equilibrium is given as

  $d_B{T}_F\sin \theta -d_B{T}_B\sin {45}^{\circ }=0$

Here, $T_F$ is the tension in the forestay, $T_B$ is the tension in the backstay, and $d_B$ is the distance of the backstay.

Substitute $4.88\text{ m}$ for $d_B$ and $1000\text{ N}$ for $T_F$ in equation of net torque

  $\begin{array}{l}\left(4.88\text{ m}\right)\left(1000\text{ N}\right)\sin {\theta }_F-\left(4.88\text{ m}\right)T_B\sin {45}^{\circ }=0\\ T_B=\frac{\left(1000\text{ N}\right)\sin {\theta }_F}{\sin {45}^{\circ }}\end{array}$

From the figure, angle of forestay is obtained as follows:

  ${\theta }_F={\tan }^{-1}\left(\frac{d_F}{d_B}\right)$

Here, $d_F$ is the distance of the forestay.

Substitute $2.74\text{ m}$ for $d_F$ and $4.88\text{ m}$ for $d_B$

  $\begin{array}{l}{\theta }_F={\tan }^{-1}\left(\frac{2.74\text{ m}}{4.88\text{ m}}\right)\\ ={29.3}^{\circ }\end{array}$

Determine the tension in the backstay as below.

Substitute the ${29.3}^{\circ }$ for ${\theta }_F$ in equation $T_B=\frac{\left(1000\text{ N}\right)\sin {\theta }_F}{\sin {45}^{\circ }}$

  $\begin{array}{l}{T}_B=\frac{\left(1000\text{ N}\right)\sin {29.3}^{\circ }}{\sin {45}^{\circ }}\\ =692\text{ N}\end{array}$

Therefore, the tension in the backstay is $692\text{ N}$ .

Torque due to the left side of the pole is given as,

  $T_F\cos {\theta }_F=0$

Torque due to right side of the pole is given as,

  $T_B\cos {45}^{\circ }=0$

Since the torque due to right side of the wire is in clock wise direction, it is negative.

Thus, the net torque of vertical components according to the condition of rotational equilibrium is given as,

  $F_D-T_F\cos {\theta }_F-T_B\cos {45}^{\circ }-mg=0$

Here, $T_F$ is the tension in the forestay, $T_B$ is the tension in the backstay, $d_B$ is the distance of the backstay, $F_D$ is the normal force by the deck, $m$ is the mass of the pole, and g is the acceleration due to gravity.

Re-arrange the above equation for $F_D$ .

  $F_D=T_F\cos {\theta }_F+T_B\cos {45}^{\circ }+mg$

Substitute $1000\text{ N}$ for $F_D$ , ${29.3}^{\circ }$ for ${\theta }_F$ , $692\text{ N}$ for $T_F$ , $120\text{ kg}$ for $m$ and $9.81{\text{ m/s}}^2$ for g in equation in above re-arranged equation.

  $\begin{array}{c}{F}_D=\left(1000\text{ N}\right)\left(\cos {29.3}^{\circ }\right)+\left(692\text{ N}\right)\left(\cos {45}^{\circ }\right)+\left(120\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\\ =2.54\text{ kN}\end{array}$

Conclusion:

The normal force exerted by the deck on the mast is $2.54\text{ kN}$ .