Chapter 12, Problem 23P

(a)

To determine

To Find:The force exerted on the strut by the hinge if the strut is weightless.

Answer to Problem 23P

  $\overrightarrow{F}=\left(30.0N\right)\widehat{i}+\left(30.0N\right)\widehat{j}$

Explanation of Solution

Given data:

The strut is weightless.

Weight of the block, 60 N

Formula used:

Static translational equilibrium condition:

  ${\displaystyle \sum \overrightarrow{F}=0}$

Calculation:

  

The force exerted on the strut at the hinge:

  $\overrightarrow{F}=F_h\widehat{i}+f_v\widehat{j}\left(1\right)$

Ignore the weight of the strut and use ${\displaystyle \sum \overrightarrow{\tau }=0}$ at the hinge:

  $LT-\left(L\cos {45}^{\circ }\right)W=0$

  $\begin{array}{c}T=W\cos {45}^{\circ }\\ =\left(60\text{N}\right)\cos {45}^{\circ }\\ =42.43\text{N}\end{array}$

Use ${\displaystyle \sum \overrightarrow{F}=0}$ for the strut:

  ${\displaystyle \sum F_x=F_h}-T\cos {45}^{\circ }=0$

  ${\displaystyle \sum F_y=F_v}+T\cos {45}^{\circ }-Mg=0$

Solve for $F_h$ :

  $\begin{array}{c}{T}_h=T\cos {45}^{\circ }\\ =\left(42.43\text{N}\right)\cos {45}^{\circ }\\ =30\text{N}\end{array}$

Solve for $F_v$ :

  $\begin{array}{l}{F}_v=Mg-T\cos {45}^{\circ }\\ =60N-\left(42.43\text{N}\right)\cos {45}^{\circ }\\ =30.0\text{N}\end{array}$

Subtitute in equation $\left(1\right)$ to obtain:

  $\overrightarrow{F}=\left(30.0N\right)\widehat{i}+\left(30.0N\right)\widehat{j}$

Conclusion:

  $\overrightarrow{F}=\left(30.0N\right)\widehat{i}+\left(30.0N\right)\widehat{j}$

(b)

To determine

The force exerted on the strut by the hinge if the strut weighs $20\text{ N}$ .

Answer to Problem 23P

  $\overrightarrow{F}=\left(35.0N\right)\widehat{i}+\left(45.0N\right)\widehat{j}$

Explanation of Solution

Given data:

Strut weighs, $20\text{ N}$

Weight of the block, 60 N

Formula used:

Translational and rotational equilibrium:

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

Calculation:

Including the weight of the strut, apply ${\displaystyle \sum \overrightarrow{\tau }=0}$

  $LT-\left(L\cos {45}^{\circ }\right)W-\left(\frac{L}{2}\cos 45\right)w=0$

  $\begin{array}{l}T=\left(\cos {45}^{\circ }\right)W+\left(\frac{1}{2}\cos {45}^{\circ }\right)w\\ =\left(\cos {45}^{\circ }\right)\left(60\text{N}\right)+\left(\frac{1}{2}\cos {45}^{\circ }\right)\left(2\text{0N}\right)\\ =49.5\text{N}\end{array}$

Apply ${\displaystyle \sum \overrightarrow{F}=0}$ to the strut:

  ${\displaystyle \sum F_x=F_h-T\cos {45}^{\circ }=0}$

And

  ${\displaystyle \sum F_y=F_v+T\cos {45}^{\circ }-W-w=0}$

Solve for $F_h$ :

  $\begin{array}{l}{T}_h=T\cos {45}^{\circ }=\left(\text{49}\text{.5N}\right)\cos {45}^{\circ }\\ =35.0\text{N}\end{array}$

Solver for $F_v$ :

  $\begin{array}{l}{F}_v=W+w-T\cos 5^{\circ }\\ =\text{60}\text{N+20}\text{N-}\left(\text{49}\text{.5}\text{N}\right)\cos {45}^{\circ }\\ =45.0\text{2}\text{N}\end{array}$

Substitute in equation $\left(1\right)$ to obtain:

  $\overrightarrow{F}=\left(35.0N\right)\widehat{i}+\left(45.0N\right)\widehat{j}$

Conclusion:

  $\overrightarrow{F}=\left(35.0N\right)\widehat{i}+\left(45.0N\right)\widehat{j}$