Chapter 10, Problem 10.19P

Determine the force F and the angle a required to hold the linkage in the position ${\theta }_1=60°,\theta =15°.$ Each bar of the linkage is homogeneous and of weight W.

To determine

Force F and angle $\alpha$ required to hold the linkage.

Answer to Problem 10.19P

The force F is equal to $0.753W$ and angle $\alpha$ is $24.89°$ with horizontal.

Explanation of Solution

  $\begin{array}{l}{\theta }_1=60°\\ {\theta }_2=15°\end{array}$

Virtual work $\delta U$ done by force F is defined as:

  $\delta U=F.\delta r=F\cos \alpha \delta r$

In above equation:

  $F$ - Force

  $\delta r$ - Virtual force

  $\alpha$ - Angle between $F$ and $\delta r$

Virtual displacement $\delta r_B$ of point B is defined as:

  $\delta r_B=r_{AB}\delta \theta$

In above equation:

  $\delta \theta$ - Virtual rotation

  $r_{AB}$ - Magnitude of vector drawn from A to B:

Calculation:

The length $y_1$ :

  $y_1=\frac{L}{2}\sin {\theta }_1$

The virtual displacement $\delta y_1$ :

  $\delta y_1=\frac{L}{2}cos{\theta }_1\delta {\theta }_1$

The length $y_2$ :

  $y_2=L\sin {\theta }_1+\frac{L}{2}\sin {\theta }_2$

The virtual displacement $\delta y_2$

  $\delta y_2=L\cos {\theta }_1\delta {\theta }_1+\frac{L}{2}\cos {\theta }_2\delta {\theta }_2$

The length $y_C$ :

  $y_C=L\sin {\theta }_1+L\sin {\theta }_2$

The virtual displacement $\delta y_C$ :

  $\delta y_C=L\cos {\theta }_1\delta {\theta }_1+L\cos {\theta }_2\delta {\theta }_2$

The length $x_C$ :

  $x_C=L\cos {\theta }_1+L\cos {\theta }_2$

The virtual displacement $\delta x_C$ :

  $\delta x_C=-L\sin {\theta }_1\delta {\theta }_1-L\sin {\theta }_2\delta {\theta }_2$

Virtual work $\delta U$ done by acting forces:

  $\begin{array}{l}\delta U=W\delta y_1+W\delta y_2+F_x\delta x_C-F_y\delta y_C\\ =W\frac{L}{2}cos{\theta }_1\delta {\theta }_1+W\left(L\cos {\theta }_1\delta {\theta }_1+\frac{L}{2}\cos {\theta }_2\delta {\theta }_2\right)\\ +F_x\left(-L\sin {\theta }_1\delta {\theta }_1-L\sin {\theta }_2\delta {\theta }_2\right)-F_y\left(L\cos {\theta }_1\delta {\theta }_1+L\cos {\theta }_2\delta {\theta }_2\right)\\ =\left(\frac{3}{2}Wcos{\theta }_1-F_x\sin {\theta }_1-F_y\cos {\theta }_1\right)L\delta {\theta }_1+\left(\frac{1}{2}W\cos {\theta }_2-F_x\sin {\theta }_2-F_y\cos {\theta }_2\right)L\delta {\theta }_2\end{array}$

We know that:

  $\delta U=0$

Therefore:

  $\frac{3}{2}Wcos{\theta }_1-F_x\sin {\theta }_1-F_y\cos {\theta }_1=0$

Substitute:

  $\begin{array}{l}\frac{3}{2}Wcos60°-F_x\sin 60°-F_y\cos 60°=0\\ 0.866F_x+0.5F_y=0.75W\to \left(1\right)\end{array}$

  $\begin{array}{l}\frac{1}{2}W\cos 15°-F_x\sin 15°-F_y\cos 15°=0\\ 0.2588F_x+0.9659F_y=0.4829W\to \left(2\right)\end{array}$

Solve above equations:

  $\begin{array}{l}{F}_x=0.683W\\ F_y=0.317W\end{array}$

The magnitude and angle $\alpha$ of force F:

  $\begin{array}{l}F=\sqrt{F_x{}^2+F_y{}^2}=0.753W\\ \alpha ={\tan }^{-1}\left(\frac{0.317}{0.683}\right)=24.89°\end{array}$

Conclusion:

The force F is equal to $0.753W$ and angle $\alpha$ is $24.89°$ with horizontal.