Chapter 2, Problem 2.80P

The figure shows one-half of a universal coupling known as the Hooke's joint. The coupling is acted on by the three couples shown: (a) the input couple consisting of forces of magnitude P, (b) the output couple $C_0,$ and (c) the couple formed by bearing reactions of magnitude R. If the resultant of these couples is zero, compute R and $C_0$ for $P=750\text{lb}\text{.}$

To determine

(i)

The input couple consisting of forces of magnitude P.

Answer to Problem 2.80P

  $TheinputcoupleconsistingofforcesofmagnitudePis4500\mathrm{l}b-in.$

Explanation of Solution

Given:

P= 750 lb

Concept Used:

  $Couple=r\times P.\widehat{n}$

Calculation:

  $\begin{array}{l}TheinputCouplefromtheforces=r\times P.\widehat{n}\\ C=6P\widehat{i}\\ C=6\times 750\widehat{i}\\ C=4500\mathrm{l}b-in\end{array}$

Conclusion:

  $TheinputcoupleconsistingofforcesofmagnitudePis4500\mathrm{l}b-in.$

To determine

(ii)

The output couple C_o.

Answer to Problem 2.80P

The output couple C_ois 3900 lb-ft.

Explanation of Solution

Given:

P= 750 lb

Calculation:

  $\begin{array}{l}{\overrightarrow{C}}_o=C_o(-\cos 30°\widehat{i}+\sin 30°\widehat{k})\\ {\overrightarrow{C}}_o=C_o(-0.8660\widehat{i}+0.5\widehat{k})\\ C_R=2R(-\sin 30°\widehat{i}-\cos 30°\widehat{k})\\ C_R=R(-1\widehat{i}-1.732\widehat{k})\\ Now,weknowthatthesumofallthereactionwillbe0.\\ {\displaystyle \sum C}=C_i\widehat{i}+C_o(-0.8660\widehat{i}+0.5\widehat{k})+R(-1\widehat{i}-1.732\widehat{k})=0\\ \left(4500-0.8660C_o-R\right)\widehat{i}+\left(0.5C_o-1.732R\right)\widehat{k}=0\\ \left(4500-0.8660C_o-R\right)=0\\ \left(0.5C_o-1.732R\right)=0\\ Solvingtheabovetwoequation,\\ C_o=3900lb-in\\ R=1125lb\end{array}$

Conclusion:

The output couple C_o is 3900 lb-ft.

To determine

(iii)

The magnitude of the reaction forces.

Answer to Problem 2.80P

The magnitude of the reaction forces is 1125 lb.

Explanation of Solution

Given:

P= 750 lb

Calculation:

  $\begin{array}{l}{\overrightarrow{C}}_o=C_o(-\cos 30°\widehat{i}+\sin 30°\widehat{k})\\ {\overrightarrow{C}}_o=C_o(-0.8660\widehat{i}+0.5\widehat{k})\\ C_R=2R(-\sin 30°\widehat{i}-\cos 30°\widehat{k})\\ C_R=R(-1\widehat{i}-1.732\widehat{k})\\ Now,weknowthatthesumofallthereactionwillbe0.\\ {\displaystyle \sum C}=C_i\widehat{i}+C_o(-0.8660\widehat{i}+0.5\widehat{k})+R(-1\widehat{i}-1.732\widehat{k})=0\\ \left(4500-0.8660C_o-R\right)\widehat{i}+\left(0.5C_o-1.732R\right)\widehat{k}=0\\ \left(4500-0.8660C_o-R\right)=0\\ \left(0.5C_o-1.732R\right)=0\\ Solvingtheabovetwoequations.\\ C_o=3900lb-in\\ R=1125lb\end{array}$

Conclusion:

The magnitude of the reaction forces is 1125 lb.