Chapter 2, Problem 2.14P

Find the forces $Q_1,Q_2,$ and $Q_3$ so that the two forces systems are equivalent.

To determine

The forces $Q_1,Q_2,Q_3$.

Answer to Problem 2.14P

The forces in right hand systems are $Q_1=121.74\text{ lb, }{Q}_2=0\text{, }{Q}_3=202.9\text{ lb}\text{. }$

Explanation of Solution

Given Information:

The following system of forces is given,

  

and the resultant of both force systems are the same.

Calculation:

The forces applied can be written in vector notation as,

  $\begin{array}{l}{\overrightarrow{P}}_1=100\left(\frac{(3-0)i+(0-0)j+(4-0)k}{\sqrt{{(3-0)}^2+{(0-0)}^2+{(4-0)}^2}}\right)=60i+0j+80k\\ {\overrightarrow{P}}_2=120\left(\frac{(3-0)i+(3-0)j+(4-0)k}{\sqrt{{(3-0)}^2+{(3-0)}^2+{(4-0)}^2}}\right)=61.74i+61.74j+82.32k\\ {\overrightarrow{P}}_3=60\left(\frac{(0-0)i+(3-0)j+(0-0)k}{\sqrt{{(0-0)}^2+{(3-0)}^2+{(0-0)}^2}}\right)=0i+60j+0k\end{array}$

  $\begin{array}{l}{\overrightarrow{Q}}_1=Q_1\left(\frac{(3-0)i+(0-0)j+(0-0)k}{\sqrt{{(3-0)}^2+{(0-0)}^2+{(0-0)}^2}}\right)=Q_{1}i\\ {\overrightarrow{Q}}_2=Q_2\left(\frac{(0-3)i+(0-3)j+(0-0)k}{\sqrt{{(0-3)}^2+{(3-0)}^2+{(0-0)}^2}}\right)=Q_2(-0.71i-0.71j)\\ {\overrightarrow{Q}}_3=Q_3\left(\frac{(0-0)i+(3-0)j+(4-0)k}{\sqrt{{(0-0)}^2+{(3-0)}^2+{(4-0)}^2}}\right)=Q_3(0.6j+0.8k)\end{array}$

Since resultant of both the force systems is the same, hence,

  $\begin{array}{l}{\overrightarrow{Q}}_1+{\overrightarrow{Q}}_2+{\overrightarrow{Q}}_3={\overrightarrow{P}}_1+{\overrightarrow{P}}_2+{\overrightarrow{P}}_3\\ \Rightarrow Q_{1}i+Q_2(-0.71i-0.71j)+Q_3(0.6j+0.8k)\\ =60i+0j+80k+61.74i+61.74j+82.32k+60j\end{array}$

Compare the components to get the following equations,

  $\begin{array}{l}{Q}_1-0.71Q_2=60+61.74\mathrm{...}\text{(1)}\\ -0.71Q_2+0.6Q_3=60+61.74\mathrm{..}\text{(2)}\\ 0.8Q_3=80+82.32\mathrm{...}\text{(3)}\end{array}$

Solve equation (1-3) to get $Q_1=121.74\text{ lb, }{Q}_2=0\text{, }{Q}_3=202.9\text{ lb}\text{. }$

Conclusion:

Therefore, the forces in the right hand systems are $Q_1=121.74\text{ lb, }{Q}_2=0\text{, }{Q}_3=202.9\text{ lb}\text{. }$