Chapter 3, Problem 3.30P

The values of ${\displaystyle \sum F_z,{\displaystyle \sum M_x,}}$ and ${{\displaystyle \sum M}}_y$ for three force systems that are parallel to the z-axis are

Determine the resultant of each force system and show it on a sketch of the coordinate system.

To determine

(a)

The resultant of force system and show it on a sketch of the co-ordinate system.

Answer to Problem 3.30P

The resultant force is -50 k lb.

The resultant moment of the system is $-250\widehat{i}+200\widehat{j}lb-ft.$

The force system is shown in the co-ordinate system.

Explanation of Solution

Given:

ΣF_z = -50 lb.

ΣM_X = -250 lb-ft.

ΣM_Y = 200 lb-ft.

Concept Used:

  $Momentaboutanyaxis=\overrightarrow{F}\times \overrightarrow{r}.\widehat{n}$

Calculation:

Let a and b be the distances from the x and y- axes respectively.

$\begin{array}{l}{\displaystyle \sum M_X=-250lb.ft}\\ {\displaystyle \sum M_X=F_z\times x}\\ -250=-50\times x\\ x=5ft\\ {\displaystyle \sum M_Y=200lb.ft}\\ 200=F_z\times y\\ y=-4ft\\ \mathrm{Re}sul\tan toftheforcesystem:\\ F_{\mathrm{Re}sul\tan t}=-50\widehat{k}lb\\ Momen{t}_{\mathrm{Re}aul\tan t}=-250\widehat{i}+200\widehat{j}lb-ft\end{array}$

Conclusion:

The resultant force is -50 k lb.

The resultant moment of the system is $-250\widehat{i}+200\widehat{j}lb-ft$

The force system is shown in the co-ordinate system.

To determine

(c)

The resultant of force system and show it on a sketch of the co-ordinate system.

Answer to Problem 3.30P

The resultant force is$50\widehat{k}kN$

The resultant moment of the system is $-250\widehat{y}kN-m.$

The force system is shown in the co-ordinate system.

Explanation of Solution

Given:

ΣF_z = 50 kN.

ΣM_X = 0

ΣM_Y = -250 kN-m.

Concept Used:

  $Momentaboutanyaxis=\overrightarrow{F}\times \overrightarrow{r}.\widehat{n}$

Calculation:

Let a and b be the distances from the x and y- axes respectively.

  $\begin{array}{l}{\displaystyle \sum M_X=0}\\ {\displaystyle \sum M_X=F_z\times x}\\ x=0m\\ {\displaystyle \sum M_Y=-250kN.m}\\ -250=50kN\times y\\ y=-5m\\ \mathrm{Re}sul\tan toftheforcesystem:\\ F_{\mathrm{Re}sul\tan t}=50\widehat{k}kN\\ Momen{t}_{\mathrm{Re}aul\tan t}=-250\widehat{j}kN.m\end{array}$

Conclusion:

The resultant force is $50\widehat{k}kN.$.

The resultant moment of the system is $-250\widehat{y}kN-m.$

The force system is shown in the co-ordinate system.

To determine

(c)

The resultant of force system and show it on a sketch of the co-ordinate system.

Answer to Problem 3.30P

The resultant force is$40\widehat{k}N.$

The resultant moment of the system is $320\widehat{i}-400\widehat{j}N-m.$

The force system is shown in the co-ordinate system.

Explanation of Solution

Given:

ΣF_z = 40 N.

ΣM_X = 320 N.m.

ΣM_Y = -400 N-m.

Concept Used:

  $Momentaboutanyaxis=\overrightarrow{F}\times \overrightarrow{r}.\widehat{n}$

Calculation:

Let a and b be the distances from the x and y- axes respectively.

  $\begin{array}{l}{\displaystyle \sum M_X=320N.m}\\ {\displaystyle \sum M_X=F_z\times x}\\ 320=40\times x\\ x=8m\\ {\displaystyle \sum M_Y=-400N.m}\\ -400N.m=40N\times y\\ y=-10m\\ \mathrm{Re}sul\tan toftheforcesystem:\\ F_{\mathrm{Re}sul\tan t}=40\widehat{k}N\\ Momen{t}_{\mathrm{Re}aul\tan t}=320\widehat{i}-400\widehat{j}N.m\end{array}$

Conclusion:

The resultant force is $40\widehat{k}N.$.

The resultant moment of the system is $320\widehat{i}-400\widehat{j}N-m.$

The force system is shown in the co-ordinate system.