Chapter 3.1, Problem 3.5P

A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D.

To determine

Moment of $300\text{N}$ force about D.

Answer to Problem 3.5P

Moment of $300\text{N}$ force about D is $41.7\text{N}\cdot \text{m}$ in counter clockwise direction.

Explanation of Solution

Refer Figure 1.

Write an expression to calculate the force vector acting at point A

$F=\left(F\cos \theta \right)i+\left(F\sin \theta \right)j$ (I)

Here, $F$ is the magnitude of force acting at point A, $\theta$ is the angle between the force vector and AB.

Write an expression to calculate the position vector along line connecting A and D.

$r=-\left(AO\right)i-\left(AE\right)j$ (II)

Here, $r$ is the position vector along line connecting A and D, AO is the horizontal distance between A and D, AE is the vertical distance between A and D.

Write an expression to calculate the moment about point D.

$M_D=r\times F$ (III)

Here, $M_D$ is the moment about point D.

Substitute equation (I) and (II) in equation (III) to rewrite equation (III).

$M_D=\left(-\left(AO\right)i-\left(AE\right)j\right)\times \left(\left(F\cos \theta \right)i+\left(F\sin \theta \right)j\right)$ (IV)

Conclusion:

Substitute $300\text{N}$ for $F$ , $25°$ for $\theta$, $100\text{mm}$ for $AO$ and $200\text{mm}$ for $AE$ in equation (IV) to find $M_B$.

$\begin{array}{c}{M}_D=\left(\begin{array}{c}\left(-\left(\left(100\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)i-\left(\left(200\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)j\right)\times \\ \left(\left(\left(300\text{N}\right)\cos \left(25°\right)\right)i+\left(\left(300\text{N}\right)\sin \left(25°\right)\right)j\right)\end{array}\right)\\ =\left(-\left(0.001\text{m}\right)i-\left(0.002\text{m}\right)j\right)\times \left(\left(271.89\text{N}\right)i+\left(126.785\text{N}\right)j\right)\\ =\left(41.7\text{N}\cdot \text{m}\right)k\end{array}$

Therefore, the of $300\text{N}$ force about D is $41.7\text{N}\cdot \text{m}$ in counter clockwise direction.

To determine

The magnitude and sense of horizontal force applies at C that creates the same moment about D.

Answer to Problem 3.5P

The magnitude and sense of horizontal force applies at C that creates the same moment about D is $333.60\text{N}$.

Explanation of Solution

Refer Figure 2.

Write an expression to calculate the position vector along line connecting C and D.

$r_{CD}=\left(OB\right)i-\left(CF\right)j$ (V)

Here, $r_{C/D}$ is the position vector D, BO is the horizontal distance between C and D, CF is the vertical distance between C and D.

Write an expression to calculate the moment about point D.

$M_D=r_{C/D}\times C$ (VI)

Here, $C$ is the force acting at point C.

Write an expression to calculate the force acting at point C.

$C=Ci$ (VII)

Here, $C$ is the magnitude of force acting at point C.

Substitute equation (V) and (VII) in equation (VI) to rewrite equation (VI).

$\begin{array}{c}{M}_D=\left(\left(OB\right)i-\left(CF\right)j\right)\times \left(Ci\right)\\ =\left(\left(C\right)\left(-CF\right)\right)\left(-k\right)\end{array}$ (VIII)

Write an expression to determine the magnitude of the moment about D.

$M_D=\left(C\right)\left(CF\right)$ (IX)

Here, $M_D$ is the magnitude of moment about D.

Rearrange the equation (IX) to determine the magnitude of force acting at point C.

$C=\frac{M_D}{CF}$ (X)

Conclusion:

Substitute $41.7\text{N}\cdot \text{m}$ for $M_D$ and $125\text{mm}$ for $CF$ in equation (X) to find $C$.

$\begin{array}{c}C=\frac{41.7\text{N}\cdot \text{m}}{\left(125\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\\ =333.60\text{N}\end{array}$

Therefore, the magnitude and sense of horizontal force applies at C that creates the same moment about D is $333.60\text{N}$.

To determine

The smallest force applied at C that creates the same moment about D.

Answer to Problem 3.5P

The smallest force applied at C that create the same moment about D is $176.8\text{N}$ at an angle $58°$ before horizontal.

Explanation of Solution

Refer Figure 3.

For minimum force, the force vector $C_{\min }$   will be perpendicular to the line DC.

Write an expression to calculate the smallest magnitude of force vector $C_{\min }$ that creates the same moment about D.

$C_{\min }=\frac{M_D}{DC\sin {\theta }_{\perp }}$ (XI)

Here, $C_{\min }$ is the smallest magnitude of force vector $C_{\min }$ that creates the same moment about D, ${\theta }_{\perp }$ is the perpendicular angle between DC and force vector $C_{\min }$.

Write an expression to calculate the length of DB.

$DC=\sqrt{{\left(OB\right)}^2+{\left(CF\right)}^2}$ (XII)

Substitute equation (XII) in (XI) to rewrite the equation (XI).

$Q=\frac{M_D}{\sqrt{{\left(OB\right)}^2+{\left(CF\right)}^2}\sin {\theta }_{\perp }}$ (XIII)

Write an expression to calculate the angel $\alpha$.

$\alpha ={\tan }^{-1}\left(\frac{CF}{OB}\right)$ (XIV)

Write an expression to calculate the direction of force.

$\varphi ={\theta }_{\perp }-\alpha$ (XV)

Substitute equation (XIV) in (XV) to rewrite find (XV).

$\varphi ={\theta }_{\perp }-{\tan }^{-1}\left(\frac{CF}{OB}\right)$

Conclusion:

Substitute $41.7\text{N}\cdot \text{m}$ for $M_D$, $90°$ for ${\theta }_{\perp }$, $200\text{mm}$ for $OB$ and $125\text{mm}$ for $CF$ in equation (XIII) to find $C_{\min }$.

$\begin{array}{c}{C}_{\min }=\frac{41.7\text{N}\cdot \text{m}}{\sqrt{{\left(\left(200\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2+{\left(\left(125\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}\sin \left(90°\right)}\\ =176.8\text{N}\end{array}$

Write an expression to calculate the angel $\alpha$.

$\alpha ={\tan }^{-1}\left(\frac{CF}{OB}\right)$ (XIV)

Write an expression to calculate the direction of force.

$\varphi ={\theta }_{\perp }-\alpha$ (XV)

Substitute equation (XIV) in (XV) to rewrite find (XV).

$\varphi ={\theta }_{\perp }-{\tan }^{-1}\left(\frac{CF}{OB}\right)$

Substitute $90°$ for ${\theta }_{\perp }$, $200\text{mm}$ for $OB$ and $125\text{mm}$ for $CF$ in equation (XV) to find $\varphi$.

$\begin{array}{c}\varphi =90°-{\tan }^{-1}\left(\frac{\left(125\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}{\left(200\text{mm}\right)\left(\frac{1\text{m}}{1000\text{mm}}\right)}\right)\\ =58°\end{array}$

Therefore, the smallest force applied at C that create the same moment about D is $176.8\text{N}$ at an angle $58°$ before horizontal.