Chapter 6.4, Problem 6.159P

The gears A and D are rigidly attached to horizontal shafts that are held by frictionless bearings. Determine (a) the couple M₀ that must be applied to shaft DEF to maintain equilibrium, (b) the reactions at G and H.

Fig. P6.159

To determine

The couple $M_0$ that must be applied to shaft $DEF$ to maintain equilibrium.

Answer to Problem 6.159P

The couple $M_0$ that must be applied to shaft $DEF$ to maintain equilibrium is $\left(12.50\text{N}\text{.m}\right)i$.

Explanation of Solution

Take all vectors along the $x$ axis and $y$ axis as positive.

Radius of gear $A$ is $80\text{mm}$ and radius of gear $D$ is $50\text{mm}$.

Consider the projection of the gears on $yz$ plane.

The free body diagram of the Gear $A$ and $D$  is shown below figure1.

Here,$M_D$ is the moment of force acting at gear $D$ and $J$ is the tangential force.

Write the expression for the moment at $A$

${\displaystyle \sum M_A}={\displaystyle \sum F\times D_{\perp }}$ (I)

Here, ${\displaystyle \sum M_A}$ is the net moment at $A$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $A$ and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at $A$ is due to the external torque of $20\text{N}\text{.m}$ and due to the tangential force $J$.

Thus, write the complete expression of anticlockwise moment $M_A$ at $A$.

${\displaystyle \sum M_A}=20\text{N}\text{.m}-J\left(r_A\right)$ (II)

Here, $r_A$ is the radius of gear $A$.

At equilibrium, the sum of the moment acting at $A$ will be zero.

Write the expression for the total anticlockwise moment acting at $A$.

${\displaystyle \sum M_A}=20\text{N}\text{.m}-J\left(r_A\right)=0$ (III)

Write the expression for the moment at $D$

${\displaystyle \sum M_D}={\displaystyle \sum F\times D_{\perp }}$ (IV)

Here, ${\displaystyle \sum M_D}$ is the net moment at $D$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $D$ and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at $D$ is due to the couple $M_0$ and tangential force $J$.

Thus, write the complete expression of anticlockwise moment $M_D$ at $D$.

${\displaystyle \sum M_D}=M_0-J\left(r_D\right)$ (V)

Here, $M_0$ is the magnitude of couple at applied at gear $A$ and $r_D$ is the radius of gear $D$.

At equilibrium, the sum of the moment acting at $D$ will be zero.

Write the expression for the total anticlockwise moment acting at $A$.

${\displaystyle \sum M_D}=M_0-J\left(r_D\right)=0$ (VI)

Calculation:

Substitute $80\text{mm}$ for $r_A$ in equation (III) to get $J$.

$20\text{N}\text{.m}-J\left(80\text{mm}\right)=0$

$\begin{array}{c}J=\frac{20\text{N}\text{.m}}{80\text{mm}\times \frac{1\text{m}}{1000\text{mm}}}\\ =250\text{N}\end{array}$

Substitute $250\text{N}$ for $J$ and $50\text{mm}$ for $r_D$ in equation (VI) to get $M_0$.

$\begin{array}{l}\left(M_0\right)-\left(250\text{N}\right)\left(50\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)=0\\ M_0=12.50\text{N}\text{.m}\end{array}$

Since the rotation is in the yz plane , the direction of couple is in $x$ direction.

Therefore, the couple $M_0$ that must be applied to shaft $DEF$ to maintain equilibrium is $\left(12.50\text{N}\text{.m}\right)i$.

To determine

The reaction at $G$ and $H$.

Answer to Problem 6.159P

The point $G$ experiences a moment of force , $M_G=-\left(45.5\text{N}\text{.m}\right)i$ and the point $H$ experiences a moment of force, $M_H=\left(13.00\text{N}\text{.m}\right)i$.

Explanation of Solution

Free body diagram of Projection on $xz$ plane Gear $A$ and axle $AC$ is shown in figure 2.

Here, is the tangential force acting on the gear, $B$ is the force acting at point $B$ , $C$ is the force acting at point $C$.

From figure 2, write the equation of net moment about $B$.

${\displaystyle \sum M_B}={\displaystyle \sum F\times D_{\perp }}$ (VII)

Here, ${\displaystyle \sum M_B}$ is the net moment at $B$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $B$ and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at $B$ is due to the tangential $250\text{N}\text{.m}$ on the gear and the force $C$.

Thus, write the complete expression of anticlockwise moment $M_B$ at $B$.

${\displaystyle \sum M_B}=-C\left(d_{\perp \text{BC}}\right)+\left(250\text{N}\right)\left(d_{\perp \text{JB}}\right)$ (VIII)

Here, $d_{\perp \text{BC}}$ is the perpendicular distance between the points $B$ and $C$ and $d_{\perp \text{JB}}$ is the perpendicular distance between the force $J$ and point $B$.

At equilibrium, the sum of the moment acting at $B$ will be zero.

Write the expression for the total anticlockwise moment acting at $A$.

${\displaystyle \sum M_B}=-C\left(d_{\perp \text{BC}}\right)+\left(250\text{N}\right)\left(d_{\perp \text{JB}}\right)=0$ (IX)

Write the expression for the moment at $D$

${\displaystyle \sum M_C}={\displaystyle \sum F\times D_{\perp }}$ (X)

Here, ${\displaystyle \sum M_C}$ is the net moment at $C$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $C$  and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

At equilibrium, the sum of the moment acting at $C$ will be zero.

Write the expression for the total anticlockwise moment acting at $A$.

${\displaystyle \sum M_C}=-B\left(d_{\perp \text{BC}}\right)+\left(250\text{N}\right)\left(d_{\perp \text{JC}}\right)=0$ (XI)

Here, $d_{\perp \text{BC}}$ is the perpendicular distance between the points $B$ and $C$ and $d_{\perp \text{JC}}$ is the perpendicular distance between the force $J$ and point $C$.

The free body diagram of the projection on $xz$ plane with Gear $D$ and axle $DF$ is shown in figure3.

Here, $E$ is the force acting at point $E$ and $F$ is the force acting at point $F$.

From figure 3, write the equation of net moment about $E$.

${\displaystyle \sum M_E}={\displaystyle \sum F\times D_{\perp }}$ (XII)

Here, ${\displaystyle \sum M_E}$ is the net moment at $E$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $E$ and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

The moment at $E$ is due to the tangential $250\text{N}\text{.m}$ on the gear and the force $F$.

Thus, write the complete expression of anticlockwise moment $M_E$ at $E$.

${\displaystyle \sum M_E}=-F\left(d_{\perp \text{FE}}\right)+\left(250\text{N}\right)\left(d_{\perp \text{JF}}\right)$ (XIII)

Here, $d_{\perp \text{FE}}$ is the perpendicular distance between the points $F$ and $E$ and $d_{\perp \text{JF}}$ is the perpendicular distance between the force $J$ and point $F$.

At equilibrium, the sum of the moment acting at $E$ will be zero.

Write the expression for the total anticlockwise moment acting at $E$.

${\displaystyle \sum M_E}=+F\left(d_{\perp \text{FE}}\right)-\left(250\text{N}\right)\left(d_{\perp \text{JF}}\right)=0$ (XIV)

Write the expression for the moment at $F$

${\displaystyle \sum M_F}={\displaystyle \sum F\times D_{\perp }}$ (XV)

Here, ${\displaystyle \sum M_F}$ is the net moment at $C$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $F$  and the point where force is experienced.

Above equation implies that net moment at any point is the sum of product of each force acting on the system and perpendicular distance of the force and the point.

At equilibrium, the sum of the moment acting at $F$ will be zero.

Write the expression for the total anticlockwise moment acting at $A$.

${\displaystyle \sum M_F}=E\left(d_{\perp \text{EF}}\right)-\left(250\text{N}\right)\left(d_{\perp \text{JF}}\right)=0$ (XVI)

Here, $d_{\perp \text{BC}}$ is the perpendicular distance between the points $B$ and $C$ and $d_{\perp \text{JC}}$ is the perpendicular distance between the force $J$ and point $C$.

Consider the projection at $yz$ plane.

The free body diagram of the Bracket $BG$.

Here, $B$ is the force at point $B$, $E$ is the force at point $E$ , $M_G$ is the moment at point $G$ and $G$ is the force at point $G$.

Write the expression for the total force along $z$ direction.

${\displaystyle \sum F_z}=G-E+B$ (XVII)

Since in this direction net force is equal to zero. Equate above equation to zero.

${\displaystyle \sum F_z}=G-E+B=0$ (XVIII)

Since total moment of force about $x$ axis is zero, moment of force about $E$ is zero.

Write the equilibrium moment of force about $E$.

$M_G+B\left(130\text{mm}\right)=0$ (XIX)

The free body diagram of the Bracket $CH$.

Here, $C$ is the force at point $C$, $F$ is the force at point $F$ , $M_H$ is the moment at point $H$ and $H$ is the force at point $H$.

Write the expression for the total force along $z$ direction.

${\displaystyle \sum F_z}=H-F+C$ (XX)

Since in this direction net force is equal to zero. Equate above equation to zero.

${\displaystyle \sum F_z}=H-F+C=0$ (XXI)

Since total moment of force about $x$ axis is zero, moment of force about $E$ is zero.

Write the equilibrium moment of force about $E$.

$M_H+C\left(130\text{mm}\right)=0$ (XXII)

Calculation:

Substitute $100\text{mm}$ for $d_{\perp \text{BC}}$ and $40\text{mm}$ for $d_{\perp \text{JB}}$ in equation (VIII) to get $C$.

$-C\left(100\text{mm}\right)+\left(250\text{N}\right)\left(40\text{mm}\right)=0$

$\begin{array}{c}C=\frac{\left(250\text{N}\right)\left(40\text{mm}\right)}{\left(100\text{mm}\right)}\\ =100\text{N}\end{array}$

Substitute $140\text{mm}$ for $d_{\perp \text{JC}}$ and $100\text{mm}$ for $d_{\perp \text{BC}}$ in equation (XI) to get $B$.

$-B\left(100\text{mm}\right)+\left(250\text{N}\right)\left(140\text{mm}\right)=0$

$\begin{array}{c}B=\frac{\left(250\text{N}\right)\left(140\text{mm}\right)}{\left(100\text{mm}\right)}\\ =350\text{N}\end{array}$

Substitute $100\text{mm}$ for $d_{\perp \text{FE}}$ and $40\text{mm}$ for $d_{\perp \text{JE}}$ in equation (XIV) to get $E$.

$F\left(100\text{mm}\right)-\left(250\text{N}\right)\left(40\text{mm}\right)=0$

$\begin{array}{c}F=\frac{\left(250\text{N}\right)\left(40\text{mm}\right)}{100\text{mm}}\\ =100\text{N}\end{array}$

Substitute $140\text{mm}$ for $d_{\perp \text{JF}}$ and $100\text{mm}$ for $d_{\perp \text{FE}}$ in equation (XI) to get $E$.

$E\left(100\text{mm}\right)-\left(250\text{N}\right)\left(140\text{mm}\right)=0$

$\begin{array}{c}E=\frac{\left(250\text{N}\right)\left(140\text{mm}\right)}{100\text{mm}}\\ =350\text{N}\end{array}$

Substitute $350\text{N}$ for $E$ and $B$ in equation (XVIII)to get $G$.

$\begin{array}{c}G-350\text{N}+350\text{N}=0\\ G=0\text{N}\end{array}$

Substitute $350\text{N}$ for $B$ in equation (XIX) to get $M_G$.

$\begin{array}{l}{M}_G=\left(350\text{N}\right)\left(130\text{mm}\times \frac{1\text{m}}{1000\text{m}}\right)\\ =-45.5\text{N}\cdot \text{m}\end{array}$

The negative sign indicate that it is directed along $x$ direction.

Substitute $100\text{N}$ for $F$ and $C$ to get $H$.

$\begin{array}{c}H-100\text{N}+100\text{N}=0\\ H=0\text{N}\end{array}$

Substitute $100\text{N}$ for $C$ in equation (XXII) to get $M_H$.

$\begin{array}{c}{M}_H=\left(100\text{N}\right)\left(130\text{mm}\times \frac{1\text{m}}{1000\text{m}}\right)\\ =13.00\text{N}\cdot \text{m}\end{array}$

The positive value indicate that it is directed along $x$ direction.

Substitute $100\text{N}$ for $F$ and $C$ to get $H$.

$\begin{array}{c}H-100\text{N}+100\text{N}=0\\ H=0\text{N}\end{array}$

Therefore, the net force at $G$ and $H$ is zero. The moment of force about $G$$M_G=-\left(45.5\text{N}\cdot \text{m}\right)i$ and moment of force about $H$ , $M_H=\left(13.00\text{N}\text{.m}\right)i$