Chapter 6.3, Problem 6.89P

The 48-lb load is removed and a 288-lb · in. clockwise couple is applied successively at A, D, and E. Determine the components of the reactions at Band F if the couple is applied (a) at A, (b) at D, (c) at E.

To determine

The component of reactions at point B and F when the couple is applied at A.

Answer to Problem 6.89P

The x component of the reaction force at point B is $\underset{\_}{24\text{lb}}$ towards left direction, and the y component is $\underset{\_}{7.5\text{lb}}$ downward direction.

The x component of force applied is $\underset{\_}{24\text{lb}}$ in the positive x direction, and the y component of force is $\underset{\_}{7.5\text{lb}}$ in the positive y direction.

Explanation of Solution

The free body diagram of the problem 6.89P is shown in figure 1 below.

A clockwise couple is applied at A,D, and E. Due to this couple, resultant reaction forces are experienced in points A, D, and E.

First consider the couple applied at point A.

Write the equation to find the sum of moments of force at point F.

$\sum M_F=-288\text{lb}\cdot \text{in-}{B}_x\left(12\text{in}\right)$

Here, $\sum M_F$ is the sum of moments, $B_x$ is the x component of reaction force at point B.

Since the sum of moments of force at a point of a system in equilibrium is zero, rewrite the equation for the sum of moments.

$\begin{array}{c}-288\text{lb}\cdot \text{in-}{B}_x\left(12\text{in}\right)=0\\ B_x=-24\text{lb}\end{array}$

Write the equation to find the x components of force.

$\sum F_x=B_x+F_x$

Here, $\sum F_x$ is the sum of forces, $B_x$ is the x component of reaction force at point B, and $F_x$ is the x component of force applied at point B.

Since the sum of forces at a point is zero in equilibrium, the above equation is rewritten.

Substitute $-24\text{lb}$ for $B_x$ in equation.

$\begin{array}{c}-24\text{lb}+{\text{F}}_x=0\\ F_x=24\text{lb}\end{array}$

Write the equation to find the sum of y component of forces.

$\sum F_y=B_y+F_y$

Here, $\sum F_y$ is the sum of forces, $B_y$ is the y component of reaction force at a point, and $F_y$ is the y component of force applied at a point.

No force is applied in the y direction, therefore there will be no reaction also.

$B_y+F_y=0$ (I)

Consider figure 2.

Write the equation to find the y component of reaction force at point B.

$\frac{B_y}{B_x}=\frac{r_{BY}}{r_{BX}}$ (II)

Here, $B_y$ is the y component of reaction force at point B, $B_x$ is the x component of reaction force at point B, $r_{BY}$ is the y component of distance between point B and C, and $r_{BX}$ is the x component of distance between point B and C.

Rewrite equation (I) to find the value of $F_y$.

$B_y+F_y=0$ (III)

Conclusion:

Observe figure 2.

Substitute $24\text{lb}$ for $B_x$ , $5\text{in}$ for $r_{BY}$ , and $16\text{in}$ for $r_{BX}$ in equation (II) to find $B_y$.

$\begin{array}{c}\frac{B_y}{24\text{lb}}=\frac{5\text{in}}{16\text{in}}\\ B_y=7.5\text{lb}\end{array}$

The y component of reaction force at point B is having a magnitude of $7.5\text{lb}$ and is directed in the negative y direction.

Substitute $-7.5\text{lb}$ for $B_y$ in equation (III) to get $F_y$.

$\begin{array}{c}-7.5{\text{lb+F}}_y=0\\ F_y=7.5\text{lb}\end{array}$

The y component of the force applied at point b is $7.5\text{lb}$ in the positive y direction.’

Therefore, the x component of the reaction force at point B is $\underset{\_}{24\text{lb}}$ in the negative direction, and the y component is $\underset{\_}{7.5\text{lb}}$ in the negative y direction.

The x component of force applied is $\underset{\_}{24\text{lb}}$ in the positive x direction, and the y component of force is $\underset{\_}{7.5\text{lb}}$ in the positive y direction.

To determine

The component of reactions at point B and F when the couple is applied at D.

Answer to Problem 6.89P

The x component of the reaction force at point B is $24\text{lb}$ in the negative x direction, and the y component is $10.50\text{lb}$ in the positive y direction.

The x component of force applied is $24\text{lb}$ in the positive x direction, and the y component of force is $10.50\text{lb}$ in the negative y direction.

Explanation of Solution

The free body diagram of the problem 6.89P is shown in figure 1.

A clockwise couple is applied at A,D, and E. Due to this couple, resultant reaction forces are experienced in points A, D, and E.

Consider the couple applied at point D.

Write the equation to find the sum of moments of force at point F.

$\sum M_F=-288\text{lb}\cdot \text{in-}{B}_x\left(12\text{in}\right)$

Here, $\sum M_F$ is the sum of moments, $B_x$ is the x component of reaction force at point B.

Since the sum of moments of force at a point of a system in equilibrium is zero, rewrite the equation for the sum of moments.

$\begin{array}{l}-288\text{lb}\cdot \text{in-}{B}_x\left(12\text{in}\right)=0\\ B_x=-24\text{lb}\end{array}$

Write the equation to find the x components of force.

$\sum F_x=B_x+F_x$

Here, $\sum F_x$ is the sum of forces, $B_x$ is the x component of reaction force at point B, and $F_x$ is the x component of force applied at point B.

Since the sum of forces at a point is zero in equilibrium, the above equation is rewritten.

Substitute $-24\text{lb}$ for $B_x$ in equation.

$\begin{array}{l}-24{\text{lb+F}}_x=0\\ F_x=24\text{lb}\end{array}$

Write the equation to find the sum of y component of forces.

$\sum F_y=B_y+F_y$

Here, $\sum F_y$ is the sum of forces, $B_y$ is the y component of reaction force at a point, and $F_y$ is the y component of force applied at a point.

No force is applied in the y direction, therefore there will be no reaction also.

$B_y+F_y=0$ (IV)

Consider figure 3.

Write the equation to find the y component of reaction force at point B.

$\frac{F_y}{F_x}=\frac{r_{FY}}{r_{FX}}$ (V)

Here, $F_y$ is the y component of force at point B, $F_x$ is the x component of reaction force at point B, $r_{FY}$ is the y component of distance between point F and C, and $r_{FX}$ is the x component of distance between point F and C.

Rewrite equation (I) to find the value of $B_y$.

$B_y+F_y=0$ (VI)

Conclusion:

Observe figure 3.

Substitute $24\text{lb}$ for $F_x$ , $7\text{in}$ for $r_{FY}$ , and $16\text{in}$ for $r_{FX}$ in equation (V) to find $F_y$.

$\begin{array}{c}\frac{F_y}{24\text{lb}}=\frac{7\text{in}}{16\text{in}}\\ F_y=10.5\text{lb}\end{array}$

The y component of force at point B is having a magnitude of $10.5\text{lb}$ and is directed in the negative y direction.

Substitute $10.5\text{lb}$ for $F_y$ in equation (VI) to get $B_y$.

$\begin{array}{c}10.5{\text{lb+B}}_y=0\\ B_y=10.5\text{lb}\end{array}$

The y component of the reaction force applied at point B is $10.5\text{lb}$ in the positive y direction.’

Therefore, the x component of the reaction force at point B is $\underset{\_}{24\text{lb}}$ in the negative x direction, and the y component is $\underset{\_}{10.5\text{lb}}$ in the positive y direction.

The x component of force applied is $\underset{\_}{24\text{lb}}$ in the positive x direction, and the y component of force is $\underset{\_}{10.5\text{lb}}$ in the negative y direction.

To determine

The component of reactions at point B and F when the couple is applied at E.

Answer to Problem 6.89P

The x component of the reaction force at point B is $\underset{\_}{24\text{lb}}$ in the negative x direction, and the y component is $\underset{\_}{10.50\text{lb}}$ in the positive y direction.

The x component of force applied is $\underset{\_}{24\text{lb}}$ in the positive x direction, and the y component of force is $\underset{\_}{10.50\text{lb}}$ in the negative y direction.

Explanation of Solution

The free body diagram of the problem 6.89P is shown in figure 1.

A clockwise couple is applied at A,D, and E. Due to this couple, resultant reaction forces are experienced in points A, D, and E.

First consider the couple applied at point E.

Write the equation to find the sum of moments of force at point F.

$\sum M_F=-288\text{lb}\cdot \text{in-}{B}_x\left(12\text{in}\right)$

Here, $\sum M_F$ is the sum of moments, $B_x$ is the x component of reaction force at point B.

Since the sum of moments of force at a point of a system in equilibrium is zero, rewrite the equation for the sum of moments.

$\begin{array}{c}-288\text{lb}\cdot \text{in-}{B}_x\left(12\text{in}\right)=0\\ B_x=-24\text{lb}\end{array}$

Write the equation to find the x components of force.

$\sum F_x=B_x+F_x$

Here, $\sum F_x$ is the sum of forces, $B_x$ is the x component of reaction force at point B, and $F_x$ is the x component of force applied at point B.

Since the sum of forces at a point is zero in equilibrium, the above equation is rewritten.

Substitute $-24\text{lb}$ for $B_x$ in equation.

$\begin{array}{l}-24{\text{lb+F}}_x=0\\ F_x=24\text{lb}\end{array}$

Write the equation to find the sum of y component of forces.

$\sum F_y=B_y+F_y$

Here, $\sum F_y$ is the sum of forces, $B_y$ is the y component of reaction force at a point, and $F_y$ is the y component of force applied at a point.

No force is applied in the y direction, therefore there will be no reaction also.

$B_y+F_y=0$ (VII)

Consider figure 4.

Write the equation to find the y component of reaction force at point B.

$\frac{B_y}{B_x}=\frac{r_{BY}}{r_{BX}}$ (VIII)

Here, $B_y$ is the y component of reaction force at point B, $B_x$ is the x component of reaction force at point B, $r_{BY}$ is the y component of distance between point B and C, and $r_{BX}$ is the x component of distance between point B and C.

Rewrite equation (VII) to find the value of $F_y$.

$B_y+F_y=0$ (IX)

Conclusion:

Observe figure 4.

Substitute $24\text{lb}$ for $B_x$ , $5\text{in}$ for $r_{BY}$ , and $16\text{in}$ for $r_{BX}$ in equation (VIII) to find $B_y$.

$\begin{array}{c}\frac{B_y}{24\text{lb}}=\frac{5\text{in}}{16\text{in}}\\ B_y=7.5\text{lb}\end{array}$

The y component of reaction force at point B is having a magnitude of $7.5\text{lb}$ and is directed in the negative y direction.

Substitute $-7.5\text{lb}$ for $B_y$ in equation (IX) to get $F_y$.

$\begin{array}{c}-7.5{\text{lb+F}}_y=0\\ F_y=7.5\text{lb}\end{array}$

The y component of the force applied at point b is $7.5\text{lb}$ in the positive y direction.’

Therefore, the x component of the reaction force at point B is $\underset{\_}{24\text{lb}}$ in the negative direction, and the y component is $\underset{\_}{7.5\text{lb}}$ in the negative y direction.

The x component of force applied is $\underset{\_}{24\text{lb}}$ in the positive x direction, and the y component of force is $\underset{\_}{7.5\text{lb}}$ in the positive y direction.