Chapter 6.4, Problem 6.158P

Solve Prob. 6.157 assuming that the 2-kip force P acts horizontally to the right (θ = 0).

Fig. P6.157

To determine

The force exerted by each cylinder shown in figure $\text{P6}\text{.157}$ if $\theta =0°$ .

Answer to Problem 6.158P

The force exerted by the cylinder $EF$ is $3.35\text{kips}$ compression, force exerted by the cylinder $AD$ is $11.03\text{kips}$ compression and the force exerted by the cylinder $CG$ is $2.29\text{kips}$ compression.

Explanation of Solution

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

Let P is the force exerted on the bucket at J.

The magnitude of force $P$ is equal to $2\text{kips}$ and angle that force $P$ makes with $x$ axis equal to, $\theta =0°$.

The free body diagram of the bucket is sketched below as figure 1.

 

Here, $F_{CG}$ is the magnitude of force exerted by the cylinder $CG$ on the bucket ,$H$ is the joint about which bucket rotate and $\alpha$ is the angle that direction of force $F_{CG}$ makes with $y$ axis.

Write the expression for the moment at $H$

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

Here, ${\displaystyle \sum M_H}$ is the net moment at $H$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $H$ 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 $H$ is due to the $x$ and $y$ component of force $F_{CG}$ and sine and cosine component of force $P$.

Thus, the complete expression of net anticlockwise moment $M_H$ at point $H$.

${\displaystyle \sum M_H}=F_{CG}\cos \alpha \left(10\text{in}\text{.}\right)+F_{CG}\sin \alpha \left(10\text{in}\text{.}\right)+P\cos \theta \left(16\text{in}\text{.}\right)+P\sin \theta \left(8\text{in}\text{.}\right)=0$ (II)

Here, ${\displaystyle \sum M_H}$ is the sum of all moment of force at $H$.

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

Write the expression for the total moment acting at $H$.

${\displaystyle \sum M_H}=F_{CG}\cos \alpha \left(10\text{in}\text{.}\right)+F_{CG}\sin \alpha \left(10\text{in}\text{.}\right)+P\cos \theta \left(16\text{in}\text{.}\right)+P\sin \theta \left(8\text{in}\text{.}\right)=0$ (III)

From figure 1 , write the expression for the $\cos \alpha$.

$\cos \alpha =\frac{4}{5}$

From figure 1 , write the expression for the $\sin \alpha$.

$\sin \alpha =\frac{3}{5}$

The free body diagram of the bucket and arm $ABH$ is sketched below as figure 2.

Here, $F_{AD}$ is the magnitude of force exerted by the cylinder $AD$ on the arm $AH$ ,$B$ is the joint about which the arm $AH$ rotate and $\beta$ is the angle that direction of force $F_{AD}$ makes with $-x$ axis.

Write the expression for the moment at $B$

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

Here, ${\displaystyle \sum M_B}$ is the net moment at $J$, $F$ 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 $x$ component of force $F_{AD}$ , $y$ component of force $F_{AD}$, sine and cosine component of force $P$.

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

${\displaystyle \sum M_B}=F_{AD}\cos \beta \left(12\text{in}\text{.}\right)+F_{AD}\sin \beta \left(10\text{in}\text{.}\right)+P\cos \theta \left(86\text{in}\text{.}\right)-P\sin \theta \left(42\text{in}\text{.}\right)=0$ (V)

Here, ${\displaystyle \sum M_B}$ is the sum of all moment of force at $B$.

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

Write the expression for the total moment acting at $J$.

${\displaystyle \sum M_B}=F_{AD}\cos \beta \left(12\text{in}\text{.}\right)+F_{AD}\sin \beta \left(10\text{in}\text{.}\right)+P\cos \theta \left(86\text{in}\text{.}\right)-P\sin \theta \left(42\text{in}\text{.}\right)=0$ (VI)

From figure 2 , write the expression for the $\cos \beta$ and $\sin \beta$.

$\cos \beta =\frac{4}{5}$

$\sin \beta =\frac{3}{5}$

Geometry of cylinder $EF$ is shown in figure 4.

The free body diagram of the bucket and both arms is sketched below as figure 4.

Here, $F_{EF}$ is the magnitude of force exerted by the cylinder $EF$ on the arm $BI$ ,$I$ is the joint about which arm $BI$ rotate and $\gamma$ is the angle that direction of force $F_{EF}$ makes with $x$ axis.

Write the expression for the moment at $I$

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

Here, ${\displaystyle \sum M_I}$ is the net moment at $I$, $F$ is the force, and $D_{\perp }$ is the perpendicular distance between the $I$ 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 $I$ is due to the $x$ component of force $F_{EF}$ and sine and cosine component of force $P$.

Thus, the complete expression of net anticlockwise moment $M_K$ at point $K$.

${\displaystyle \sum M_I}=F_{EF}\cos \gamma \left(18\text{in}\text{.}\right)+P\cos \theta \left(28\text{in}\text{.}\right)-P\sin \theta \left(120\text{in}\text{.}\right)=0$ (VIII)

Here, ${\displaystyle \sum M_I}$ is the sum of all moment of force at $I$.

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

Write the expression for the total moment acting at $I$.

${\displaystyle \sum M_I}=F_{EF}\cos \gamma \left(18\text{in}\text{.}\right)+P\cos \theta \left(28\text{in}\text{.}\right)-P\sin \theta \left(120\text{in}\text{.}\right)=0$ (IX)

From figure 3 , write the expression for the $\tan \gamma$

$\tan \gamma =\frac{16\text{in}\text{.}}{40\text{in}\text{.}}$ (X)

Calculation:

Substitute $\frac{4}{5}$ for $\cos \alpha$ , $\frac{3}{5}$ for $\sin \alpha$ , $2\text{kips}$ for $P$ and $0°$ for $\theta$ rearrange the equation (III) to get $F_{EF}$.

$F_{CG}\left(\frac{4}{5}\right)\left(10\text{in}\text{.}\right)+F_{CG}\left(\frac{3}{5}\right)\left(10\text{in}\text{.}\right)+P\cos 0°\left(16\text{in}\text{.}\right)+P\sin 0°\left(8\text{in}\text{.}\right)=0$

$\begin{array}{c}{F}_{CG}=-\frac{\left(2\text{kips}\right)\left(16\text{in}\text{.}\right)+\left(2\text{kips}\right)\left(\frac{1}{\sqrt{2}}\right)\left(8\text{in}\text{.}\right)}{\left(\frac{4}{5}\right)\left(10\text{in}\text{.}\right)+\left(\frac{3}{5}\right)\left(10\text{in}\text{.}\right)}\\ =-\frac{\left(2\text{kips}\right)\left(\left(16\text{in}\text{.}\right)+\left(8\text{in}\text{.}\right)\right)}{\left(\frac{4}{5}\right)\left(10\text{in}\text{.}\right)+\left(\frac{3}{5}\right)\left(10\text{in}\text{.}\right)}\\ =-2.29\text{kips}\end{array}$

The negative sign indicate that the cylinder undergoes compression.

Substitute $\frac{4}{5}$ for $\cos \beta$ , $\frac{3}{5}$ for $\sin \beta$ , $2\text{kips}$ for $P$ and $0°$ for $\theta$ rearrange the equation (VI) to get $F_{AD}$.

$F_{AD}\left(\frac{4}{5}\right)\left(12\text{in}\text{.}\right)+F_{AD}\left(\frac{3}{5}\right)\left(10\text{in}\text{.}\right)+\left(2\text{kips}\right)\cos 0°\left(86\text{in}\text{.}\right)-\left(2\text{kips}\right)\sin 0°\left(42\text{in}\text{.}\right)=0$

$\begin{array}{c}{F}_{AD}=\frac{\left(2\text{kips}\right)\left(\left(42\text{in}\text{.}\right)-\left(86\text{in}\text{.}\right)\right)}{\left(\frac{4}{5}\right)\left(12\text{in}\text{.}\right)+\left(\frac{3}{5}\right)\left(10\text{in}\text{.}\right)}\\ =-11.03\text{kips}\end{array}$

The negative sign indicate that the cylinder undergoes compression.

Rearrange the equation (X) to get $\gamma$ .

$\begin{array}{c}\tan \gamma =\frac{16\text{in}\text{.}}{40\text{in}\text{.}}\\ \gamma =21.801°\end{array}$

Substitute $21.801°$ for $\gamma$ and $2\text{kips}$ for $P$ and $0°$ for $\theta$ rearrange the equation (IX) to get $F_{EF}$.

$F_{EF}\cos \left(21.801°\right)\left(18\text{in}\text{.}\right)+\left(2\text{kips}\right)\cos 0°\left(28\text{in}\text{.}\right)-\left(2\text{kips}\right)\sin 0°\left(120\text{in}\text{.}\right)=0$

$\begin{array}{c}{F}_{EF}=\frac{\left(2\text{kips}\right)\left(\left(120\text{in}\text{.}\right)-\left(28\text{in}\text{.}\right)\right)}{\cos \left(21.801°\right)\left(18\text{in}\text{.}\right)}\\ =-3.35\text{kips}\end{array}$

The positive sign indicate that the cylinder $EF$ undergoes compression.

Therefore, the force exerted by the cylinder $EF$ is $3.35\text{kips}$ compression, force exerted by the cylinder $AD$ is $11.03\text{kips}$ compression and the force exerted by the cylinder $CG$ is $2.29\text{kips}$ compression.