Chapter 4, Problem 4.142RP

(a)

To determine

The reaction at each of the two front wheels A.

Answer to Problem 4.142RP

The reaction at each of the two front wheels A is $1761\text{ lb}\uparrow$ .

Explanation of Solution

The free-body diagram is shown in figure 1.

The weight of the truck is $3200\text{ lb}$ and it is lifting $1700\text{lb}$ crate. The distance at which force acting from the rear wheels are marked in figure 1. Let $\overrightarrow{A}$ be the reaction at one of the front wheel hence, the total reaction across two wheels is $2\overrightarrow{A}$.

The reaction at $A$ is acting in the upward direction. The wheel is rotating and there is a torque, which is the product of force acting and perpendicular distance. The moment about $B$ is responsible for the reaction at the front wheel.

Write the expression for the moment at $B$

${\overrightarrow{M}}_B=\overrightarrow{F}\times D_{\perp }$ (I)

Here, ${\overrightarrow{M}}_B$ is the moment at $B$, $\overrightarrow{F}$ is the force, and ${\overrightarrow{D}}_{\perp }$ is the perpendicular distance between the $B$ and the point where force is experienced.

Under equilibrium, the moment about $B$ will balances and sum of all forces will be zero.

$\begin{array}{c}\Sigma {\overrightarrow{M}}_B=\left(W_1\times {\overrightarrow{d}}_1\right)+\left(W_2\times {\overrightarrow{d}}_2\right)-\left(2A\times d_3\right)\\ =0\end{array}$ (II)

Here, $W_1$ is the magnitude of the weight of the crate, $W_2$ is the magnitude of the weight of the truck, $d_1$ is the distance from crate to rear wheels, $d_2$ is the distance from center of gravity to rear wheels, $d_3$ is the distance between front and rear wheels, and $A$ is the magnitude of the reaction at a front wheel.

Calculation:

Substitute $1700\text{lb}$ for $W_1$, $52\text{in}\text{.}$ for $d_1$, $3200\text{lb}$ for $W_2$ , $110\text{in}\text{.}$ for $d_2$, and $36\text{in}\text{.}$ for $d_3$ in equation (II) and rearrange to obtain $A$

$\begin{array}{c}\Sigma M_B=\left(1700\text{ lb}\right)\left(52\text{ in}\right)+\left(3200\text{ lb}\right)\left(12\text{ in}\right)-2A\left(36\text{ in}\right)\\ =0\\ A=\frac{\left(1700\text{lb}\times 52\text{in}\text{.}\right)+\left(3200\text{lb}\times 12\text{in}\text{.}\right)}{\left(36\text{in}\right)\times 2}\\ =+1761\text{lb}\\ \overrightarrow{A}=1761\text{ lb}\uparrow \end{array}$

Therefore, the reaction on each front wheel is $\underset{\_}{\overrightarrow{A}=+1761\text{lb}}\uparrow$.

(b)

To determine

The reaction at each of the two rear wheels.

Answer to Problem 4.142RP

The reaction at each of the rear wheels is $\underset{\_}{+689\text{lb}\uparrow }$.

Explanation of Solution

Refer to figure 1.

Let $\overrightarrow{B}$ be the reaction at one of the rear wheel hence, the total reaction across two wheels is $2\overrightarrow{B}$.

The reaction at $B$ is acting in the upward direction. Under equilibrium, the net force in the $y$ direction will be equal to zero.

$+\uparrow \Sigma F_y=0$ (III)

Here, $\Sigma F_y$ is the net force in the $y$ direction.

Write the expression for the net force.

$+\Sigma F_y=-W_1-W_2+2A+2B$

Here, $W_1$ is the magnitude of the weight of the crate, $W_2$ is the magnitude of weight of the truck, $A$ is the magnitude of reaction at the front wheel and $B$ is the magnitude of reaction at the rear wheels.

Put the above equation in equation (III).

$-W_1-W_2+2A+2B=0$ (IV)

Calculation:

Substitute $1700\text{lb}$ for $W_1$, $3200\text{lb}$ for $W_2$ and $1761\text{lb}$ for $A$ in the equation (IV) and rearrange to obtain $B$ .

$\begin{array}{c}-\left(1700\text{lb}\right)-\left(3200\text{lb}\right)+2\left(1761\text{lb}\right)+2B=0\\ B=\frac{1378\text{ lb}}{2}\\ =689\text{ lb}\\ \overrightarrow{B}=689\text{ lb}\uparrow \end{array}$

Therefore, the reaction on each rear wheel is $\underset{\_}{+689\text{lb}\uparrow }$.