Chapter 4, Problem 4.142RP

A 3200-lb forklift truck is used to lift a 1700-lb crate. Determine the reaction at each of the two (a) front wheels A, (b) rear wheels B.

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 $\underset{\_}{\overrightarrow{A}=+1761\text{lb}}\uparrow$.

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 $\underset{\_}{+689\text{lb}\uparrow }$.