Chapter 3, Problem 1RP

The boom has a length of 30 ft, a weight of 800 lb, and mass center at G. If the maximum moment that can be developed by a motor at A is M = 20(10³) lb · ft, determine the maximum load W, having a mass center at Gʹ, that can be lifted.

Prob. R3-1

To determine

The maximum load having a mass center $G^{\prime }$, which can be lifted.

Answer to Problem 1RP

The maximum load W having a mass center $G^{\prime }$, which can be lifted is $\underset{\_}{319\text{lb}}$.

Explanation of Solution

Given information:

• The moment M developed by motor A is $20\left({10}^3\right)\text{lb}\cdot \text{ft}$.
• The weight $W_1$ of the boom is 800 lb.

Calculation:

Determine the maximum load having a mass center $G^{\prime }$, which can be lifted using the principles of moments.

$M=W_1\left(l_1\cos \theta \right)+W\left(l\cos \theta +l_2\right)$ (1)

Here, the length between motor and the center of the boom is $l_1$, the angle of the boom is $\theta$, the total length of the boom is l, and the length between the end of the boom and $G^{\prime }$ is $l_2$.

Conclusion:

Substitute $20\left({10}^3\right)\text{lb}\cdot \text{ft}$ for M, 800 lb for $W_1$, 16 ft for $l_1$, $30°$ for $\theta$, 30 ft for l, and 2.0 ft for $l_2$ in Equation (1).

$\begin{array}{l}20\left({10}^3\right)=800\left(16\cos 30°\right)+W\left(30\cos 30°+2.0\right)\\ 20\left({10}^3\right)-800\left(16\cos 30°\right)=27.981W\\ W=\frac{8,914.875}{27.98}\\ W=318.62\text{lb}\end{array}$

$W\approx 319\text{lb}$

Thus, the maximum load having a mass center $G^{\prime }$, which can be lifted is $\underset{\_}{319\text{lb}}$.