# Problem 343 | Equilibrium of Parallel Force System

**Problem 343**

The weight W of a traveling crane is 20 tons acting as shown in Fig. P-343. To prevent the crane from tipping to the right when carrying a load P of 20 tons, a counterweight Q is used. Determine the value and position of Q so that the crane will remain in equilibrium both when the maximum load P is applied and when the load P is removed.

**Solution 343**

When load P is removed

$\Sigma M_A = 0$

$\Sigma M_A = 0$

$Qx = 20(5 + 1)$

$Qx = 120$ → Equation (1)

When load P is applied

$\Sigma M_B = 0$

$Q(x + 5) = 20(1) + 20(10)$

$Qx + 5Q = 220$

From Equation (1), Qx = 120, thus,

$120 + 5Q = 220$

$5Q = 100$

$Q = 20 \, \text{ tons}$ *answer*

Substitute Q = 20 tons to Equation (1)

$20x = 120$

$x = 6 \, \text{ ft}$ *answer*

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