For a standard 42-inch guardrail,a 200 pound horizontal force applied at the top of the guardrail imparts which of the following counterbalancing forces at the bottom of the stanchion offset from the pivot point by 3.75 inches?
A)40 pounds
B)200 pounds
C)240 pounds
D)2240 pounds

Question

Quizplus · Accepted Answer

This problem can be solved using the principle of moments (torque). The moment caused by the 200-pound force is equal to the force multiplied by the distance from the pivot point (the top of the guardrail) to where the force is applied. This distance is 42 inches (the height of the guardrail). The counterbalancing moment at the bottom of the stanchion must be equal to this moment for equilibrium. The distance from the pivot point to where the counterbalancing force acts is 3.75 inches. Setting the moments equal to each other gives us:$$200 \text{ pounds} \times 42 \text{ inches} = F \times 3.75 \text{ inches}$$Solving for $F$ (the counterbalancing force) gives:$$F = \frac{200 \times 42}{3.75} = 2240 \text{ pounds}$$Therefore, the counterbalancing force at the bottom of the stanchion is 2240 pounds.

For a Standard 42-Inch Guardrail,a 200 Pound Horizontal Force Applied