Question

A 10 m long uniform iron bar pivoted at the center of the bar is subjected to the horizontal forces F1 and F2 = 200 N as shown in the picture. If the iron bar stays in equilibrium with zero net torque under these two forces, what is the magnitude of the force F1 to the nearest newton?

a) 100 N
b) 150 N
c) 200 N
d) 250 N

Answer 1

The magnitude of force F1 needed to balance a pivoted iron bar against force F2 of 200 N can be determined using the principle of moments, requiring that the moments due to both forces about the pivot are equal.

The student's question revolves around determining the magnitude of force F1 required to keep a uniform iron bar in equilibrium, considering that it is subjected to another horizontal force F2 of 200 N at a specified point. To solve for F1, we need to use the principle of moments which states that for a system to be in equilibrium, the sum of clockwise moments about any pivot must equal the sum of counterclockwise moments around that pivot.

Since the iron bar is pivoted at the center, the force F1 will create a moment that is the product of its magnitude and its distance from the pivot. Similarly, the force F2 will create a moment which is the product of 200 N and its distance from the pivot. For equilibrium, these two moments must be equal.

To find the magnitude of F1, we use the equation:

• Moment due to F1 = Moment due to F2
• (F1 × distance from pivot to F1) = (200 N × distance from pivot to F2)

Without the distances provided, we cannot calculate the exact value; however, we can say that the distances will determine the required magnitude of F1 to maintain equilibrium.