To analyze the equilibrium of the uniform beam AB pivoted at its geometrical center and subjected to various weights, we need to consider the moments about the pivot point to ensure that the sum of moments is zero (rotational equilibrium).
Let's designate the following:
- The center of the beam (pivot point) is at 2 meters from either end since the total length of the beam is 4 meters.
- \( x_1 \), \( x_2 \), etc., represent the distances from the pivot point to the location where forces are applied, with positive distances to the right of the pivot and negative distances to the left.
Based on the information provided:
- A 70 N weight is applied at one end of the beam (let's assume this is point A, which is 2 meters to the left of the pivot point).
- A 10 N weight is applied 1.5 meters from the pivot point (on the right side).
- A 20 N weight is another applied force (position not specified yet; let's assume we need to find this position).
- A 60 N weight is applied at the other end of the beam (at point B, which is 2 meters to the right of the pivot point).
Let's define the unknown position of the 20 N weight as \( x \) meters from the pivot point. We need to ensure that the sum of moments about the pivot point is zero.
**Calculation of Moments:**
- The moment due to 70 N weight: \( 70 \text{ N} \times (-2 \text{ m}) = -140 \text{ Nm} \).
- The moment due to 10 N weight: \( 10 \text{ N} \times 1.5 \text{ m} = 15 \text{ Nm} \).
- The moment due to 20 N weight: \( 20 \text{ N} \times x \text{ m} \).
- The moment due to 60 N weight: \( 60 \text{ N} \times 2 \text{ m} = 120 \text{ Nm} \).
Sum of moments about the pivot point should be zero:
\[
-140 \, \text{Nm} + 15 \, \text{Nm} + 20 \, \text{N} \times x \, \text{m} + 120 \, \text{Nm} = 0
\]
Simplify the equation:
\[
-140 + 15 + 20x + 120 = 0
\]
Combine like terms:
\[
-140 + 135 + 20x = 0
\]
\[
-5 + 20x = 0
\]
Solve for \( x \):
\[
20x = 5
\]
\[
x = \frac{5}{20}
\]
\[
x = 0.25 \, \text{m}
\]
Therefore, the 20 N weight must be located 0.25 meters to the right of the pivot point for the beam to be in equilibrium.
Auniform beam AB4m along is pivoted at its geometrical center and is held in equilibrium by a number of weight suspended 70N10N1.5mx 20N 60N
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