Consider a beam of length L with a fulcrum x feet from one end (see figure). There are objects with weights W₁ and W₂ placed on opposite ends of the beam. In order to move a 600 pound rock, a person weighing 100 pounds wants to balance it on a beam that is 5 feet long. Find x such that the system is in equilibrium. W₂ W L-x Step 1 Assign the values. W₂ = 600 W₁ = 100✔ L = 5 ft Step 2 Assume that the person has to sit x feet away from the fulcrum in order to balance the system. Assume the fulcrum is located at the origin. Therefore, if x₂ = x, then X1 = L-X - X. Step 3 For equilibrium, the resultant moment should be 0 Therefore, Mo = 0 = W₁x1 + W₂x2 0 = 100 x x X = 600 X = 100 X (5-x) x= 30 x = 30 x= 30 X = X = 100 100 X (5-x) - 600 ✓ (5-x) Enter an exact number. 30

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**Problem: Balancing a Beam with Weights**

Consider a beam of length \( L \) with a fulcrum \( x \) feet from one end (see figure). There are objects with weights \( W_1 \) and \( W_2 \) placed on opposite ends of the beam.

In order to move a 600-pound rock, a person weighing 100 pounds wants to balance it on a beam that is 5 feet long. Find \( x \) such that the system is in equilibrium.

![Diagram Description]
The diagram consists of a beam balanced on a fulcrum. Weight \( W_1 \) (100 pounds) is placed on one end of the beam and weight \( W_2 \) (600 pounds) on the other end. The distances from the fulcrum are noted as \( x \) and \( L - x \) respectively.

### Step 1: Assign the Values

\[
W_2 = 600
\]

\[
W_1 = 100
\]

\[
L = 5 \text{ ft}
\]

### Step 2: Determine the Distance from the Fulcrum

Assume the person has to sit \( x \) feet away from the fulcrum in order to balance the system. Assume the fulcrum is located at the origin. Therefore, if \( x_2 = x \), then:

\[
x_1 = L - x
\]

\[
x_1 = 5 - x
\]

### Step 3: Ensure Equilibrium

For equilibrium, the resultant moment should be 0.

\[
M_0 = 0 = W_1 x_1 + W_2 x_2
\]

Therefore,

\[
0 = 100 (5 - x) - 600 x
\]

Simplify:

\[
100 (5 - x) = 600 x
\]

\[
500 - 100x = 600x
\]

\[
500 = 700x
\]

\[
x = \frac{500}{700}
\]

\[
x = \frac{5}{7} \text{ ft}
\]

Thus, the person must sit \( \frac{5}{7} \) feet away from the fulcrum to balance the system.

---

### Explanation of Diagram

The diagram shows a horizontal beam (5 feet in length) balanced on
Transcribed Image Text:**Problem: Balancing a Beam with Weights** Consider a beam of length \( L \) with a fulcrum \( x \) feet from one end (see figure). There are objects with weights \( W_1 \) and \( W_2 \) placed on opposite ends of the beam. In order to move a 600-pound rock, a person weighing 100 pounds wants to balance it on a beam that is 5 feet long. Find \( x \) such that the system is in equilibrium. ![Diagram Description] The diagram consists of a beam balanced on a fulcrum. Weight \( W_1 \) (100 pounds) is placed on one end of the beam and weight \( W_2 \) (600 pounds) on the other end. The distances from the fulcrum are noted as \( x \) and \( L - x \) respectively. ### Step 1: Assign the Values \[ W_2 = 600 \] \[ W_1 = 100 \] \[ L = 5 \text{ ft} \] ### Step 2: Determine the Distance from the Fulcrum Assume the person has to sit \( x \) feet away from the fulcrum in order to balance the system. Assume the fulcrum is located at the origin. Therefore, if \( x_2 = x \), then: \[ x_1 = L - x \] \[ x_1 = 5 - x \] ### Step 3: Ensure Equilibrium For equilibrium, the resultant moment should be 0. \[ M_0 = 0 = W_1 x_1 + W_2 x_2 \] Therefore, \[ 0 = 100 (5 - x) - 600 x \] Simplify: \[ 100 (5 - x) = 600 x \] \[ 500 - 100x = 600x \] \[ 500 = 700x \] \[ x = \frac{500}{700} \] \[ x = \frac{5}{7} \text{ ft} \] Thus, the person must sit \( \frac{5}{7} \) feet away from the fulcrum to balance the system. --- ### Explanation of Diagram The diagram shows a horizontal beam (5 feet in length) balanced on
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