Question

A 1 m long metal rod is used to lift a heavy object. The rod is supported at 10 cm from one end, the object is placed at this end of the rod (10 cm from pivot point). The other end of the rod is pushed down by a force of 200 N (90 cm from the pivot point). Please calculate the weight of the heavy object.

Answer 1

To solve for the weight of the heavy object using the principle of moments (or torque), we can use the formula:

\[
\text{Moment} = \text{Force} \times \text{Distance from pivot}
\]

In this scenario, we have:

- The pivot point is 10 cm from one end of the rod, meaning:
- The distance from the pivot to the heavy object (at the end of the rod) is \(10 \, \text{cm}\).
- The distance from the pivot to the force being applied (pushing down on the other end) is \(90 \, \text{cm}\).

Let \( W \) be the weight of the heavy object that we want to find.

### Step 1: Moments about the pivot
The moment due to the weight of the heavy object acting downwards at 10 cm from the pivot is:

\[
\text{Moment due to weight} = W \times 0.1 \, \text{m}
\]

The moment due to the applied force of 200 N acting downwards at 90 cm from the pivot is:

\[
\text{Moment due to applied force} = 200 \, \text{N} \times 0.9 \, \text{m}
\]

### Step 2: Setting the moments equal
For the rod to be in equilibrium (not rotating), the clockwise moments must equal the counterclockwise moments:

\[
W \times 0.1 = 200 \times 0.9
\]

### Step 3: Solve for \( W \)

First, calculate the moment due to the applied force:

\[
200 \times 0.9 = 180 \, \text{N m}
\]

Setting this equal to the moment due to the weight:

\[
W \times 0.1 = 180
\]

Now, solve for \( W \):

\[
W = \frac{180}{0.1} = 1800 \, \text{N}
\]

### Conclusion
The weight of the heavy object is \( 1800 \, \text{N} \).