One-fourth chain of mass M and length L is hanging down from a table. The work done to pull the hanging part of the chain on to the table is
mgL/32
Work Done Calculation for Hanging Chain
This problem involves calculating the work done required to pull a specific portion of a uniform chain that is hanging from a table back onto the table surface. The core principles applied here are the concept of work done against gravity and understanding the center of mass of the hanging part of the chain.
Chain Properties and Hanging Part
- The total mass of the chain is designated as \(M\).
- The total length of the chain is designated as \(L\).
- According to the problem statement, one-fourth of the entire chain is hanging down from the edge of the table.
- Consequently, the length of the hanging part of the chain can be determined as \( \frac{L}{4} \).
Mass of Hanging Chain
Given that the chain is uniform, its mass is distributed equally along its entire length. To find the mass of the hanging segment, we first determine the mass per unit length of the chain, which is \( \frac{M}{L} \). We then multiply this by the length of the hanging part:
\[ \text{Mass of hanging part} \left(m'\right) = \left(\frac{M}{L}\right) \times \left(\text{Length of hanging part}\right) \] \[ m' = \left(\frac{M}{L}\right) \times \left(\frac{L}{4}\right) = \frac{M}{4} \]Center of Mass of the Hanging Part
For any uniform object, like a straight segment of a chain, its center of mass is located precisely at its geometric midpoint. The hanging part of the chain has a length of \( \frac{L}{4} \). Therefore, the vertical position of its center of mass, measured from the table's edge (which is the reference point for lifting), is exactly half of its hanging length:
\[ \text{Distance of center of mass from table} \left(h\right) = \frac{1}{2} \times \left(\text{Length of hanging part}\right) \] \[ h = \frac{1}{2} \times \left(\frac{L}{4}\right) = \frac{L}{8} \]Work Done to Pull the Chain
The work done against the force of gravity to lift an object is calculated using the formula that relates the mass of the object, the acceleration due to gravity, and the vertical distance through which the object's center of mass is lifted:
\[ \text{Work Done} \left(W\right) = \text{mass of object} \times \text{acceleration due to gravity} \times \text{vertical displacement of its center of mass} \] \[ W = m' \times g \times h \]Now, by substituting the values we have derived for \(m'\) (the mass of the hanging part) and \(h\) (the vertical displacement of its center of mass):
\[ W = \left(\frac{M}{4}\right) \times g \times \left(\frac{L}{8}\right) \] \[ W = \frac{MgL}{32} \]Summary of Calculation
To provide a clear overview of the steps involved in this calculation, let's summarize the key parameters and results in the table below:
| Parameter | Value |
|---|---|
| Total Chain Mass | \(M\) |
| Total Chain Length | \(L\) |
| Hanging Length | \(L/4\) |
| Mass of Hanging Part (\(m'\)) | \(M/4\) |
| Center of Mass Depth (\(h\)) | \(L/8\) |
| Work Done (\(W = m'gh\)) | \( (M/4) \times g \times (L/8) = MgL/32 \) |
Therefore, the work done to pull the hanging part of the chain back onto the table is \( \frac{MgL}{32} \).
Important Questions from Work
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A load of 16.5 kg is lifted through a height of 3.4 metres. Find the work done in kg metre.
______ efforts are where our eyes direct the movement of our bodies.
Task simplifications represent _______ work methods.
The energy equivalence of 1 eV is
