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Answer :
Consider a uniform beam that weighs [tex]$56\,\text{N}$[/tex] and has a [tex]$100\,\text{N}$[/tex] weight suspended at one end. We assume the beam is [tex]$2.4\,\text{m}$[/tex] long, so its center of gravity is at the midpoint, which is [tex]$1.2\,\text{m}$[/tex] from either end.
Let [tex]$x$[/tex] be the distance from the [tex]$100\,\text{N}$[/tex] weight to the support point. Since the beam is in equilibrium when it is horizontal, the clockwise moments must equal the anticlockwise moments about the support.
The [tex]$100\,\text{N}$[/tex] weight creates a clockwise moment about the support given by:
[tex]$$
\text{Moment from load} = 100\,\text{N} \times x.
$$[/tex]
The weight of the beam (which is [tex]$56\,\text{N}$[/tex]) acts at its center, and its moment arm with respect to the support is [tex]$(1.2 - x)$[/tex] (assuming the support is between the [tex]$100\,\text{N}$[/tex] weight and the beam's center). Thus, the anticlockwise moment is:
[tex]$$
\text{Moment from beam} = 56\,\text{N} \times (1.2 - x).
$$[/tex]
Setting these moments equal gives:
[tex]$$
100x = 56(1.2 - x).
$$[/tex]
Now, solve this equation step by step:
1. Expand the right side:
[tex]$$
100x = 67.2 - 56x.
$$[/tex]
2. Add [tex]$56x$[/tex] to both sides:
[tex]$$
100x + 56x = 67.2,
$$[/tex]
which simplifies to:
[tex]$$
156x = 67.2.
$$[/tex]
3. Divide both sides by [tex]$156$[/tex] to solve for [tex]$x$[/tex]:
[tex]$$
x = \frac{67.2}{156} \approx 0.4308\,\text{m}.
$$[/tex]
Thus, the support must be placed approximately [tex]$0.43\,\text{m}$[/tex] from the end where the [tex]$100\,\text{N}$[/tex] weight is hung.
To verify, we can compute the moments:
- Moment from the load: [tex]$100 \times 0.4308 \approx 43.08\,\text{Nm}$[/tex].
- Moment from the beam: [tex]$56 \times (1.2 - 0.4308) \approx 56 \times 0.7692 \approx 43.08\,\text{Nm}$[/tex].
Since these moments are equal, the beam balances horizontally.
The final answer is that the support should be located approximately [tex]$0.43\,\text{m}$[/tex] from the end where the [tex]$100\,\text{N}$[/tex] weight is attached.
Let [tex]$x$[/tex] be the distance from the [tex]$100\,\text{N}$[/tex] weight to the support point. Since the beam is in equilibrium when it is horizontal, the clockwise moments must equal the anticlockwise moments about the support.
The [tex]$100\,\text{N}$[/tex] weight creates a clockwise moment about the support given by:
[tex]$$
\text{Moment from load} = 100\,\text{N} \times x.
$$[/tex]
The weight of the beam (which is [tex]$56\,\text{N}$[/tex]) acts at its center, and its moment arm with respect to the support is [tex]$(1.2 - x)$[/tex] (assuming the support is between the [tex]$100\,\text{N}$[/tex] weight and the beam's center). Thus, the anticlockwise moment is:
[tex]$$
\text{Moment from beam} = 56\,\text{N} \times (1.2 - x).
$$[/tex]
Setting these moments equal gives:
[tex]$$
100x = 56(1.2 - x).
$$[/tex]
Now, solve this equation step by step:
1. Expand the right side:
[tex]$$
100x = 67.2 - 56x.
$$[/tex]
2. Add [tex]$56x$[/tex] to both sides:
[tex]$$
100x + 56x = 67.2,
$$[/tex]
which simplifies to:
[tex]$$
156x = 67.2.
$$[/tex]
3. Divide both sides by [tex]$156$[/tex] to solve for [tex]$x$[/tex]:
[tex]$$
x = \frac{67.2}{156} \approx 0.4308\,\text{m}.
$$[/tex]
Thus, the support must be placed approximately [tex]$0.43\,\text{m}$[/tex] from the end where the [tex]$100\,\text{N}$[/tex] weight is hung.
To verify, we can compute the moments:
- Moment from the load: [tex]$100 \times 0.4308 \approx 43.08\,\text{Nm}$[/tex].
- Moment from the beam: [tex]$56 \times (1.2 - 0.4308) \approx 56 \times 0.7692 \approx 43.08\,\text{Nm}$[/tex].
Since these moments are equal, the beam balances horizontally.
The final answer is that the support should be located approximately [tex]$0.43\,\text{m}$[/tex] from the end where the [tex]$100\,\text{N}$[/tex] weight is attached.
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