A 200 kg steel bar sits horizontally on two supports as shown in the figure. The force on each support is (\( g = 9.8 \, \text{ms}^{-2} \))
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For a uniform bar in equilibrium on two supports, the total weight is equally distributed if the supports are symmetric. Use \( F_{\text{each}} = \frac{m g}{2} \) to find the force on each support.
The steel bar of mass 200 kg rests horizontally on two supports, with the supports symmetrically placed at the ends of the bar (as implied by the problem setup). The total weight of the bar is:
\[
W = m g = 200 \times 9.8 = 1960 \, \text{N}
\]
Since the bar is in equilibrium and the supports are symmetrically placed, the weight is equally distributed between the two supports. Therefore, the force on each support is:
\[
F_{\text{each support}} = \frac{W}{2} = \frac{1960}{2} = 980 \, \text{N}
\]
So, the force on each support is 980 N.
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Approach Solution -2
Given:
Mass of the steel bar = 200 kg
Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)
The total weight of the steel bar is:
\[
W = m \cdot g = 200 \cdot 9.8 = 1960 \, \text{N}
\]
Since the bar is symmetrically supported on two supports and is in equilibrium, the weight is equally distributed between the two supports.
Therefore, the force on each support is:
\[
\frac{1960}{2} = 980 \, \text{N}
\]
Final Answer:
980 N
