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Stels [109]
2 years ago
12

Draw the free-body diagram of the beam which supports the 80-kg load and is supported by the

Engineering
1 answer:
Sauron [17]2 years ago
5 0

The free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.

The first image shows the diagram of the beam and the second image shows the free-body diagram of the beam.

The resolution of forces in the system is well understood by the principle of equilibrium where a stationary body will remain balanced when subject to parallel forces provided that the total sum of the overall external forces is zero.

The free-body diagram is a graphical representation used to visualize the forces applied to an object.

The equilibrium of forces on the x-axis is:

\mathbf{\sum F_x  = 0}

The equilibrium of forces on the y-axis is:

\mathbf{\sum F_y = 0}

The equilibrium condition at any point is:

\mathbf{\sum M = 0}

From the free body diagram attached in the second image below,

  • the horizontal reaction is located at point A as \mathbf{ A_x}
  • the vertical reaction is  located at point A  as \mathbf{A_y}
  • the tension =  T
  • the weight = W

Therefore, we can conclude that the free-body diagram of the beam which supports the 80-kg load and is supported by the pin at A can be seen in the image attached below.

Learn more about the free-body diagram here:

brainly.com/question/19345060?referrer=searchResults

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Answer:

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As per the question:

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Specific heat of Aluminium, C_{p} = 0.949\ kJ/kgK

At room temperature

Specific heat of iron, C_{p} = 0.45\ kJ/kgK

Now,

To calculate the final equilibrium temperature:

Amount of heat loss by Aluminium = Amount of heat gain by Iron

MC_{p}\Delta T = mC_{p}\Delta T

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\Delta s = \Delta s_{a} + \Delta s_{i}

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For Aluminium:

\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}

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\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}

\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K

Thus

\Delta s =-2.025 + 2.253 = 0.228\ kJ/K

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