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3 years ago

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

1 answer:

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:

#include <iostream>

#include <string>

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string studentFName;

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char getStudentGrade(int testScore){

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}

else if(testScore >= 60) {

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outputFile.open ("outputStudentData.txt");

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4 years ago

Which items are NOT found on a

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3 years ago

An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the

Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe $T_p$ = -15° C = (-15 +273)K =258 K

Temperature of water $T_w$ = 3° C = (3 + 273)K = 276 K

Temperature of ice $T_i$ = 0° C = (0 +273)K =273 K

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convection coefficient $Lh_l$ =2000 W/m².K

The energy balance can be expressed as:

$q_{conduction} =q_{convention}$

where;

$q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}$ ------------- equation (1)

$q_{convention} = \pi DLh_l(T_w-T_i)$ ------------ equation(2)

Equating both equation (1) and (2); we have;

$\frac{2\pi LK(T_i-T_p)}{In(R/r)}$ $= \pi DLh_l(T_w-T_i)$

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$In(\frac{0.05}{r})*1884.96 = 182.84$

$In(\frac{0.05}{r}) = \frac{182.84}{1884.96}$

$In(\frac{0.05}{r}) =0.0970$

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t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

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4 years ago