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2 years ago

Documentation of a flow chart?

Engineering

1 answer:

jonny [76]2 years ago

8 0

Answer:

A Document in a flowchart represents a printout, for instance a document or a report.

Flowcharts are used in planning and documenting basic process or projects. Like different sorts of outlines, they help imagine what is happening and along these lines help comprehend a process, and maybe additionally find more subtle highlights inside the process.

Explanation:

A flowchart is essentially a graphical portrayal of steps. It shows steps in consecutive request and is generally used in introducing the progression of calculations or work process. Regularly, a flowchart shows the means as boxes of different sorts, and their request by interfacing them with bolts.

It is likewise a kind of graph that speaks to a work process or cycle. A flowchart can likewise be characterized as a diagrammatic portrayal of a calculation, a bit by bit way to deal with settling an assignment.

The flowchart shows the means as boxes of different sorts, and their request by associating the containers with bolts. This diagrammatic portrayal outlines an answer model to a given issue. Flowcharts are used in investigating, planning, reporting or dealing with a process or program in different fields.

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Explanation:

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2 years ago

A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

$F_f= \mu N_2$

where :

$F_f =$ friction force

$\mu$ = coefficient of friction

$N_2$ = normal friction

Given that:

$\mu$ = 0.3

$F_f =$ 0.3 $N_2$

Using the equation of equilibrium along horizontal direction.

$\sum f_x = 0$

P - $F_f =$ 0

P = 0.3 $N_2$ ----- Equation (1)

To determine the moment about point B ; we have the expression

$\sum M_B = 0$

0 = $N_2*70-W*35-P*100$

where;

P = horizontal force

$N_2$ = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = $N_2*70-650*35-100*P$

$P = \frac{70 N_2-22750}{100} ----- equation (2)$

Replacing $\frac{70 N_2-22750}{100}$ for P in equation (1)

$\frac{70N_2 -22750}{100} =0.3 N_2$

$N_2 = \frac{22750}{40}$ $N_2 = 568.75 \ N$

Plugging the value of $N_2 = 568.75 \ N$ in equation (2)

$P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}$

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b) From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

$\sum M_A = 0$