Frame: A-frame Difficulty 3

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William Haynes

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2 years, 11 months ago

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William Haynes 2 years, 11 months ago

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William Haynes 6 years, 4 months ago

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William Haynes 6 years, 4 months ago

Created this as a copy of Frame: A-frame Difficulty 2.

Name	Status	Author	Last Modified
Frame: A-frame Difficulty 1	Ready to use	William Haynes	28/02/2023 23:29
Frame: A-frame Difficulty 2	Ready to use	William Haynes	21/03/2021 14:25
Frame: A-frame Difficulty 3	Ready to use	William Haynes	30/03/2022 19:14
Frame: A-frame Difficulty 4	Ready to use	William Haynes	31/03/2022 01:05
Keith's copy of Frame: A-frame Difficulty 1	draft	Keith Tarnowski	03/06/2019 11:52
Katrin's copy of Frame: A-frame Difficulty 1	draft	Katrin Thomson	03/06/2019 11:53

There are 7 other versions that do you not have access to.

Question statement

{geogebra_applet('qe7xhnse',[["B",B],["C",C],['units','"'+units[1]+'"']]+ggb_load )}

The A-frame shown supports a {display(abs(F_d))} force $D$ at the midpoint of member AB and a {display(abs(M_E)) } {if(me>0,"counter-","")}clockwise moment $M_E$ at E as shown.

Determine the scalar components of the reactions at pins A and C.

Note: the sign of a scalar component indicates the sense of the force. A positive sense indicates the force points towards the positive end of the corresponding axis.

Give any introductory information the student needs.

			Name	Type	Generated Value

F_w	quantity	quantity(0, "lbs")
F_d	quantity	quantity(475, "lbs")
M_E	quantity	quantity(-1800, "lbs*ft")
alpha	number	38.6598082541
beta	number	38.6598082541
x1	quantity	quantity(5, "ft")
y1	quantity	quantity(4, "ft")
x2	quantity	quantity(5, "ft")
y2	quantity	quantity(4, "ft")
description	string	475 lbs force $D$ at the midpo
Cy	quantity	quantity(275, "lbs")
Cx	quantity	quantity(-106.25, "lbs")

			Name	Type	Generated Value

vecW	vector	vector(0,0)
vecA	vector	vector(-581.25,-275)
vecC	vector	vector(106.25,275)
sigma_f	vector	vector(0,0)
vecD	vector	vector(475,0)

			Name	Type	Generated Value

B	vector	vector(5,4)
C	vector	vector(10,0)
fd	integer	475
me	integer	-1800
w	integer	0
units	list	[ "lb", "ft" ]
ggb_load	list	Nested 3×2 list
debug	boolean	false

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
qty(w,units[0])

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `quantity`

quantity(0, "lbs")

← Depends on:

units
w

This variable doesn't seem to be used anywhere.

Parts

Gap-fill

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Ask the student a question, and give any hints about how they should answer this part.

$A_x = $ $A_x$ $A_y = $ $A_y$ $C_x = $ $C_x$ $C_y = $ $C_y$

Advice

Strategy:

Draw a free body diagram of the entire frame and each of its legs. Note that members AB or BC are not two-force bodies; therefore the direction of the forces at A and C don't align with the members and should be represented with x and y components. In this case all three of the free body diagrams have four unknowns. Fortunately, the lines of action of three of the unknowns on FBD I intersect at points A and C.

{geogebra_applet('p6rz9bqx',[["B",B],["C",C]] + ggb_load )}

Known:

$M_E = \var{display(abs(M_E))}$ and $D = \var{display(abs(F_D))}$

Determine necessary angles:

$\alpha = \tan^{-1}\left(\frac{\var{y1}}{\var{x1}}\right) = \var{siground(alpha,4)}$°

$\beta = \tan^{-1}\left(\frac{\var{y2}}{\var{x2}}\right) = \var{siground(beta,4)}$°

Start with FBD I and take moments at point A to find force $C_y$.

$\begin{align} \textrm{I: } \Sigma M_A &= 0\\ \simplify{{sign(-fd)} D ({y1/2}) + {sign(me)} M_E } + C_y ( \var{x1 + x2}) &= 0\\ C_y &= \dfrac{\simplify{({sign(fd)} D ({y1/2}) + {-sign(me)} M_E )}} { \simplify{({x1+x2})}}\\ &= \var{display(Cy)}\\ &= \var{display(abs(Cy))} \var{if(VecC[1]<0,latex('\\downarrow'),latex('\\uparrow'))} \end{align}$

Use FBD III and take moment about B to find $C_x$.

$\begin{align} \textrm{III: } \Sigma M_B &= 0\\ \simplify{- C_x ({y1}) + C_y ({x2}) + {sign(me)} M_E } &= 0\\ C_x &= \dfrac{\simplify{( C_y ({x2}) + {-sign(me)} M_E )}} {\var{y2}}\\ &= \var{display(Cx)}\\ &= \var{display(abs(Cx))} \var{if(VecC[0]<0,latex('\\leftarrow'),latex('\\rightarrow'))} \end{align}$

Apply $\Sigma F_x = 0$ and $\Sigma F_y=0$ to FBD I to find the reactions at A.

$\begin{align} \Sigma F_x &= 0\\ \simplify{ A_x + {sign(fd)} D } - C_x &= 0\\ A_x &= \simplify{C_x - {sign(fd)} D } \\&= \simplify[!collectNumbers]{({-siground(vecC[0],4)}) + ({-vecD[0]})}\\ &= \simplify{({-siground(vecC[0],4)}) + ({-vecD[0]})}\\&= \var{display(abs(qty(vecA[0],units[0])))} \var{if(vecA[0]<0,latex('\\leftarrow'),latex('\\rightarrow'))} \end{align}$

$\begin{align}\Sigma F_y &= 0\\ A_y + C_y &=0\\A_y &= -C_y \\ &= -( \var{display(qty(vecC[1],units[0]))}) \\ &= \var{display(abs(qty(vecA[1],units[0])))} \var{if(vecA[1]<0,latex('\\downarrow'),latex('\\uparrow'))} \end{align}$

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Taxonomy: mathcentre

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William Haynes 2 years, 11 months ago

William Haynes 2 years, 11 months ago

William Haynes 5 years, 9 months ago

William Haynes 5 years, 10 months ago

William Haynes 5 years, 10 months ago

William Haynes 6 years, 4 months ago

William Haynes 6 years, 4 months ago

Error in variable testing condition

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Generated value: `quantity`

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Frame: A-frame Difficulty 3

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

William Haynes 2 years, 11 months ago

William Haynes 2 years, 11 months ago

William Haynes 5 years, 9 months ago

William Haynes 5 years, 10 months ago

William Haynes 5 years, 10 months ago

William Haynes 6 years, 4 months ago

William Haynes 6 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: quantity

Parts

Gap-fill

Generated value: `quantity`