Particle Equilibrium: Use slopes to find components

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Question statement

{applet1("G", 300)}

A $\var{qty(F,"N")}$ force $F$ is applied in the direction shown to a small ring, which is held in equilibrium by two inextensible ropes attached to fixed points $A$ and $B$ .

Give any introductory information the student needs.

			Name	Type	Generated Value

alpha	number	-144.462322208
beta	number	78.690067526
phi	number	-45
dxA	number	-7
dyA	number	-5
dxB	number	1
dyB	number	5
dxF	number	3
dyF	number	-3
rA2	number	74
rB2	number	26
rF2	number	18

			Name	Type	Generated Value

applet	ggbapplet	HTML node
A	vector	vector(0,-10)
B	vector	vector(8,0)
P	vector	vector(7,-5)
C	vector	vector(10,-8)
F	integer	300
FA	number	364.9657512391
FB	number	432.6661530861
soln	matrix	Matrix of size 2×3
vecA	vector	Vector with 2 components
vecB	vector	Vector with 2 components
vecF	vector	vector(212.1,-212.1)
Check	vector	Vector with 2 components

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
degrees(atan2(A[1]-P[1], A[0]-P[0]))

JME function reference

Description

Direction of Force A will only point in quadrant 2 or 4

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

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Generated value: `number`

-144.462322208

← Depends on:

→ Used by:

soln
vecA

Advice
Variable testing condition

Parts

Gap-fill
1. Gap A Engineering Accuracy with units
  Add
2. Gap B Engineering Accuracy with units
  Add

Gap-fill

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

Determine the tensions in the two ropes.

$A = $ A

$B = $ B

Advice

1. Draw FBD of the Particle

{applet1("D", 250)}

2. Determine slope triangles from geometry.

Force	Rise	Run	Hypotenuse
$F$	$\var{dyF}$	$\var{dxF}$	$\sqrt{\var{rF2}}$
$A$	$\var{dyA}$	$\var{dxA}$	$\sqrt{\var{rA2}}$
$B$	$\var{dyB}$	$\var{dxB}$	$\sqrt{\var{rB2}}$

3. Determine scalar components in terms of $A$, $B$, and $F$ from slopes and given $F = \var{qty(F,'N')}$

$\begin{align}
F_x &= \left(\frac{\var{dxF}}{\sqrt{\var{rF2}}}\right) F = \var{precround(qty(vecF[0], 'N'),4)} & F_y &= \left(\frac{\var{dyF}}{\sqrt{\var{rF2}}}\right) F = \var{precround(qty(vecF[1], 'N'),4)}\\
A_x &= \left(\frac{\var{dxA}}{\sqrt{\var{rA2}}}\right) A = \var{precround(cos(radians(alpha)),3)} A & A_y &= \left(\frac{\var{dyA}}{\sqrt{\var{rA2}}}\right) A = \var{precround(sin(radians(alpha)),3)} A\\
B_x &= \left(\frac{\var{dxB}}{\sqrt{\var{rB2}}}\right) B = \var{precround(cos(radians(beta )),3)} B & B_y &= \left(\frac{\var{dyB}}{\sqrt{\var{rB2}}}\right) B = \var{precround(sin(radians(beta )),3)} B
\end{align}$

4. Set up two equations of equilibrium.

$\begin{align}
\Sigma F_x &= 0 & \Sigma F_y &= 0\\
A_x + B_x + F_x &=0& A_y + B_y +F_y &=0\\
\simplify[!noLeadingMinus]{{precround(cos(radians(alpha)),3)} A + {precround(cos(radians(beta)),3)} B} & =\var{precround(qty(- vecF[0], 'N'),4)} &
\simplify[!noLeadingMinus]{{precround(sin(radians(alpha)),3)} A + {precround(sin(radians(beta)),3)} B} & =\var{precround(qty(- vecF[1], 'N'),4)}\\
\end{align}$

5. Solve simultaneously to get:

$\begin{align}
A &= \var{siground(qty(abs(FA),'N'),4)} & B &= \var{siground(qty(abs(FB),'N'),4)}
\end{align}$

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This question is used in the following exam:

Chapter 3 Exercises by William Haynes in Engineering Statics.

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Particle Equilibrium: Use slopes to find components

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William Haynes 2 years, 3 months ago

William Haynes 2 years, 3 months ago

William Haynes 2 years, 3 months ago

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Error in variable testing condition

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Particle Equilibrium: Use slopes to find components

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

William Haynes 2 years, 3 months ago

William Haynes 2 years, 3 months ago

William Haynes 2 years, 3 months ago

William Haynes 2 years, 3 months ago

Error in variable testing condition

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Generated value: number

Parts

Gap-fill

Generated value: `number`