Equilibrium of a rigid body: Car on a hill

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William Haynes

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Question statement

{geogebra_applet('w4xb6n95',['alpha': radians(alpha), 'fbd': [visible: false] ])}

A car is parked on a {alpha}° slope, with the rear wheels locked by the parking brake. Find the components of the forces acting on each of the front and rear wheels, parallel and perpendicular to the roadbed.

Vehicle Specifications

Wheelbase	$L =$ {wb} mm
Curb weight	$m =$ {m} kg
Center of Gravity	$d =$ {d} mm $h =$ {h} mm

Give any introductory information the student needs.

			Name	Type	Generated Value

w_y	number	11815.3590417719
A	number	7147.6688875647
w_x	number	2083.3665858221
B_x	number	2083.3665858221
B_y	number	4667.6901542072
W	number	11997.63
debug	boolean	false

			Name	Type	Generated Value

alpha	integer	10
m	integer	1223
wb	integer	2527
units	list	[ "N", "mm" ]
d	integer	859
h	decimal	dec("790")
g	number	9.81

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Name

Data type

Value

xxxxxxxxxx
 
W cos(radians(alpha))

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Generated value: `number`

11815.3590417719

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Parts

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$A_{||} =$ Gap 0. $\qquad A_\perp = $ Gap 3. $0 B_{\perp} = \var{qty(A/2 ,'N')}$

$B_{||} =$ Gap 1. $\qquad B_\perp = $ Gap 2. $B_{||}= \var{qty(B_x/2,'N')} B_{\perp} = \var{qty(B_y/2,'N')}$

Advice

{geogebra_applet('w4xb6n95',['alpha': radians(alpha), 'fbd': [definition: 'true', visible: true] ])}

First, draw a free body diagram of the vehicle.

For convenience, let the x- and y- directions be parallel and perpendicular to the roadbed and define $\alpha$ as the the angle the road makes with the horizontal. Note that $\alpha$ is also the angle between the weight and the y-direction. Treat the front wheels like a roller, the rear wheels like a rough surface.

$\alpha = \var{alpha}°$

Define $L$ as the distance between the front and rear wheels:

$L= \var{qty(wb,'mm')}$

Note that the curb 'weight' is actually a mass, so let:

$\begin{align}W &= m g\\ &= (\var{qty(m,'kg')}) \cdot (\var{qty(g, 'm/s^2')}) =\var{show(w)}\\ W_x& = W \sin \alpha = \var{show(w_x)}\\W_y& = W \cos \alpha = \var{show(w_y)}\end{align}$

Apply the equations of equilibrium, starting with the sum of the moments about either A or B, since two unknown forces intersect there.

$\begin{align}\Sigma M_A &= 0\\ B_y (L) &= W_y(d) + W_x (h)\\ B_y &= W \left(\dfrac{d \cos \alpha + h \sin \alpha}{L}\right)\\&= \var{show(B_y)}\\\\ \Sigma F_x &= 0\\ B_x &= W_x\\ B_x &= \var{show(w_x)}\\\\ \Sigma F_y &= 0\\A + B_y &= W_y \\A &= W_y-B_y\\&=\var{show(A)}\end{align}$

Since there are two front wheels and two rear wheels, the forces acting on each wheel are half of these values:

$A_{||} = 0 \text{ N} \qquad A_\perp = A/2 = \var{show(A/2)} \qquad B_{||} = B_x/2 = \var{show(B_x/2)} \qquad B_\perp = B_y/2 = \var{show(B_y/2)}$

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This question is used in the following exams:

Chapter 5 Exercises by William Haynes in Engineering Statics.
Elise's copy of 14. More Equilibrium Problems by Elise Agra in Elise's workspace.
Chapter 05 Exercises by Michael Proudman in Engineering Statics.
14. More Equilibrium Problems by Michael Proudman in Engineering Statics.
studio 4 - Chapter 5 Exercises by Jay McCormack in studio 4.
studio 4 - Chapter 5 Exercises by Jay McCormack in Jay's workspace.
14. More Equilibrium Problems by William Haynes in Engineering Statics.

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Equilibrium of a rigid body: Car on a hill

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

William Haynes 5 years, 11 months ago

William Haynes 5 years, 11 months ago

William Haynes 5 years, 11 months ago

Error in variable testing condition

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Warning

Generated value: number

Parts

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Generated value: `number`