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Deck of Cards
idbigdeck
h3.

Part

A

A

3.0

m

long

ladder

is

leaned

against

a

vertical

wall

at a 70° angle above the horizontal. The wall is frictionless, but the (level) floor beneath the ladder does have friction. Assuming the center of mass of the 15.0 kg ladder is in its exact center, find the forces exerted on the ladder by the wall and by the floor.

Solution

Card
labelPart A
Wiki Markup
Toggle Cloak
idsysa
System:
Cloak
idsysa

The ladder as a .

Toggle Cloak
idinta
Interactions:
Cloak
idinta

External influences from the earth (gravity) the wall (normal force) and the floor (normal force and frictional force).

Toggle Cloak
idmoda
Model:
Cloak
idmoda

and .

Toggle Cloak
idappa
Approach:

Cloak
idappa

Toggle Cloak
iddiaga
Diagrammatic Representation

Cloak
iddiaga

This is a statics problem, so the object is to ensure that the ladder does not rotate or translate. Thus, we know that all accelerations and angular accelerations are zero. We first sketch the situation and set up coordinates, which includes selecting an axis. Selecting the position of the axis is technically arbitrary, but often it is actually specified by the unknowns in the system. In this case, as we shall see, the forces in the x-direction are less constrained than the y-direction forces. Thus, it is a good idea to position the axis to cancel the torque from one of the x forces. We choose to put the axis at the point of contact with the floor.

Image Added

Cloak
diaga
diaga

Toggle Cloak
idmatha
Mathematical Representation

Cloak
idmatha

We then write the equations of linear and rotational equilibrium:

Latex
 a 70° angle above the horizontal.   The wall is frictionless, but the (level) floor beneath the ladder does have friction.  Assuming the center of mass of the 15.0 kg ladder is in its exact center, find the forces exerted on the ladder by the wall and by the floor.  

h4. Solution

{toggle-cloak:id=sysa} *System:*  {cloak:id=sysa} The ladder as a [rigid body].{cloak}

{toggle-cloak:id=inta} *Interactions:* {cloak:id=inta}External influences from the earth (gravity) the wall (normal force) and the floor (normal force and frictional force).{cloak}

{toggle-cloak:id=moda} *Model:*  {cloak:id=moda}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appa} *Approach:*  

{cloak:id=appa}

{toggle-cloak:id=diaga} {color:red} *Diagrammatic Representation*{color}

{cloak:id=diaga}
This is a [statics] problem, so the object is to ensure that the ladder does _not_ rotate or translate.  Thus, we know that all accelerations and angular accelerations are zero.  We first sketch the situation and set up coordinates, which includes selecting an axis.  Selecting the position of the axis is technically arbitrary, but often it is actually specified by the unknowns in the system.  In this case, as we shall see, the forces in the x-direction are less constrained than the y-direction forces.  Thus, it is a good idea to position the axis to cancel the torque from one of the x forces.  We choose to put the axis at the point of contact with the floor.

!ladder1.jpg!

{cloak:diaga}
{toggle-cloak:id=matha} {color:red} *Mathematical Representation* {color}
{cloak:id=matha}

We then write the equations of linear and rotational equilibrium:

{latex}\begin{large}\[ \sum F_{x} = F_{w} + F_{f,x} = 0 \]
\[ \sum F_{y} = - mg + F_{f,y} = 0 \]
\[ \sum \tau = + mg (L/2) \cos\theta - F_{w} L \sin\theta \]\end{large}{latex}

{note}Note that our selection of the axis point guaranteed that the floor's force would contribute zero torque.{note}

The y-direction equation and the torque equation are immediately solvable to obtain:

{latex}
Note

Note that our selection of the axis point guaranteed that the floor's force would contribute zero torque.

The y-direction equation and the torque equation are immediately solvable to obtain:

Latex
\begin{large} \[ F_{f,y} = mg = \mbox{147 N}\]
\[ F_{w} = \frac{mg}{2}\cot\theta = \mbox{27 N}\]\end{large}{latex}

Once

we

have

_

F

_~w~

w,

we

can

find

the

x-component

of

the

floor's

force:

{
Latex
}\begin{large}\[ F_{f,x} = -F_{w} = - \mbox{27 N}\]\end{large}{latex}

We

can

now

construct

the

total

floor

force:

Image Added

Which gives a total force of 150 N at 80° above the horizontal.

Cloak
matha
matha

!ladderfloorforce.jpg! Which gives a total force of 150 N at 80° above the horizontal. {cloak:matha} {cloak:appa}
Cloak
appa
appa

unmigrated-wiki-markup
Card
labelPart B
h3.

Part

B

Consider

the

same

basic

situation

as

in

Part

A,

but

now

suppose

that

the

wall

has

friction.

If

the

coefficient

of

static

friction

between

the

floor

and

the

ladder

is

only

0.15,

what

is

the

minimum

coefficient

of

static

friction

needed

between

the

wall

and

the

ladder

to

prevent

the

ladder

from

slipping?

h4. Solution {

Solution

Toggle Cloak

:

id

=

sysb

} *

System:

* {

Cloak

:

id

=

sysb

}

The

ladder

as

a

[rigid body].{cloak} {

.

Toggle Cloak

:

id

=

intb

} *

Interactions:

* {

Cloak

:

id

=

intb

}

External

influences

from

the

earth

(gravity)

the

wall

(normal

force

and

friction)

and

the

floor

(normal

force

and

frictional

force).

{cloak} {

Toggle Cloak

:

id

=

modb

} *

Model:
Cloak
idmodb

and .

Toggle Cloak
idappb
Approach:

Cloak
idappb

Toggle Cloak
idprelim
An Important Assumption

Cloak
idprelim

The minimum coefficient of friction will occur when both friction forces are maxima. Thus, we have the important relationships:

Latex
*  {cloak:id=modb}[Single-Axis Rotation of a Rigid Body] and [Point Particle Dynamics].{cloak}

{toggle-cloak:id=appb} *Approach:*  
{cloak:id=appb}

{toggle-cloak:id=prelim} {color:red} *An Important Assumption* {color}

{cloak:id=prelim}
The minimum coefficient of friction will occur when both friction forces are maxima.  Thus, we have the important relationships:

{latex}\begin{large}\[ |F_{f,x}| = \mu_{s,f} |F_{f,y}| \]
\[ |F_{w,y}| = \mu_{s,w} |F_{w,x}| \]\end{large}{latex}
{cloak:prelim}

{
Cloak
prelim
prelim

Toggle Cloak

:

iddiagb
Diagrammatic Representation

Cloak
iddiagb

Because we know the floor's static friction coefficient, we can streamline the solution by switching our choice of axis.

Image Added

Cloak
diagb
diagb

Toggle Cloak
idmathb
Mathematical Representation

Cloak
idmathb

The resulting equations are:

Latex
=diagb} {color:red} *Diagrammatic Representation* {color}

{cloak:id=diagb}

Because we know the floor's static friction coefficient, we can streamline the solution by switching our choice of axis.

!ladder2.jpg!

{cloak:diagb}

{toggle-cloak:id=mathb} {color:red} *Mathematical Representation* {color}

{cloak:id=mathb}

The resulting equations are:

{latex}\begin{large}\[ \sum F_{x} = F_{w,x} - \mu_{s,f}F_{f,y} = 0 \]
\[ \sum F_{y} = - mg + F_{f,y} + \mu_{s,w}F_{w,x}= 0 \]
\[ \sum \tau = - mg (L/2) \cos\theta + F_{f,y} L\cos\theta - \mu_{s,f}F_{f,y} L \sin\theta \]\end{large}{latex}

{note}Note that the 
Note

Note that the y-component

of

the

floor

force

is

the

normal

force

from

the

floor

on

the

ladder,

and

the

x-component

of

the

wall

force

is

the

normal

force

from

the

wall

on

the

ladder.

Since

we

are

assuming

static

friction

is

maximized,

we

have

rewritten

_

F

_~f

f,

x~

x and

_

F

_~w

w,

y~

y as

the

coefficient

of

friction

times

the

relevant

normal

force.

{note}

The

only

equation

that

can

be

solved

immediately

is

torque

balance,

which

gives:

{
Latex
}\begin{large}\[ F_{f,y} = \frac{mg}{2\left(1-\mu_{s,f}\tan\theta\right)} = \mbox{125 N} \]\end{large}{latex}

We

can

then

substitute

this

result

into

the

equation

of

x-force

balance

to

find:

{
Latex
}\begin{large}\[ F_{w,x} = \mu_{s,f}F_{f,y} = \mbox{18.8 N} \]\end{large}{latex}

Finally,

we

can

find

the

required

coefficient

of

static

friction

to

be:

{
Latex
}\begin{large}\[ \mu_{s,w} = \frac{mg-F_{f,y}}{F_{w,x}} = 1.19 \] \end{large}{latex}

{tip}We found in Part A that the ladder is stable as long as the coefficient of friction with the ground is 0.19 or greater, even if there is no friction at all from the wall.  Does it make sense that in Part B the wall's friction coefficient has to be so much larger than 0.19 if the wall is only partly responsible for keeping the ladder from sliding?{tip}

{cloak:mathb}
{cloak:appb}
Tip

We found in Part A that the ladder is stable as long as the coefficient of friction with the ground is 0.19 or greater, even if there is no friction at all from the wall. Does it make sense that in Part B the wall's friction coefficient has to be so much larger than 0.19 if the wall is only partly responsible for keeping the ladder from sliding?

Cloak
mathb
mathb

Cloak
appb
appb