Physics For Scientists And Engineers 6E - part 93

S E C T I O N 1 2 . 3 • Examples of Rigid Objects in Static Equilibrium

369

This positive value indicates that our estimate of the direc-
tion of

R was accurate.

Finally,

To finalize this problem, note that if we had selected

some other axis for the torque equation, the solution might
differ in the details, but the answers would be the same. For
example, if we had chosen an axis through the center of
gravity of the beam, the torque equation would involve both
T and R. However, this equation, coupled with Equations
(1) and (2), could still be solved for the unknowns. Try it!

When many forces are involved in a problem of this na-

ture, it is convenient in your analysis to set up a table. For in-
stance, for the example just given, we could construct the
following table. Setting the sum of the terms in the last col-
umn equal to zero represents the condition of rotational
equilibrium.

580 N

R !

188 N

cos )

188 N

cos 71.1*

71.1*

) !

What If?

What if the person walks farther out on the beam?

Does T change? Does R change? Does

change?

Answer T must increase because the weight of the person ex-
erts a larger torque about the pin connection, which must be
countered by a larger torque in the opposite direction due to
an increased value of T. If T increases, the vertical component
of

R decreases to maintain force equilibrium in the vertical di-

rection. But force equilibrium in the horizontal direction re-
quires an increased horizontal component of

R to balance the

horizontal component of the increased

T. This suggests that )

will become smaller, but it is hard to predict what will happen
to R. Problem 26 allows you to explore the behavior of R.

Example 12.4 The Leaning Ladder

A uniform ladder of length ! rests against a smooth, vertical
wall (Fig. 12.11a). If the mass of the ladder is m and the co-
efficient of static friction between the ladder and the
ground is +

0.40, find the minimum angle )

min

at which

the ladder does not slip.

Solution The  free-body  diagram  showing  all  the  external
forces  acting  on  the  ladder  is  illustrated  in  Figure  12.11b.
The force exerted by the ground on the ladder is the vector
sum  of  a  normal  force

n and the force of static friction f

The reaction force

P exerted by the wall on the ladder

is horizontal because the wall is frictionless. Notice how we
have  included  only  forces  that  act  on  the  ladder.  For
example,  the  forces  exerted  by  the  ladder  on  the  ground
and on the wall are not part of the problem and thus do not
appear  in  the  free-body  diagram.  Applying  the  first  condi-
tion for equilibrium to the ladder, we have

(2)

n ' mg ! 0

(1)

P ! 0

At the Interactive Worked Example link at http://www.pse6.com, you can adjust the position of the person and observe
the effect on the forces.

53.0

8.00 m

(a)

Figure 12.10 (Example 12.3) (a) A uniform beam supported

by a cable. A person walks outward on the beam. (b) The free-

body diagram for the beam. (c) The free-body diagram for the

beam showing the components of R and T.

(c)

200 N

600 N

(b)

53.0

200 N

600 N

4.00 m

2.00 m

R cos

R sin

T cos 53.0

T sin 53.0

Force

Moment arm

component

relative to O (m)

Torque about O (N $ m)

T sin 53.0°

8.00

(8.00)T sin 53.0°

T cos 53.0°

200 N

4.00

(4.00)(200)

600 N

2.00

(2.00)(600)

R sin

)

R cos

)

Interactive

370

C H A P T E R 1 2 • Static Equilibrium and Elasticity

The first equation tells us that P ! f

. From the second

equation we see that n ! mg. Furthermore, when the lad-
der is on the verge of slipping, the force of friction must be
a maximum, which is given by f

s, max

n. (Recall Eq. 5.8:

n.) Thus, we must have P ! f

n ! +

mg.

To find )

min

, we must use the second condition for equi-

librium. When we take the torques about an axis through
the origin O at the bottom of the ladder, we have

This expression gives

What If?

What if a person begins to climb the ladder when

the angle is 51°? Will the presence of a person on the ladder
make it more or less likely to slip?

Answer The presence of the additional weight of a person
on  the  ladder  will  increase  the  clockwise  torque  about  its
base  in  Figure  12.11b.  To  maintain  static  equilibrium,  the
counterclockwise torque must increase, which can occur if P
increases.  Because  equilibrium  in  the  horizontal  direction
tells us that P ! f

, this would suggest that the friction force

rises above the maximum value f

s, max

and the ladder slips.

However, the increased weight of the person also causes n to
increase, which increases the maximum friction force f

s, max

Thus, it is not clear conceptually whether the ladder is more
or less likely to slip.

Imagine that the person of mass M is at a position d that

is measured along the ladder from its base (Fig. 12.11c).
Equations (1) and (2) can be rewritten

51*

)

min

tan )

min

1.25

(3)

(

sin ) ' mg

!
2

cos ) ! 0

Equation (3) can be rewritten

Solving this equation for tan ), we find

Incorporating Equations (4) and (5), and imposing the con-
dition that the ladder is about to slip, this becomes

When  the  person  is  at  the  bottom  of  the  ladder,  d ! 0.  In
this case, there is no additional torque about the bottom of
the ladder and the increased normal force causes the max-
imum  static  friction  force  to  increase.  Thus,  the  ladder  is
less likely to slip than in the absence of the person. As the
person climbs and d becomes larger, however, the numera-
tor  in  Equation  (6)  becomes  larger.  Thus,  the  minimum
angle  at  which  the  ladder  does  not  slip  increases.  Eventu-
ally,  as  the  person  climbs  higher,  the  minimum  angle  be-
comes  larger  than  51° and  the  ladder  slips.  The  particular
value  of  d at  which  the  ladder  slips  depends  on  the  coeffi-
cient  of  friction  and  the  masses  of  the  person  and  the
ladder.

(6)

tan )

min

m(!/2) $ Md

(m $ M)

tan ) !

mg(!/2) $ Mgd

(

sin ) ' mg

!
2

cos ) ' Mgd cos ) ! 0

(5)

n ' (m $ M )g ! 0

(4)

P ! 0

(a)

(b)

Figure 12.11 (Example 12.4) (a) A uniform ladder at rest, leaning against a smooth

wall. The ground is rough. (b) The free-body diagram for the ladder. (c) A person of

mass M begins to climb the ladder when it is at the minimum angle found in part (a) of

the example. Will the ladder slip?

At the Interactive Worked Example link at http://www.pse6.com, you can adjust the angle of the ladder and watch what
happens when it is released.

(c)

S E C T I O N 1 2 . 3 • Examples of Rigid Objects in Static Equilibrium

371

Example 12.5 Negotiating a Curb

(A)

Estimate the magnitude of the force

F a person must ap-

ply to a wheelchair’s main wheel to roll up over a sidewalk
curb (Fig. 12.12a). This main wheel that comes in contact
with the curb has a radius r, and the height of the curb is h.

Solution Normally, the person’s hands supply the required
force to a slightly smaller wheel that is concentric with the
main wheel. For simplicity, we assume that the radius of the
smaller  wheel  is  the  same  as  the  radius  of  the  main  wheel.
Let us estimate a combined weight of mg ! 1 400 N for the
person and the wheelchair and choose a wheel radius of r !
30 cm. We also pick a curb height of h ! 10 cm. We assume
that  the  wheelchair  and  occupant  are  symmetric,  and  that
each wheel supports a weight of 700 N. We then proceed to
analyze  only  one  of  the  wheels.  Figure  12.12b  shows  the
geometry for a single wheel.

When the wheel is just about to be raised from the street,

the reaction force exerted by the ground on the wheel at
point B goes to zero. Hence, at this time only three forces act
on the wheel, as shown in the free-body diagram in Figure
12.12c. However, the force

R, which is the force exerted by the

curb on the wheel, acts at point A, and so if we choose to have
our axis of rotation pass through point A, we do not need to
include

R in our torque equation. From the triangle OAC

shown in Figure 12.12b, we see that the moment arm d of the
gravitational force mg acting on the wheel relative to point A is

The moment arm of

F relative to point A is 2r ' h (see Fig-

ure 12.12c). Therefore, the net torque acting on the wheel
about point A is

(Notice that we have kept only one digit as significant.) This
result indicates that the force that must be applied to each
wheel is substantial. You may want to estimate the force re-
quired to roll a wheelchair up a typical sidewalk accessibility
ramp for comparison.

(B)

Determine the magnitude and direction of

Solution We use the first condition for equilibrium to de-
termine the direction:

Dividing the second equation by the first gives

tan ) !

700 N
300 N

R sin ) ' mg ! 0

F ' R cos ) ! 0

3 - 10

(700 N)

√

2(0.3 m)(0.1 m)'(0.1 m)

2(0.3 m) ' 0.1 m

F !

√

2rh ' h

2r ' h

√

2rh 'h

F(2r 'h) ! 0

mgd ' F(2r ' h) ! 0

d !

√

(r ' h)

√

2rh ' h

Figure 12.12 (Example 12.5) (a) A person in a wheelchair at-

tempts to roll up over a curb. (b) Details of the wheel and curb.

The person applies a force

F to the top of the wheel. (c) The

free-body diagram for the wheel when it is just about to be

raised. Three forces act on the wheel at this instant:

F, which is

exerted by the hand;

R, which is exerted by the curb; and the

gravitational force m

g. (d) The vector sum of the three external

forces acting on the wheel is zero.

(a)

(d)

(c)

2r – h

r – h

(b)

We can use the right triangle shown in Figure 12.12d to obtain
R:

800 N

R !

√

(mg)

√

(700 N)

(300 N)

) !

70*

372

C H A P T E R 1 2 • Static Equilibrium and Elasticity

Application Analysis of a Truss

Roofs,  bridges,  and  other  structures  that  must  be  both
strong and lightweight often are made of trusses similar to
the  one  shown  in  Figure  12.13a.  Imagine  that  this  truss
structure represents part of a bridge. To approach this prob-
lem,  we  assume  that  the  structural  components  are  con-
nected  by  pin  joints.  We  also  assume  that  the  entire  struc-
ture is free to slide horizontally because it rests on “rockers”
on each end, which allow it to move back and forth as it un-
dergoes thermal expansion and contraction. We assume the
mass of the bridge structure is negligible compared with the
load. In this situation, the force exerted by each of the bars
(struts)  on  the  hinge  pins  is  a  force  of  tension  or  of  com-
pression and must be along the length of the bar. Let us cal-
culate the force in each strut when the bridge is supporting
a 7 200-N load at its center. We will do this by determining
the forces that act at the pins.

The force notation that we use here is not of our usual

format. Until now, we have used the notation F

to mean

“the force exerted by A on B.” For this application, however,
the first letter in a double-letter subscript on F indicates the
location of the pin on which the force is exerted. The combi-
nation of two letters identifies the strut exerting the force on
the pin. For example, in Figure 12.13b, F

is the force ex-

erted by strut AB on the pin at A. The subscripts are symmet-
ric in that strut AB is the same as strut BA and F

First, we apply Newton’s second law to the truss as a

whole in the vertical direction. Internal forces do not enter
into this accounting. We balance the weight of the load with
the normal forces exerted at the two ends by the supports
on which the bridge rests:

7 200 N

Next, we calculate the torque about A, noting that the over-
all length of the bridge structure is L ! 50 m:

Although we could repeat the torque calculation for the
right end (point E), it should be clear from symmetry argu-
ments that n

3 600 N.

Now let us balance the vertical forces acting on the pin

at point A. If we assume that strut AB is in compression,
then the force F

that the strut exerts on the pin at point A

has a negative y component. (If the strut is actually in ten-
sion, our calculations will result in a negative value for the
magnitude of the force, still of the correct size):

The positive result shows that our assumption of compres-
sion was correct.

We can now find the force F

by considering the hori-

zontal forces acting on the pin at point A. Because point A is
not accelerating, we can safely assume that F

must point

toward the right (Fig. 12.13b); this indicates that the bar be-
tween points A and C is under tension:

Now consider the vertical forces acting on the pin at point
C. We shall assume that strut CB is in tension. (Imagine the
subsequent motion of the pin at point C if strut CB were to

(7 200 N) cos 30* ! 6 200 N

cos 30* ! 0

7 200 N

sin 30* ! 0

/2 ! 3 600 N

! !

(L/2)

50 m

(a)

(b)

Load: 7 200 N

Figure 12.13 (a) Truss structure for a bridge. (b) The forces acting on

the pins at points A, B, C, and E. Force vectors are not to scale.

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