Taller

Chapter 12 Solutions
12.51

Choosing torques about R, with ∑τ = 0,
–

1

Ry

L
2L
(350 N) + (T sin 12.0°)  – (200 N)L = 0
2
3

T

Rx

12.0°

From which, T = 2 .71 kN

200 N
350 N

Let Rx = compression force along spine, and from ∑Fx = 0,
Rx = Tx = T cos 12.0° = 2.65 kN
12.52

(a)

Using the first diagram, ∑Fx = 0 gives
–T1 cos θ1 + T2 cos θ2 = 0
or

T1T2

t
θ11

t2
θ2

T3

cos θ1
T
cos θ2 1

T2 = 

If θ1 = θ2,
then T 2 = T 1
( b)

Since θ1 = θ2, T2 = T1
T1

Using the second diagram and ∑Fy gives:
T1 sin 8.00° – mg = 0
T1 =

t1
θ1

so

T3

200 N
= 1.44 kN
sin 8.00°

Then, T 2 = T 1 = 1 .44 kN
Also, ∑Fx = 0 gives –T1 cos 8.00° + T3 = 0, or
400 N

(200 N) sin 37.0°

T3 = (1.44 kN) cos 8.00° = 1 .42kN

12.53

(a)

Locate the origin at the bottom left corner of the
cabinet and let x = distance between the resultant
normal force and the front of the cabinet. Then we
have

h

(200 N) cos 37.0°

f
n
60.0 cm

(1) ∑Fx = 200 cos(37.0°) – µn = 0
(2) ∑Fy = 200 sin(37.0°) + n – 400 = 0, and
(3) ∑τ = n(0.600 – x) – 400(0.300) + 200 sin 37.0°(0.600) – 200 cos 37.0°(0.400) = 0

©2000 by Harcourt College Publishers. All rights reserved.

2

Chapter 12 Solutions

From (2), n = 400 – 200 sin 37.0° = 280 N
From (3), x =

[72.2 – 120 + 260(0.600) – 64.0]
= 20.1 cm
280

Then from (1),
( b)

µk =

to the left of the front edge.

200 cos 37.0°
= 0.571
280

In this case, locate the origin x = 0 at the bottom right corner of the cabinet. Since the
cabinet isabout to tip, we can use Στ = 0 to find h:

Στ = 400(0.300) – 300 cos 37.0°(h) = 0
h=
12.54

120
= 0.501 m
300 cos 37.0°

(a) & (b) Use the first diagram and sum the torques about the lower front corner of the
cabinet.
0.300 m

F’
F

θ

φ

400 N
1.00 m

1.00 m

400 N

θ
f
f

n
0.600 m

n

∑τ = 0 ⇒ – F(1.00 m) + (400 N)(0.300 m) = 0
yielding F =

(400N)(0.300 m)
= 120 N
1.00 m

∑Fx = 0 ⇒ – f + 120 N = 0,

or f = 120 N

∑Fy = 0 ⇒ – 400 N + n = 0, so n = 400 N
Thus, µs =
( c)

f
120 N
=
= 0.300
n 400 N

Apply F' at the upper rear corner and directed so θ + φ = 90.0° to obtain the largest
possible lever arm.

θ = tan–1 

1.00 m 
= 59.0°
0.600 m

Thus, φ = 90.0° – 59.0° = 31.0°

© 2000 by Harcourt College Publishers. Allrights reserved.

Chapter 12 Solutions

3

Sum the torques about the lower front corner of the cabinet:
–F'
so

(1.00 m)2 + (0.600 m)2 + (400 N)(0.300 m) = 0

F' =

120 N · m
= 103 N
1.17 m

Therefore, the minimum force required to tip the cabinet is
103 N applied at 31.0 ° a bove the horizontal at the upper left corner .
12.55

(a)

(b)

Just three forces act on therod: forces perpendicular to the sides of the trough at A and B,
and its weight. The lines of action of A and B will intersect at
a point above the rod. They will have no torque about this
point. The rod’s weight will cause a torque about the point of
intersection as in Figure 1, and the rod will not be in
equilibrium unless the center of the rod lies vertically below
B
the intersectionpoint, as in Figure 2. All three forces must be
A
Fg
concurrent. Then the line of action of the weight is a diagonal
of the rectangle formed by the trough and the normal forces,
and the rod's center of gravity is vertically above the bottom
O
of the trough.
—
—
In Figure 2, AO cos 30.0° = BO cos 60.0° and
—
—
—
—
cos2 30.0°
L2 = AO 2 + BO 2 = AO 2 + AO 2  2
cos 60.0°

B

(1)ph = I ω

(2)

60.0°
O

—
So cos θ = AO /L = 1/2 and θ = 60.0°
12.56

Fg

θ
A
30.0°

—
A O = L/ 1 + cos2 30.0°/cos2 60.0° = L/2

p = MvCM

ω
p

If the ball rolls without slipping, R ω = vCM
So, h =

h

Iω
Iω
I
2
=
=
=
R
p
MvCM
MR
5

© 2000 by Harcourt College Publishers. All rights reserved.

vCM

Chapter 12 Solutions

4

12.57

(a)

We...