Problem 1: What is the tension T exerted by the hamstring muscles in the back of the thigh and the
compressive force Fc in the knee joint due to the application of a horizontal force of 100 N to the ankle as
shown in the figure?
Solution: Require that Στ = 0 about an axis through the knee
joint and perpendicular to the page. This gives
100 N
Στ = Fc ( 0 ) − T ( 4.00 cm ) + ( 100 N )( 42.0 cm ) = 0 ,
or the tension in the muscle is
T = 1.05 × 103 N = 1.05 kN
Then, ΣFx = − Fc − 100 N + 1.05 × 103 N=0 gives the
compression force as Fc = 950 N
42.0 cm
T
4.00 cm
Fc
Problem2: The principal forces acting on a foot when a person is squatting are shown in the figure.
Determine the magnitude of the force FH exerted by the Achilles tendon on the heel at point H and the
magnitude of the force FJ exerted on the ankle joint at point J.
Solution: Require that Στ = 0 about an axis through point J
and perpendicular to the page. This gives
FH
− FH d1 + FJ ( 0 ) + ( 350 N )( 4.00 cm ) = 0 , or
FH =
52.0°
52.0°
( 350 N )( 4.00 cm )
H
d1
=
3.20 cm
( 350 N )( 4.00 cm ) 711 N
=
( 3.20 cm ) cos 52.0°
( )x = 0 ,
Then, ΣFx = 0 ⇒ ( 711 N ) sin 52.0° − FJ
or
( FJ )x = 560 N .
( )y + 350 N = 0 , giving ( FJ )y = 788 N .
( FJ )x + ( FJ )y =
2
2
J
4.00 cm
350 N
Also, ΣFy = 0 ⇒ ( 711 N ) cos 52.0° − FJ
Hence, FJ =
FJ
d1
( 560 N )2 + ( 788 N )2 = 966 N
Problem 3: The figure shows a person using both hands to lift a 30.0-kg barbell. When the lifter’s back is
horizontal, what is the magnitude of the tension T in the back muscles and the hinge force Fc which the hips
exert on the base of the spine? The weight w1 = 380 N is the weight of the upper torso, w2 = 60 N is the
weight of the head, and w3 = 394 N is the weight of the arms (100 N) plus the weight being lifted (294 N).
The tension T in the back muscles is directed 12° above the horizontal.
Solution: We consider the torques about
(Fc)y
an axis perpendicular to the
page through the base of the
spine, where the compression
force Fc acts.
Ty
12 cm 12 cm
(Fc)x
36 cm
Tx
380 N
394 N
Στ = − ( 380 N )( 36 cm ) + ( T sin 12° )( 48 cm ) − ( 394 N )( 60 cm ) − ( 60 N )( 78 cm ) = 0
or
T=
4.2 × 10 4 N ⋅ cm
= 4.2 × 103 N = 4.2 kN .
( 48 cm ) sin 12°
Then, ΣFx = 0 ⇒ ( Fc )x − Tx = 0, or
( Fc )x = T cos 12° = ( 4.2 kN ) cos 12° = 4.1 kN
and
ΣFy = 0 ⇒ ( Fc )y + Ty − ( 380 + 394 + 60 ) N=0,
or
( Fc )y = 834 N − ( 4.2 × 103
Thus, Fc =
and
( Fc )x + ( Fc )y
2
2
=
N ) sin 12° = − 41 N .
( 4.1 × 10
3
N ) + ( −41 N ) = 4.1 × 103 N
2
18 cm
2
⎡ ( Fc )y ⎤
−41 N ⎞
⎥ = tan −1 ⎛⎜
⎟ = −0.57° .
3
⎝ 4.1 × 10 N ⎠
⎢⎣ ( Fc )y ⎥⎦
θ = tan −1 ⎢
The compression force is, Fc = 4.1 × 103 N at 0.57° below the horizontal
60 N
Problem4: A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold
weighs 200 N and is 3.00 m long. What is the tension in each rope when the 700-N worker stands 1.00 m
from one end?
Solution: Consider the torques about an axis perpendicular to the page through the left
end of the scaffold.
Στ = 0 ⇒ T1 ( 0 ) − ( 700 N )( 1.00 m ) − ( 200 N )( 1.50 m ) + T2 ( 3.00 m ) = 0 .
From which,
T2 = 333 N .
Then, from ΣFy = 0 , we have
T1
700 N
1.00 m
T1 + T2 − 700 N-200 N=0 ,
T2
2.00 m
1.50 m
or
200 N
T1 = 900 N-T2 = 900 N-333 N= 567 N
Problem 5: A uniform semicircular sign 1.00 m in diameter and of weight w is supported by two wires as
shown in the figure. What is the tension in each of the wires supporting the sign?
Solution: We call the tension in the cord at the left end of the
sign, T1 and the tension in the cord near the right end
T2 . Consider the torques about an axis perpendicular
to the page and through the left end of the sign.
Στ = − w ( 0.50 m ) + T2 ( 0.75 m ) = 0 , so T2 =
From ΣFy = 0 , T1 + T2 − w = 0 , or
2
3
T1 = w − T2 = w − w =
1
w.
3
T1
0.75 m
2
w
3
0.50 m w
T2
Problem 6: A hungry 700-N bear walks out on a beam in an attempt to retrieve some “goodies” hanging at
the end, see the figure. The beam is uniform, weighs 200 N, and is 6.00 m long; the goodies weigh 80.0 N.
(a) Draw a free-body diagram for the beam. (b) When the bear is at x = 1.00 m, find the tension in the wire
and the components of the reaction force at the hinge. (c) If the wire can withstand a maximum tension of
900 N, what is the maximum distance the bear can walk before the wire breaks?
Solution: (a) See the diagram below:
700 N
x
T
60.0°
H
V
3.00 m
3.00 m
200 N
80.0 N
(b) If x = 1.00 m, then
Στ )left end = 0 ⇒ − ( 700 N )( 1.00 m ) − ( 200 N )( 3.00 m ) −
( 80.0 N )( 6.00 m ) + ( T sin 60.0° ) ( 6.00 m ) = 0 ,
giving
T = 343 N
Then, ΣFx = 0 ⇒ H − T cos 60.0° = 0 , or H = ( 343 N ) cos 60.0° = 171 N
and
ΣFy = 0 ⇒ V − 980 N+ ( 343 N ) sin 60.0° = 0 , or V = 683 N
(c) When the wire is on the verge of breaking, T = 900 N and
Στ )left end = − ( 700 N ) xmax − ( 200 N )( 3.00 m ) −
( 80.0 N )( 6.00 m ) + ⎡⎣( 900 N ) sin 60.0°⎤⎦ ( 6.00 m ) = 0 ,
which gives xmax = 5.14 m
Problem 7: A 20.0-kg floodlight in a park is supported at the end of a horizontal beam of negligible mass
that is hinged to a pole, as shown in the figure. A cable at an angle of 30.0° with the beam helps to support
the light. Find (a) the tension in the cable and (b) the horizontal and vertical forces exerted on the beam by
the pole.
Solution: (a) Consider the torques about an axis
perpendicular to the page and through the
left end of the horizontal beam.
Στ = + ( T sin 30.0° ) d − ( 196 N ) d = 0 ,
T
30.0°
H
giving T = 392 N
V
(b) From ΣFx = 0 ,
H − T cos 30.0° = 0 , or
H = ( 392 N ) cos 30.0° = 339 N to the right .
From ΣFy = 0 , V + T sin 30.0° − 196 N = 0 ,
or V = 196 N- ( 392 N ) sin 30.0° = 0 .
d
196 N
Problem 8: A uniform plank of length 2.00 m and mass 30.0 kg is supported by three ropes, as indicated by
the blue vectors in the figure below. Find the tension in each rope when a 700-N person is 0.500 m from the
left end.
T2
Solution: Consider the torques about an axis
perpendicular to the page and through
the left end of the plank.
Στ = 0 gives
700 N
T1
0.500 m
T3
1.00 m
40.0°
1.00 m
mg = 294 N
− ( 700 N )( 0.500 m ) − ( 294 N )( 1.00 m ) + ( T1 sin 40.0° )( 2.00 m ) = 0
or
T1 = 501 N
Then, ΣFx = 0 gives − T3 + T1 cos 40.0° = 0 , or
T3 = ( 501 N ) cos 40.0° = 384 N
From ΣFy = 0 , T2 − 994 N + T1 sin 40.0° = 0 ,
or
T2 = 994 N − ( 501 N ) sin 40.0° = 672 N
Problem 9: A 15.0-m, 500-N uniform ladder rests against a frictionless wall, making an angle of 60.0° with
the horizontal. (a) Find the horizontal and vertical forces exerted on the base of the ladder by Earth when an
800-N fire fighter is 4.00 m from the bottom. (b) If the ladder is just on the verge of slipping when the fire
fighter is 9.00 m up, what is the coefficient of static friction between ladder and ground?
Solution: First, we compute some needed dimensions:
F2
d1 = ( 7.50 m ) cos 60.0° = 3.75 m
d2 = d cos 60.0° = ( 0.500 ) d
15.0 m
800 N
d
d3
d3 = ( 15.0 m ) sin 60.0° = 13.0 m
Using an axis perpendicular to the page and through the
lower end of the ladder, Στ = 0 gives
− ( 500 N ) d1 − ( 800 N ) d2 + F2 d3 = 0 ,
or
F2 =
1875 N ⋅ m + ( 800 N ) ⎣⎡( 0.500 ) d ⎦⎤
13.0 m
.
Then, ΣFx = 0 gives f − 267 N = 0 , or f = 267 N to the right ,
and ΣFy = 0 yields F1 − 500 N − 800 N=0 , or F1 = 1.30 kN upward .
(b) When d = 9.00 m, equation (1) gives F2 = 421 N to the left .
Then, ΣFx = 0 gives f = 421 N to the right , while
ΣFy = 0 yields F1 = 1.30 × 103 N = 1.30 kN as before .
If the ladder is ready to slip under these conditions, then f = ( fs )max ,
μs =
( fs )max ( fs )max
n
=
F1
=
60.0°
f
d1
d2
(a) When d = 4.00 m, equation (1) gives F2 = 267 N to the left .
and
500 N
F1
421 N
= 0.324 .
1.30 × 103 N
(1)
Problem10: The large quadriceps muscle in the upper leg terminates at its lower end in a tendon attached to
the upper end of the tibia, see the figure a below. The forces on the lower leg when the leg is extended are
modeled as in figure b below, where T is the tension in the tendon, w is the force of gravity acting on the
lower leg, and F is the weight of the foot. Find T when the tendon is at an angle of 25.0° with the tibia,
assuming that w = 30.0 N, F = 12.5 N, and the leg is extended at an angle of 40.0° with the vertical ( θ =
40.0°). Assume that the center of gravity of the lower leg is at its center, and that the tendon attaches to the
lower leg at a point one fifth of the way down the leg.
Solution: First, we resolve all forces into components parallel to
d/5
wy = ( 30.0 N ) sin 40.0° = 19.3 N ,
wy
Using Στ = 0 for an axis perpendicular to the page and
through the upper end of the tibia gives
d
d
− ( 19.3 N ) − ( 8.03 N ) d = 0 ,
5
2
d
d/2
Ty = T sin 25.0° .
( T sin 25.0° )
Ty
Tx
Fy = (12.5 N ) sin 40.0° = 8.03 N ,
and
Ry
Rx
and perpendicular to the tibia, as shown. Note that
or
θ
wx
Fy
θ
T = 209 N
Fx