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Ch 12.1
#3
A uniform beam of mass mb
and length l supports blocks
of masses m1 and m2 at two
positions, as shown in the
figure. The beam rests on
two points. For what value
of x will the beam be
balanced at P such that the
normal force at O is zero?
cw
=
NO (/2 + d) + m2g (x)
=
ccw
mbg (d) + m1g (/2 + d)
x
= [(m1g + mbg) d + m1g /2] / m2g
x
= [(m1 + mb) d + m1 /2] / m2
Ch 12.3
if it balanced
#13
A 15 m uniform ladder weighing 500 N rests against a
frictionless wall. The ladder makes a 60 angle with the
horizontal.
(a) Find the horizontal and vertical forces that the ground
exerts on the base of the ladder when an 800 N firefighter
is 4 meters from the bottom.
(b) If the ladder is just on the verge of slipping when the
firefighter is 9 m up what is the coefficient of static
friction between the ladder and the ground?
a) Fx = 0 (not moving)
Fy = Nground – Wfire - Wladder = 0
Fx = Ffriction – Nwall
Ng = 800 N + 500 N
Nw = Ffriction
Ng = 1300 Newtons
Remember, the lever arm is ALWAYS perpendicular to the force.
Torque applied by the ladder, the center of mass of the ladder is at 7.5
meters …so the lever arms is 7.5 (sin ). (I’ve chosen the pivot point at the
base of the ladder.)
ccw
wall
=
cw in equilibrium
=
firefighter + ladder
Nw cos 30 15 m
= 800 N 4 m sin 30 + 500 N 7.5 m sin 30
Nw
= 268 N to the left
Ff
= 268 N to the right
Ng
= 1300 N up
b)
cos 30 Nw 15 m = 800 N 9 m sin 30 + 500 N 7.5 m sin 30
Nw
Nw
1300 N

Ch 12.3
= 421 N to the left
= Ff (where Ff =  Ng)
= 421 N
= 0.324
#24
Two identical uniform bricks of length L are placed in a stack over the edge
of a horizontal surface with the maximum overhang possible without falling.
Find the distance x.
Ch 12.2
Leg: Center of Gravity
A person is sitting with one leg outstretched so
that it makes an angle of 30 with the horizontal,
as the drawing indicates. The weight of the leg
below the knee is 44.5 N with the center of
gravity located below the knee joint. The leg is
being held in the position because of the force,
M, applied by the quadriceps muscle, which is
attached 0.100 m below the knee joint. Obtain
the magnitude of M.
τ
τ
τ
τ
τ
τ
leg
= FWeight d
leg
= 45N * cos30º 0.25
leg
= 9.7428 Nm
muscle
= Fmuscle d
muscle
= M * sin25º 0.1
muscle
= 0.04226*M
τ
leg
9.743 Nm
M = 230 N
=
τ
muscle
= 0.04226*M
Ch 12.2
#5
A carpenter’s square has the shape of an L, as in the figure. Locate the center of gravity.
Lower left hand corner: (0, 0) xCG = mixi / mi
xCG
= AAxA + ABxB / (AA + AB)
ACM = (2, 9)
BCM = (8, 1)
xCG
= 72 (2) + 48 (8) / (72 + 32)
xCG
= 3.85 cm
AA = 18*4
AB = 8*4
yCG
= AAyA + AByB / (AA + AB)
2
2
AA = 72 cm
AB = 32 cm
yCG
= 72 (9) + 48 (1) / (72 + 32)
yCG
= 6.85 cm
And since the mass is proportional to area we can use
Area instead of mass in the equation for center of gravity
Ch 12.3
#43
A hungry Bear weighing 700 N walks out on a beam in
an attempt to retrieve a basket of food hanging at the
end of the beam. The beam is uniform, weighs 200 N
and is 6 m long; the basket weighs 80 N.
(a) Draw a free body diagram for the beam
(b) When the bear is at x = 1 m, find the tension in the
wire and the components of the force exerted by the
wall on the left end of the beam
cw
=
ccw
if it balanced
700*1 + 200*3 + 80*6 = T sin60 * 6
T = 343 N
Fy-down
=
Fy-up
in equilibrium
Fx = Rx – T cos60 = 0
Rx = 171 N
Fy = Ry + T sin60 - (700+200+80) = 0
Ry = 683 N
Ch 12.3
#44
(c) If the wire can stand a maximum
tension of 900 N, what is the maximum
distance the bear can walk before the
wire breaks?
700*x + 200*3 + 80*6 = 900 sin60 * 6
x = 5.13 m
Old MacDonald had a farm, and on that farm he had a gate. The gate is 3 m wide and 1.8 m high, with
hinges attached to the top and the bottom. The guy wire makes an angle of 30˚ with the top of the
gate and is tightened by a turn buckle to a tension of 200 N. The mass of the gate is 40 kg
(a)
Determine the horizontal force exerted on the gate by the bottom hinge.
(b)
Find the horizontal force exerted by the upper hinge
(c)
Determine the combined vertical force exerted by both hinges
(d)
What must the tension in the guy wire be so that the horizontal force exerted by the
upper hinge is zero
Ch 12.2
A 3 kg particle is located on the x-axis at x = -5 m, and a 4 kg particle is located on the x axis at x =
3 m. Find the center of gravity of this two particle system.
Ch 12.2
#8
The masses and coordinates of a rod, a right triangle, and a square are given. Determine the center
of gravity for the three-object system.
Ch 12.4
#33
Ch 12.4
#35
If the shear stress in steel exceeds 4.00 x 108 N/m2, the steel ruptures. Determine the shearing
force necessary to (a) shear a steel bolt 1.00 cm in diameter and (b) punch a 1.00 cm diameter hole in
a steel plate 0.500 cm thick.
When water freezes, it expands by about 9%. What would be the
pressure increase inside your automobile’s engine block if the water
in it froze?
(Bulk modulus of ice is 2 x 109 N/m2)
(Pressure = F/A)
Ch 12.4
P = - B
(V / V)
9
P = - 2 x 10 (-.09)
P = = 1.8 x 108 N/m2
P = 26100 lb/in2
#27
A 200 kg load is hung on a wire having a length
of 4 meters, a x-sectional area of 2 x 10-5 m2,
and a Young’s Modulus of 8 x 1010 N/m2. What
is its increase in length?
Young’s M =
F/A
/ L / L
10
-5
8 x 10 N = 200”g” / 2 x 10 / L / 4
L = 0.005 m