(3) Moment arm of diver's weight is 3.0 m. Her weight is
60 kg×9.8 m/s2, giving a torque of 3.0×60×9.8 N·m = 1800 N·m.
(5) Considering the x-components of the forces:
F2 - F1 cos 45° = 0. Likewise S Fy = 0 = F1 sin 45° - M g. Additionally, we're told that
F1,max = F2,max = 1300 N. The first equation gives
F2 = 0.707 F1, so F1 is the larger of the two forces.
Thus we solve the second equation using F1 = 1300 N to give
M = 0.707×1300/9.8 kg = 94 kg or a weight of 920 N.
(7) Let L be the length of the beam,
F1 be the force acting upward on the end of the beam
closest to the piano, and F2 the similar force on the other
end of the beam. Calculate torques with respect to the F1
end. Thus F1 produces no torque (moment arm being zero) and the
torque due to F2 is in magnitude L×F2. It is opposed
by the two torques of the weights, magnitude 9.8 m/s2×L (300 kg/4 + 160 kg/2). Note that the length of the beam divides
out when we equate these to yield F2 = 1500 N. Since the sum
of the forces must also vanish (as well as the torques), we get
F1 + F2 = 9.8 (300 + 160) N, or solving for F1,
F1 = 3000 N.
(13) Letting x be horizontal and y vertical, we consider SFx = 0 and SFy = 0, since the light is in static equilibrium.
These give T1 cos 37° = T2 cos 53° and
T1 sin 37° + T2 sin 53° = 30×9.8 N.
From the first we obtain T2 = 1.327 T1, which upon substitution
into the second equation gives T1 (0.602 + 1.327×0.799) = 30×9.8 N, or T1 = 180 N. This result, back into the first
equation yields T2 = 230 N.
(17) The horizontal components of the hinge forces must be oppositely
directed; i.e., the top hinge pulls to the right on the door (FTH),
and the bottom hinge pushes to the left (FBH), to oppose
the door's weight. Calculate torques w.r.t. the lower hinge, so
that the force at that hinge has a zero moment arm. Also, the vertical
component of the force by the second hinge (FBV)has a zero moment arm.
Consequently, 1.5 FTH = 0.65 m ×13 kg×9.8 m/s2,
or FTH = - FBH = 55.2 N. Since the weight of the door is shared equally
by the two hinges, FTV = FBV = 13 kg×9.8 m/s2/2 = 64 N.
(23) Let the reference for torques be the contact point of the beam
with the wall, so that FW has no moment arm. Then the torque of the
beam's weight, m g L/2 must be balanced by that of the tension,
L FT sin 50° giving FT = 190 N. The vertical component
of the wall force must add to the vertical component of the tension
to equal 30×9.8 N (the weight). Thus FW,y + 190 N sin50° = 300 N, or FW,y = 150 N. The horizontal component
of the wall force must be equal and opposite to the horizontal
component of the tension, which is FT,x = - 190 N cos 50° = - 120 N. Thus FW,x = 120 N. The net wall force is given
by FW = (1202 + 1502)1/2 N = 190 N, and the angle made
with respect to the x-axis is tan-1(150/120) = 51°
(31) For torques, we select the contact point of the ladder with the
ground, so that FG has zero moment arm. The angle formed by the
ladder w.r.t. the ground is q = tan-1(4/3) = 53.1°,
but is not actually needed in the calculations.
The moment arm of the ladder weight (acting at its center of gravity)
is 1.5 m, and that of the painter is 2.1 m. The torque from these
two, 2.1 m×M g + 1.5 m×m g is balanced (since in static
equilibrium) by the torque of the wall, 4.0 m×FW. By equating
the two, we get FW = (2.1×60 + 1.5×12)×9.8/4 N = 350 N. From SFx = 0 and SFy = 0 we obtain
FG,x = FW = 350 N and FG,y = (12 + 60) 9.8 = 706 N.
Finally, ms FG,y = FG,x, so ms = 350/706 = 0.5.
(45) (a) The stress is F/A = 25×103×9.8/2 N/m2 = 1.2×105 N/m2.
(b) The strain is the ratio of stress to elastic modulus, given by
1.2×105/50×109 = 2.4×10-6 (the modulus obtained
from the table on page 254.
(55) The max. stress is 170×106 N/m2 from the table on
page 258. Thus Fmax = 170×106×3×10-4 N = 5×104 N (using the given area).
File translated from TEX by TTH, version 1.95.
On 10 Nov 1999, 12:18.