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gravitation and statics

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HW 08 - fa14 - statics and gravitation:
Ch. 12: 3, 6, 12, 25, 28, 37, 48; Ch. 13: 7, 8, 13, 16, 35, 40, 57.
Chapter 12, Problem 3:
In the Figure, a uniform sphere of mass
0.85
mkg
and radius
4.2
rcm
is held in place by a massless rope attached to a frictionless wall a distance
8.0
Lcm
above the center of the sphere. Find (a) the tension in the rope and (b) the force on the sphere from the wall.
Notice our introduction of the placeholder-variable
, in addition to what’s already in the text. This helps us compute the
x
-component of the tension from the
y
-component, because tan//
xy
TTrL
, so,
222222
0; 0; (tan)()(/)1;
xxxyyxy
FRTFTmgTTTmgmgmgrL
(1.1)
In (1.1),
x
R
is the reaction-force from the wall, so, plugging in numbers,
2
4.2
(0.85)(9.81)(
cmm s
Tkg
8.0
cm
2
24.2
)19.418; tan(/)(0.85)(9.81)
cmm xx s
NRTmgmgrLkg
8.0
cm
4.378;
N
(1.2)
chapter 12, problem 6:
A scaffold of mass
60
mkg
and length
5.0
Lm
is supported in a horizontal position by a vertical cable at each end. A window washer of mass
80
Mkg
stands at a point
1.5
xm
from one end. What is the tension in (a) the nearer cable and (b) the farther cable?
The weight of the window-washer exerts a torque about one end of the scaffold (which is basically a uniform stick (but 2-dimensional))
2
1.511225.02
0((/))(805260)(9.819.74);
mm L farfar m s
xMgmgLTTxLMmgkgN kg
(1.3) Newton’s 2
nd
Law in the y-direction, meanwhile, yields,
2
0()529.748()(143.66;40)(9.81)
m ynearfarnearfar s
FTTmMgTmMgTN kgN
(1.4) The symbolic expression for the “near” force is:
12
((1))
xnear L
TmMg
.
Chapter 12, problem 12:
In the Figure, trying to get his car out of mud, a man ties one end of a rope around the front bumper and the other end tightly around a utility pole
18
Lm
away. He then pushes sideways on the rope at its midpoint with a force of
550
FN
, displacing the center of the rope
0.30
ym
, but the car barely moves. What is the magnitude of the force on the car from the rope? (The rope stretches somewhat.)
Again, we use the equation tan//
yx
TTyx
, where
12
xL
, immediately yielding,
1822(0.30)
0; 0cotcot15005506
m L yyyxxxxxy ym
FFTTFFTRRTTFNN F
(1.5)
ããchapter 12, problem 25:
In the Figure, what magnitude of (constant) force applied horizontally at the axle of the wheel is necessary to raise the wheel over an obstacle of height
3.00
hcm
? The wheel’s radius is
6.00
rcm
, and its mass is
0.800
mkg
.
Assume the wheel mounts the corner at constant velocity. This is difficult to maintain physically, but necessary to solve the problem. Drawing a schematic, (1.6) Torque balance, x-force balance, and y-force balance, 0sinsin; 0; 0;
xyxy
RRxRryRrxxyy
RrRrFFRFRmg
(1.7) Trig functions are,
22
sin(90)cos()/; sinsin(90(90))sin()/;
xxyxx
RrRrRrRrRr
rrhrrhr
(1.8) Combining (1.7) and (1.8), and solving for
F
, we initially have cossin
xx
RrRr
Frmgr
(in which we used the machinations of (1.8)), which is readily rearranged into,
2
222222
()/()cot()/(0.800)(9.81)(6.00)(3.00);(3.13.9)500
x
Rr m s
rrhrmgrrh Fmgmg rhrrhkgcmcm N cm
(1.9) Fun plotting exercise suggested by the solutions manual:
The applied force here is about 1.73 times the weight of the wheel. If the height is increased, the force that must be applied also goes up. Next we plot
F
/
mg
as a function of the ratio /
hr
. The required force increases rapidly as /1
hr
.
You should get in the habit of plotting things—see if you can reproduce this above plot using my result (1.9).
chapter 12, problem 28:
In Fig. 12-43, suppose the length L of the uniform bar is 3.00 m and its weight is 200 N. Also, let the block’s weight W = 300 N and the angle
30.0
. The wire can withstand a maximum tension of 500 N. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x (pretending the string doesn’t break), what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?
(a) the torque-balance about point-A determines the max distance the block can stand on the beam,
max
1,maxmax2solve for xmaxmax
0sin()sin()sin90()sin90sin/2(500N)sin30.0(200N)/23.00 m1.50m;300N
iwireblockrodCMiiibar b
rFLTxWLW TW xLW
(1.10) (b) The x-reaction-force F
x
is due to
,
0
xi
F
, in which there are two forces: the (negative) x-component of tension, and the (positive) reaction-force,
solve for F,maxmax
0cos=cos=433N;
x
xixxxx
FTFTFFT
(1.11) (c) the y-reaction force F
y
is due to
,
0
yi
F
; unlike the x-direction (1.11), there are four forces in this balance,
solve for F,maxmax
0sin=+sin=250N;
x
yiyyblockbarybyb
FFTWWFTWWFWWT
(1.12)
chapter 12, problem 28:
In the Figure, take
3.00
Lm
and its weight is
200
bar
WN
. Also, let the block’s weight
300
WN
and the angle
30.0
. The wire can withstand a maximum tension of
max
500
TN
. (a) What is the maximum possible distance x before the wire breaks? With the block placed at this maximum x (pretending the string doesn’t break), what are the (b) horizontal and (c) vertical components of the force on the bar from the hinge at A?
(a) the torque-balance about point-A determines the max distance the block can stand on the beam,
max
1,maxmax2solve for xmaxmax
0sin()sin()sin90()sin90sin/2(500N)sin30.0(200N)/23.00 m1.50m;300N
iwireblockrodCMiiibar b
rFLTxWLW TW xLW
(1.13) (b) The x-reaction-force F
x
is due to
,
0
xi
F
, in which there are two forces: the (negative) x-component of tension, and the (positive) reaction-force,
,maxmax
0coscos500cos30.0433;
x
solveforF xixxxx
FTFTFFTNN
(1.14) (c) the y-reaction force F
y
is due to
,
0
yi
F
; unlike the x-direction (1.11), there are four forces in this balance,
,max max
0sinsin300200500sin30.0250;
x
yiyyblockbaryb solveforF ybar
FFTWWFTWW FWWTNNNN
(1.15)
chapter 12, problem 37:
In the Figure, a uniform plank, with a length
6.10
Lm
and a weight of
445
WNmg
, rests on the ground and against a frictionless roller at the top of a wall of height
3.05
hm
. The plank remains in equilibrium for any value of
70
but slips if
70
. Find the coefficient of static friction between the plank and the ground.
We choose the rotation axis to be such that the reaction force ˆˆ
xy
RRiRj
as illustrated in the schematic makes zero contribution to the torque, ; (1.16)

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