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Solution 15.4
PROBLEM STATEMENT
A 2100-lb rear-wheel drive tractor carrying 900 lb of gravel starts from rest and accelerates fonivard at2fit*lsz.
Determine the reaction at each of the two (a) rear wheels A, (b) front wheels 8.
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rn,
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Strategy: Newton's 2nd law. System: the tractor
' =P=Y6o ru 65.Zl7lb.s2lft
*, =
ffi= 27.9501b.s2lft
m,
32'2
Each rear wheel applies a force of F. So the 2 rear wheels apply a force of 2F
FA =nLoxiZF=(mt*m2)a
tLl
Plug in the 2 weights and divide the sum by gravity: 32.2ftlsz Also, use a = 2 fl/s^Z. Then divide by 2 to get F
Answer:
F=
93.L58 lb on each rear wheel.
Let NE be the normal force on each real wheel and Na the normal force on each front wheel.
So
the total normal force = 2Ne + 2Ns = 3000 lb, so
Na
+ Ns = 1500
frbo,+
lb
(1)
ff:
-2o(zroo ) + too(zNe\ (4o
'-'xlo --1--os 2r)
L
lr o
(q 0o\ =-Lrr,Al? ( L)Qo
) -zr. atto (z)(4o\
\- 21,qr(tro) + lr0cqoo ) +z-o(z roo ) -l
j
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distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whoie or part.
fr,1cArv S
V;rtrq^r
l+2
Compute Moments relative to the contact points of the wheels with the ground. This will let you compute Nn and
Ne
sepa rately.
)'Luo:
Ica *\mad,
The lastterm is becuz the acceleration of the tractor in the x-direction causes rotational motion.
d = rrertical distance from the gr"ound to the center of mass of the tractor
or"
the load, urhich
is
the perpenclicr:lar'
distancerelativetotheeffectivepointsof actionof theacceleration, a=2ftls2-Thiswasgiven.
Note that because the motion is lnear,
Ie
u=0.
AboutA(bothrearwheels):-rnexWr,".tor*IeeX2Ne-rn-LoaaxWloaa=0-Mrr"ctor*a*20-Mtora*a*40
Only Ns is unknown. Now plug in the numbers and solve for Ne. (The distances should be divided by 12 to
convert from inches to feet. However, since this factor of 12 occurs on every term, you should get the
same answer, whether or not you divide bV 12.)
'
p" = I j 1' '
,,' (Remember', this is for onlry l- of the front wheels.)
Answer: NB = 1135 lb
IVA =
'soo
-I
lj
J-
Substituteinto(1)togetNa.=@-(Remember,thisisforonly1oftherearwheels')
Reaction forces: Rear wheels:
W
F*n=Jffi=|3JJ:-l_
Forward force: 2Fn = ma = (65.22 + 27.95xtb s2/ft)*2 ft/s2) = 180 tb.
This is then divided between the 2 rear wheels, so
FA
= 90 lb
-------;
rLoB)=* (z o SY
6r
Answer: Ft = 377 lb
Then compute the angle = atan(Nn/Fa)
Then do similarly for the front wheels. Since most of the weight is on the front wheels in this problem, the reaction
S^"^^ ^t ^-^h f.^^+
ur rL ..,1^^^l
vvr rEqt ;.
tJ I^^^.1.,
tsor ty cvuqt
Lv trltrts
^^,,^l +^
Each
front wheel:
Fns
= Ne
Copyright @ McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or
distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in
whoie or pari.
"