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120% Ma w \3 gamma, Norma3 53) Given: A 50 kg crate rests on a horizontal surface for
which the kinetic friction coefﬁcient uk = 0.2. Find: The acceleration of the crate if P = 600 N. Hint — the Normal force from the ground may act at any
point along the bottom of the crate — you’ll need to ﬁgure
out the location of the normal force to see if the crate
slides or starts to tip. What type of problem? First step. 1:; Y
———§ inns ‘E ”((0, mats} , [—px
. A . _ vii ix The N is actually a distribution across the bottom. The resultant will shift to the right as the force P
increases. l have defined this distance from the center as x. EofM.
2F)" '2: mall‘s" ZFbgmaéﬂ ZMG' 7' £40k
F+P= magx (I) N’méﬁ‘; mate” (7') I .7414 4 lo ‘7' MOKGX
Assume it doesn’t tip. What would be true?
u: Np {meme , oc =0 (Re:g ~20
N  mg == 0
N '3 50161.30 == H905 \__ﬂ__\$ / «I'M v' 94410.5) +4290
06, '4 m 50 But we have to make sure it does not tip. Use moments equation.
ZM‘,’ r 136 pl :0
—' ‘P(0.3) *‘ 43(053 + N (x) =3 O (coo ( o. 3) + [o.2.(‘I%S)] (0.5) X z: X : OHi’G’i m
Li‘ios5 This is still under the box, so it is possible. If x>0.5
m, then the moment needed to be caused by N would
be too great and tippage would occur. Some people decided to place the N at the edge of the box to see if it
would prevent tipping — like a worst case scenario. ' 2: MG ”5 15. 0(
49013) , 0.10mi). 5X05“) +— #470105) == 15% 4‘28 It should make sense to you that the normal force would not be able
to cause the box to tip over in a CCW direction, so no tippage. What if the P = 1600 N? From our previous calcs, we see it would tip. How
would we approach the problem now? Do your x, y, and M equations. 25; mag . Egan/346%
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06"; m m x —— F2035) ~C{a.s) veal/9,5) 2:. got What are our unknowns? ls maGy zero? What on earth will you do?
Unknowns: 0:, am, aGy, N. How can you relate any of these? Kinematics! Do we know anything about any other point on the box? Yes,
point A (lower right corner). Write aG in terms of aA at 13A + ﬁx?»
516*? #61033 (Raﬁ 4‘ dﬂxz‘ 0.5%: +9.93) aAy,is zero (slides along the ground).
N "" 5064\5l3 =1 m<"’0:$d> — lavage) 4 N(o,:— 0.2(0593 =2 :54
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 Fall '05
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