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Unformatted text preview: PROBLEM 8.3 Determine whether the 10kg block shown is in equilibrium, and ﬁnd the
magnitude and direction of the friction force when P = 40 N and
I, 0 = 20°. SOLUTION
Tension in cord is equal to P = 40 N, from FBD of pulley.
FBD Block:
w = (10 kg)(9.81 m/sz) = 98.1 N
Y x \_ 213, = 0: N — (98.1 N)cos20° + (40 N)sin20° = 0
W ’ ,.
M N = 78.503 N
Frmx 2 MSN = (0.30)(78.503 N) = 23.551 N
5&5,“ For equilibrium: / 2F} = O: (40 N) cosZO° — (98.1 N)sin20° — F : 0 Feq = 4.0355 N < Fm, Equilibrium exists 4 q F=4.04N720°< PROBLEM 8.4 Determine whether the 10kg block shown is in equilibrium, and ﬁnd the magnitude and direction of the ﬁiction force when P: 62.5 N and
0 = 15°. Tension in cord is equal to P = 62.5 N, from FBD of pulley.
w = (10 kg)(9.81nVs2)= 98.1 N
\ 21‘}, = 0: N — (98.1 N)cos20° + (62.5 N)sin15° = 0
N = 76.008 N Fmax = ‘uSN = (0.30)(76.008 N): 22.802 N For equilibrium: / 21F} = 0: (62.5 N)coslS° — (98.1 N)sin20° — F = 0 Feq = 26.818 N > 1'7max so no equilibrium,
and block slides up the incline 4 Fslip = M = (0.25)(76.008 N): 19.00 N F=IQDON77004 PROBLEM 8.5 Knowing that 6 = 25°, detennine the range of values of P for which 11‘ = 0.30 w“
u, : 0 25 6 ‘ equilibn'um is maintained.
1 ' //i . 74‘ x
. / _ w,
, ‘\\
\ SOLUTION
Tension in cord is equal to P from FBD of pulley. w = (10 kg)(9.81m/s2)= 98.1 N
\. 2153,: 0; N — (98.1 N)cos20° + P sin25° = 0 (1) / 25;. = 0: Pcos25° — (98.1 N)sin20° + F = 0 (2) For impending slip down the incline, F: ,uSN = 0.3 N and solving
(1) and (2), PD = 7.56 N For impending slip up the incline, F=—,us.N = 0.3 N and solving
(l) and (2), PU = 59.2 N so, for equilibrium 7.56 N S P S 59.2 N 4 PROBLEM 8.41
The lZ—lb slender rod AB is pinned at A and rests on the 36lb cylinder C. r 7”,“ 5‘ Knowing that the diameter of the cylinder is 12.5 in. and that the
1 , coefﬁcient of static friction is 0.35 between all surfaces of contact,
y Y C "1 determine the largest magnitude of the force P for which equilibrium is maintained. SOLUTION
FBD Rod: {EMA =0: (20 111.)lel —(12.5 in.)(12 1b) = 0 N] = 7.5 lb. TZF),=0: N2—7.51b—361b=0, N2243.51b since [1] = #2 and NI < N2, slip will impend at top of cylinder ﬁrst, so
Fl : lLlSNl ' F, = 0.35(7.5 lb) = 2.625 lb ( 2M0 = 0: (4.25 in.)P — (12.5 in.)(2.625 lb) 2 0, P = 7.7206 lb
Pm = 7.72 lb 4 To check slip analysis above, 1 EF, = 0: N2 — 36 lb — 7.5 1b = 0 ) N2 = 43.5 lb 5m = MXNZ = 0.35(43.51b) = 15.2251b
—>ZFX=0: P—Fl—F220, 7.721b—2.6251b—F2=o
F2 = 5.095 1b < Fm, OK ...
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This note was uploaded on 02/17/2010 for the course MECH 003 taught by Professor Delph during the Fall '07 term at Lehigh University .
 Fall '07
 Delph

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