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Unformatted text preview: PROBLEM 4.67 A Tshaped bracket supports a 751b load as shown. Determine the
reactions atA and then (a) a = 90°, (b) a = 45°. 6 inr+i+—10 in. SOLUTION ..
(a) F teeBody Diagram: (06 = 90°) The bracket is a threeforce body and A is the intersection of the lines of action of the three forces. 0 = tan—1(3) = 26565"
12 From the force triangle: A = (75 lb)cot6 = (751b)cot26.565° 9
=150.0001b A 9 '
C = 7.5 1b = A = 167.705 lb
sm6 sm26.565° ‘
7C”; or A = 150.0 lb 1 4
or ‘C = 167.71b Li 63.4° 4 PROBLEM 4.67 CONTINUED
(bereeBody Diagram: (a = 45°) I 16M. Let E be the intersection of the lines of action of the three
forces acting on the bracket. Triangle ABE is isosceles and therefore AE = AB = 16 in.
km From triangle CEF
e = tan—(E) = tan—1(3) = 12.0948°
EF 28
From force triangle:
,6 = 180° —135° — 9
9 = 180° — 135° — 12.0948°
= 32.905°
A
_C_ Using the law of sines:
) A _ C = 75 1b
13:0 a sin32.905° sinl35° si1112.0948°
7: 1k
Solving for A and C:
A = 194.452 1b
C = 253.10 1b o'r A =194.51bl 4
or C = 2531b :5 779° 4 o PROBLEM 4.78 A small hoist is mounted on the back of a pickup truck and is used to lift
a 260lb crate. Determine (a) the force exerted on the hoist by the
hydraulic cylinder BC, (b) the reaction at A. SOLUTION
FreeBody Diagram: (for hoist AD)
Note that the hoist AD is a threeforce body. E is the intersection between the lines of action of the three forces
acting on the hoist. From the freebody diagram:
xAE = (48 in.)cos30° = 415692 in.
yAD = (48 in.)sin30° = 24 in.
yBE = xAD tan75° = (41.5692 in.)tan75°
= 155.1384 in. Then: a = m_, yBE — 16 in. = tan_1(139.1384)
41.5692 = 73.36588°
[3 = 75° — a = 75° — 73.36588° = 1.63412°
6 = 180° — 15° — [3 = 165° — 1.63412° =163.366° From the force triangle and using the law of sines: 2601b _ B _ A
sinﬂ sin6 sin15°
2601b B A ’ sin 1.63412° sin 163.366° sin 15° Solving for A and B:
(a) B = 2609.9 lb or B = 2.61kips 3 750° 4
(b) A = 2359.8 lb or A = 2.36 kips ‘Q 734° 4 PROBLEM 4.92 A slender rod of length L is lodged between peg C and the vertical wall.
It supports a load P at end A. Neglecting friction and the weight of the
rod, determine the angle 9 corresponding to equilibrium. SOLUTION q.
FreeBody Diagram: Note that the rod is a three—force body. In the freebody diagram, D is
the intersection between the lines of action of the three forces. Using triangle BCE:
a = BE = BC sin9
and from tn'angle BCD BC = BDsin9 Then a = BDsin29 Also from tﬁangle ABD
BD = Lsin9, so
a = Lsin3 9 ...
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This note was uploaded on 10/05/2009 for the course CE statics taught by Professor Epstein during the Spring '09 term at UConn.
 Spring '09
 Epstein

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