# The Engine Hoist Is Used To Support The 200-kg Engine. (a) Draw A Separate And Complete Free Body Diagram Of Member GAC; (b) Determine The Force Acting In The Hydraulic Cylinder AB (c) Determine The Horizontal And Vertical Components Of Force At The Pin Con GAC; (d) Determine The Reactions At The Fixed Support D.

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Question: The engine hoist is used to support the 200-kg engine. (a) Draw…
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Problem 1 (20 points) The engine hoist is used to support the 200-kg engine.
(a) Draw a separate and complete free body diagram of member GAC; (b) Determine the force acting in the hydraulic cylinder AB; (c) Determine the horizontal and vertical components of force at the pin C on GAC; (d) Determine the reactions at the fixed support D.
G
890 mm
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calcolate the weight of the engine: – Wang
= 200×9.81
= 1962N Darow the triangle ACB
Ayo 350 de
from
the triangle ACB using cosine nole a = 350 tsso -2(350) (850) Los (80) a=741679.33
az 861208 mm
busing
sine role: Sinsoo sino
a 850
Sino-8so singo
861.208 Sino 20.971.
O= sin(0.971)
0= 76.407 the force body diagram of member GAL
6 1250 2350
Donaw
Write the equilibmium fomcos acting in the sea dimection
Efx=0
Cx- FAC COS 0 0 — white the equilibrium forces acting in the y directions
{Fy=0
FAQ Sino-w7L1=— (2) by taking moments about Point ‘C’
Eme=0 -w (1250+350) + FAB Sino (350) =0 -1962 (1600) + Fee Sin (76-40) (350)=0
FAB = 9227.8N —-3) Substitute the above values in ea _(1):
Cx – Foe Cosg=0 Cx-92278 Cos (76-40) = 0
CX7 2169.84 N — 16
Substitute ea-13) Evalues in ea-12)
CycW-FAB Sine
-1962-92278 Sin (76-40)
E-7007.060. – (s) Draw the force body diagram of member CBD
V
FAO TO
850
>
DA
1
ssa
á
white the evil librium forces acting in se- direction:
EFx=0 -Dx- (x + FAB LOSO=0
Dx = -(x + FAB Coso
D2=-2169.86 +9227.8 tos (16.40)
[0x=0 write t e the Covilibrium forces acting in y-directions
EF=0 Dyr Far Sin – Cy=0
Dy = FAB Sing tly
Da =9227.8 8in176.40) – 7007 ob Dy = 1962N |
ביי-סיי – ישן
by taking the moments about D:
Cy Sin (io) (550 +850)- G (550 +850) 65100F
FAB Cos (76.40) (sso cos (100)) + FAB Sin (1640)* . (55o Sin (105)- Mo
by substituting the ea. (3) (4) & (s) we get
F-7007.06 Sin (109) (1400)–2169.8661400 COS100) 7 9227.8 Cos (76.40″) x 5so Cos (109) +9227.8 Sin (76-40) x = Mo
1 sso Sinlio)
-266 3.23 N-m – Mo
MD=2663.23 N-m . Mo= 2.66 kN.M]