Component Method Graphical Method 2x9.8 F1 =1.96 @ 60° FR = 3.5mnits m1 = FR = (200g) @ 90° 90° Part 1 · 2x9.8 F2 = \.96 @ 120° m2 = (200g) 25 x 9.8 F1 = _2 45 @ 120° m1 = FR = FR = (250g) ·2X9.8 Part 2 @ F2 =_ 1.96@ 200° m2 = (200g) .15X9,8 F, =].47 @o° FR = m1 = FR = (150g) .2X 9.8 Part 3 F2 = ]. 96 @90° m2 = (200g) ·15 X9.8 .2X9.8 F = \,47_@ 50° m1 = (150g) FR = FR = F2 = 1.96 Part 4 m2 = @ 90° (200g) 25X9.8 F3 = 2.45 m3 = @ 140° (250g) EP= ma EP =3 = in equilibrlum %3D Efx =0 on Forces acting the ring balan ce
Component Method Graphical Method 2x9.8 F1 =1.96 @ 60° FR = 3.5mnits m1 = FR = (200g) @ 90° 90° Part 1 · 2x9.8 F2 = \.96 @ 120° m2 = (200g) 25 x 9.8 F1 = _2 45 @ 120° m1 = FR = FR = (250g) ·2X9.8 Part 2 @ F2 =_ 1.96@ 200° m2 = (200g) .15X9,8 F, =].47 @o° FR = m1 = FR = (150g) .2X 9.8 Part 3 F2 = ]. 96 @90° m2 = (200g) ·15 X9.8 .2X9.8 F = \,47_@ 50° m1 = (150g) FR = FR = F2 = 1.96 Part 4 m2 = @ 90° (200g) 25X9.8 F3 = 2.45 m3 = @ 140° (250g) EP= ma EP =3 = in equilibrlum %3D Efx =0 on Forces acting the ring balan ce
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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I’ve been given this data and told to solve each part graphically and using the component method. I believe I did the graph correct for the first one but I’m confused on which numbers to plug into the equation for the component method. Some guidance would be greatly appreciated.
![Component
Method
Graphical
Method
2x9.8
Fi = 1,96 @ 60°
FR = 3.5 wnits
m1 =
FR =
(200g)
90°
90°
Part 1
.2X9.8
F2 = \. 96
@
@ 120°
m2 =
(200g)
25 x9.8
F1 = 2.45 @ 120°
FR =
m1 =
FR =
(250g)
·2×9.8
Part 2
F2 =
1.96 @ 200°
m2 =
(200g)
. 15x9.8
F1 =].47 @0°
FR =
m1 =
FR =
(150g)
2X 9.8
Part 3
F2 = \, 96 @ 90°
m2 =
(200g)
·15 X9.8
·2x 9.8
Fi = 1.47
m1 =
@ 50°
(150g)
FR =
FR =
F2 = 1.96@ 90°
Part 4
m2 =
@
(200g)
25x9.8
F3= 2.45 @ 140°
m3 =
(250g)
EP =3 = in equilibrium
Efx=0
Forces acting on
the ring balan ce](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb72c24d2-e42d-4897-97a3-bd7ae212f23e%2F4241e45b-2171-489b-af96-728767d0b357%2Fasafmvc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Component
Method
Graphical
Method
2x9.8
Fi = 1,96 @ 60°
FR = 3.5 wnits
m1 =
FR =
(200g)
90°
90°
Part 1
.2X9.8
F2 = \. 96
@
@ 120°
m2 =
(200g)
25 x9.8
F1 = 2.45 @ 120°
FR =
m1 =
FR =
(250g)
·2×9.8
Part 2
F2 =
1.96 @ 200°
m2 =
(200g)
. 15x9.8
F1 =].47 @0°
FR =
m1 =
FR =
(150g)
2X 9.8
Part 3
F2 = \, 96 @ 90°
m2 =
(200g)
·15 X9.8
·2x 9.8
Fi = 1.47
m1 =
@ 50°
(150g)
FR =
FR =
F2 = 1.96@ 90°
Part 4
m2 =
@
(200g)
25x9.8
F3= 2.45 @ 140°
m3 =
(250g)
EP =3 = in equilibrium
Efx=0
Forces acting on
the ring balan ce
Transcribed Image Text:Graphical method
3.5 units
1.96 units
120°
1.96 units
60°
Component method
0=tan
Fx = F,x +F;X = F,COS60 - F, COS 60° =
Fy = FY + F;Y =F,Sin60° – F, Sin 60°=
Fy
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