Hi, I'm confused as to how to approach this problem. I have attached the question and my work so far.To solve the problem using the Principle of Work and Energy, I would need to know the work done in moving the mass downwards and increasing the angular velocity of disk B to 6 radians/second. Since I could not figure out how to do this, I attempted the question using the Conservation of Energy principle. But due to not knowing the distance moved by mass C, I could not get any further than shown in my work so far.Thank you! The pulleys-block system consists of a 3-kg pulley A and 10-kg pulley B.Neglect the mass of the cord and treat the pulleys as thin disks. If a 2-kg blockC is suspended from the cord, and starting to descend from rest, until pulley Breaches the angular velocity of 6 rad/s.1) Apply the Principle of Work and Energy to the pulleys-block systemfrom position 1 (it is at rest) to position 2 (when pulley B reaches theangular velocity of 6 rad/s).2) Find out the linear speed Vc of block C, at position 2 (when pulley Breaches the angular velocity of 6 rad/s).3) Draw fee body diagram of block C, then apply the Principle of Workand Energy to determine the tensile force F in the cord while block C isin motion.4) Find out the linear acceleration ac of block C while it is in motion.100 mmВA30 mm B=0.1mCA :0,03m3. k3gPosirion 1: No sare dase yeh. U,-O+I6 C T= JC does worte against WB acdAWa tomove dasncwards.MOVEThin diss i I=MP²%3DTitEViz2 =<(POSITION 2)POSITION 2:TeIA =D2x3x0.032 = 0.00135 Agm²TzTi +V, = Te + VzFind In IB.2.%3D%3D%3DIB=DŽx10 x 0.12:0Find Vo at pos0.05kgm²%3D(0.00135)W7+2(0.05/W mVe)+gh13Db 2 ahere We =6 rals-"0=6.75x1042 +h dawr pom pos1→ pos 2.O.025WĞV 19.62hheight block C movesWALA= WB IBWe rB6x0.1> WA =%3D20ads-1%3D%3D0.03Assuming no sip of Hhe rope (pullegsVc = WA(0.03)WB = 10VC6%310Vcae thin disrs).,Vc = Wg(0,1) 0.03WA- 0lWB0.10B%3Dlove, WA= 33,33 VC%3D%3D

Two discs: mA = 3 kg, mB = 10 kg mass of the block, mC = 2 kg block 'C' is moving down hence the…

Answered: The pulleys-block system consists of a…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Hi, I'm confused as to how to approach this problem. I have attached the question and my work so far.

To solve the problem using the Principle of Work and Energy, I would need to know the work done in moving the mass downwards and increasing the angular velocity of disk B to 6 radians/second.

Since I could not figure out how to do this, I attempted the question using the Conservation of Energy principle. But due to not knowing the distance moved by mass C, I could not get any further than shown in my work so far.

Thank you!

The pulleys-block system consists of a 3-kg pulley A and 10-kg pulley B.
Neglect the mass of the cord and treat the pulleys as thin disks. If a 2-kg block
C is suspended from the cord, and starting to descend from rest, until pulley B
reaches the angular velocity of 6 rad/s.
1) Apply the Principle of Work and Energy to the pulleys-block system
from position 1 (it is at rest) to position 2 (when pulley B reaches the
angular velocity of 6 rad/s).
2) Find out the linear speed Vc of block C, at position 2 (when pulley B
reaches the angular velocity of 6 rad/s).
3) Draw fee body diagram of block C, then apply the Principle of Work
and Energy to determine the tensile force F in the cord while block C is
in motion.
4) Find out the linear acceleration ac of block C while it is in motion.
100 mm
В
A
30 mm

B=0.1m
CA :0,03m
3. k
3g
Posirion 1: No sare dase yeh. U,-O+
I6 C T= J
C does worte against WB acd
A
Wa to
move dasncwards.
MOVE
Thin diss i I=MP²
%3D
TitEViz2 =
<(POSITION 2)
POSITION 2:
Te
IA =D2x3x0.032 = 0.00135 Agm²
Tz
Ti +V, = Te + Vz
Find In IB.
2.
%3D
%3D
%3D
IB=DŽx10 x 0.12
:0
Find Vo at pos
0.05kgm²
%3D
(0.00135)W7+2(0.05/W mVe)+
gh
13D
b 2 ahere We =6 rals-"
0=6.75x1042 +
h dawr pom pos1→ pos 2.
O.025WĞ
V 19.62h
height block C moves
WALA= WB IB
We rB
6x0.1
> WA =
%3D20ads-1
%3D
%3D
0.03
Assuming no sip of Hhe rope (pullegs
Vc = WA(0.03)
WB = 10VC
6%310Vc
ae thin disrs).
,
Vc = Wg(0,1) 0.03WA- 0lWB
0.10B
%3D
love
, WA= 33,33 VC
%3D
%3D

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