Question

Can anyone solve this?

Answer 1

http://screencast.com/t/XyOrk7z9wr \
@Vincent-Lyon.Fr

Answer 2

Draw the forces will ya, then u can make net force zero, net torque zero!

Answer 3

can u try the question and tell me the answer, cuz i tried it and im not getting it

Answer 4

i ll walk you through it.. show me your work?

Answer 5

i found the centre of mass from O, and then found the perpendicular distance to the centre of mass from the pivot which is the clockwise moment, and then i found the antivlock wise moment by 15N acting at a perpendicular distance , and my value of sin thethas is 0.777

Answer 6

what about the normal force?

Answer 7

Oh my bad>< I think i can solve now :)

Answer 8

:D.. oki..

Answer 9

hehe

Answer 10

Are you allowed negative values of theta?

Answer 11

nope

Answer 12

where would the normal reaction force act?

Answer 13

You do not need it. Just imagine it is zero just when the equilibrium is broken.

Answer 14

yes thts the answer :)

Answer 15

I find that theta must be greater than 0.22 rad to have equilibrium.

Answer 16

can u send me the working tru email

Answer 17

Moment of F = 15 N must be greater than moment of W = 20 N

F acts R/2 away from O
W acts OG \(\cos (\dfrac \pi 3-\theta)\) away from O
with OG = \(\dfrac{\sqrt {3} R}{\pi}\)

That's all :-)

Answer 18

could u explain the OG thing?

Answer 19

anyways i did it :) i got OG sin (pi/6 + thetha) from 0

Answer 20

but could u explain me why thetha should be greater or equal to 0.224?

Answer 21

Just solve for \(\theta:\)

\(F\dfrac R2< W\dfrac{\sqrt {3} R}{\pi}\cos (\dfrac \pi 3-\theta)\)

Answer 22

what is the meaning of the sign?

Answer 23

could you explain tht part

Answer 24

< means 'less than'

that is, moment of weight must be greater than moment of F , so that the right side stays in contact with the base.

Answer 25

oh i get it, so the normal contact force is nottaken into account, because we consider the object to loose its contact once the equilbrium is broke, i mean at tht instant?

Answer 26

Exactly.

Answer 27

thanks can u help me with the new question ive posted :)