Question

Mechanics 1+2 question.

Answer 1

A load of 981 N is suspended from a steel wire of radius 1 mm. What is the maximum angle through which the wire with the load can be deflected so that it does not break when the load passes through the equilibrium position? Breaking stress is \(7.85 \times 10^8 N/m^2\).

Answer 2

This will test your knowledge of mechanics 1 (rigid body dynamics) and mechanics 2 (fluids and bulk matter)

Answer 3

stress = Y * strain

Answer 4

\[\dfrac{F}{A} = Y*\dfrac{\Delta L}{L}\]

Answer 5

just need to express \(\Delta L\) in terms of angle

Answer 6

Hmm, that's not it.

Answer 7

i don't see how this can be anything different... let me attempt :)

Answer 8

the answer is 3 XD. You can medal me your welcome

Answer 9

I think 981 N + centrifugal force cannot exceed the maximum stress

Answer 10

If that is the case, do i get :
\[981 +ma_c = 7.85*10^8*\pi*10^{-6} \]

?

Answer 11

somehow i need to express \(a_c\) as a function of the deflection angle

Answer 12

You're getting close.

Answer 13

let me grab pen and paper

Answer 14

after some rearrangement im getting

\[981(1+ h) = 7.85*10^8*\pi*10^{-6} \]

Answer 15

need to express \(h\) in terms of the deflection angle

Answer 16

but \(h\) depends on the length of wire too
so above method should lead me to a dead end hmm

Answer 17

at equilibrium, potential energy energy = 0
we have 0.5mv^2=mg h cos( )
we could eliminate h using that

Answer 18

Yeah, exactly. You need to employ a concept in rigid-body mechanics. Baru is getting there!

Answer 19

981 + \(mv^2/r\) = stress * area

Answer 20

still wondering.. not getting a clue

with that, i think baru and me are in same boat :)

Answer 21

r=h

Answer 22

brb

Answer 23

981 + mv^2/h = stress * area

0.5mv^2=mgh cos( )

substitute for v^2 in eq 1, h will get cancelled
solve for cos ( )

?

Answer 24

Yeah, hint: vertical circular motion.

Answer 25

Should I spoil it now?

Answer 26

nope, i got it

Answer 27

Oh, how?

Answer 28

eating now, will post in 10 mnts

Answer 29

http://www.wolframalpha.com/input/?i=arccos%281+-+%287.85*pi*10%5E2-981%29%2F%282*981%29%29

Answer 30

:)

Answer 31

Nice ! thanks to baru for hinting conservation of energy

Answer 32

Sure :)
Still don't get it though :(

Answer 33

\[\frac{mv^2}{2}=mg\ell (1 -\cos \theta_{max})\]\[\Rightarrow \frac{mv^2}{\ell} = 2mg(1-\cos \theta_{max})\]Now\[T_{breaking}-mg = \frac{mv^2}{\ell}\]\[\Rightarrow T_{breaking} = 3mg - 2mg \cos \theta_{max} \tag{using previous equation}\]Now use\[T_{breaking}=\rm{area \times breaking ~stress}\]

Answer 34

Oh ok..Potential energy is mgh(1-cos)....I missed that