Moment question. - Physics

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:sl:

doin sum rev... this question i completely got no idea how 2go about it... ne help would b appreciated :)

The spring loaded follower A bears against the circular portion of the cam until the lobe of the cam lifts the plunger. The force requried to lift the plunger is proportional to its vertical movement h from its lowest position. For design purposes determine the angle theta for which the moment of the contact force on the cam about the bearing O is a maximum.

In the enlarged view of the contact, neglect the small distance between the actual point B and the end C of the lobe.
Click to expand...




i think my problem is trying to model the motion of the lobe... after that it's just a matter of solving for the maxima of the function.
 
Ameeratul Layl said:
salam,

Have you managed to solve it? I'm not sure where to begin. Are we meant to consider Trig?

wasalam
Click to expand...

salams sis,

yeh in almost all the questions trig is involved.... Here's an idea how to go about it, (my problem is i don tknow how to transfer this idea into numbers)

1. Determine equation of motion fo the lobe.
2. Relate that motion to Force.
3. We know that Moment is maximum when the force is 90 degrees to the line between the force and moment center.
4. Come up with equation "M= Fd" (where F is an equation of force)... and find the maximum value of the function.

im sure das how u suppose 2go about it.. my problem is part 1 and 2.

jazaks :D :w:
 
:sl:

It's moments like this I'm so glad that I never ended up doing eng. :D

LOL I just realised my above comment had a double meaning. ;D Either way, that question looks evil. :uuh:
 
Malaikah said:
:sl:

It's moments like this I'm so glad that I never ended up doing eng. :D
Click to expand...

assuming meaning 1: did these questions make u cry? lol

meaning 2: u got questions like tht in ib/phys A :?

LOL I just realised my above comment had a double meaning. ;D Either way, that question looks evil. :uuh:
Click to expand...

it's actually the last question in the chapter xercsies.. so i guess dere's some sort of twist to it :heated:

neway atleast im sure i wont be gettin ne xam question like this :D
:w:
 
:sl:

No, we didnt have anything that evil looking for moments, but we had some pretty evil questions though...
 
:sl:

lol ok i think this is ittt...

Let l = 80 mm be the distance OC, and r = 40 mm be the cam radius.

The force F imposed by the follower A on the cam at B is vertical. The moment arm from O to the line of action of F is:

d = l cos Q

(where Q is the angle theta, we can ignore the small horizontal distance from B to C).

The compression of the spring is

h = l sin Q - r

so that the spring force is

F = k h = k l (sin Q - r).

Sooo... the moment of F about O is

M = k l cos Q (l sin Q - r) = k l r cos Q (2 sin Q - 1)

since:

l / r = 2.

This will have a stationary value when

dM/dQ = 0

i.e....

k l r [-sin Q (2 sin Q - 1) + cos Q (2 cos Q)] = 0.


That is,

4 sin^2 Q - sin Q - 2 = 0

==>

sin Q = (1/8)[1 +/- sqrt(1 + 32)]

= 0.843 or -0.593.

Hence Q = 57.5 deg.

aha! :D

:w:
 
momentz *faints* had a end of unit ting yh WHOA woz dat hard or wha.. :-\ :offended: i bet i do reali cr*p on it!
 
Ameeratul Layl said:
salam,

oh bravo! Even if you've got it wrong ATLEAST you tried. :)

I think that is a A grade type question.

wasalam
Click to expand...

:sl:
jzks sis :D im pretty sure its correct inshalah.. coz its the same answer as the top corner of that pic i posted.

the hard bit was the first 2 equations... coz i kinda was confused about the trig behaviour of the motion...
:w:
 
lol im dun doin it.. im soo happy.. now il hav 2 wait 4 da exam :offended: n i bet i get non of it corect :p
 

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