ganeshie8
  • ganeshie8
A crate, in the form of a cube with edge lengths of 1.2 m, contains a piece of machinery; the center of mass of the crate and its contents is located 0.30 m above the crate’s geometrical center. The crate rests on a ramp that makes an angle θ with the horizontal. As θ is increased from zero, an angle will be reached at which the crate will either tip over or start to slide down the ramp. If the coefficient of static friction μs between ramp and crate is 0.60, (a) does the crate tip or slide and (b) at what angle θs does this occur? If μ = 0.70, (c) does the crate tip or slide and (d) at what angle θ does this occur? (Hint: At the onset of tipping, where is the normal force located?)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Oh poop, I know how to do this but I don't have time right now to go through it with you because I'm currently studying for a Statics midterm :( I would post this in the physics section and tag IrishBoy123 or Michele_Laino!
ganeshie8
  • ganeshie8
thats fine :) i have some idea... but not full sure..
ganeshie8
  • ganeshie8
|dw:1446714808928:dw|

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More answers

ganeshie8
  • ganeshie8
It seems the normal force changes from \(N_s\) to \(N_t\) if the crate tips instead of sliding
baru
  • baru
|dw:1446717852000:dw|
baru
  • baru
i think it will tip if torque about point "P" due to T component of mg is greater than N componant
baru
  • baru
and T component also has to less than frictional force (for it to tip)
ganeshie8
  • ganeshie8
Ahh right, if it slides, it must slide at an angle \(\arctan(\mu_s)\)
ganeshie8
  • ganeshie8
How to get the angle at which it tips if the friction prevented it from sliding before tipping ?
baru
  • baru
|dw:1446718420975:dw|
baru
  • baru
when marked angle is less than 45 there is a chance of tipping,
ganeshie8
  • ganeshie8
About the point P : 1) I believe the force \(N_s\) is perfectly balanced by a component of \(mg\). so we don't need to worry about it. 2) Frictional force is parallel to the position vector, so this can be ignored too. 3) Only the tangential component of \(mg\) makes the crate rotate is it ?
ganeshie8
  • ganeshie8
|dw:1446718700633:dw|
baru
  • baru
yes, i think so. because the the box is constrained to move only along the slide
ganeshie8
  • ganeshie8
I think I am getting this... When the crate begins to tip, the normal force through center of mass moves to the right edge. So, the torque will have components in both directions and the crate tips to the side based on whatever direction the torque is greater
ganeshie8
  • ganeshie8
|dw:1446719200202:dw|
ganeshie8
  • ganeshie8
Oh, about P, with reference to above specific arrangement, the force `mg` has torque in "counter clock wise direction"... so the crate tips to the left side... Is that correct way to interpret ?
ganeshie8
  • ganeshie8
If that is correct, then the torque would be 0 when the force `mg` passes through P : |dw:1446719587649:dw|
baru
  • baru
hmm...i'm not fully following you, but anyway, in my opinion, mg has a counterclockwise torque, but as \(\theta\) increases there will be a point beyond which mg creates a clockwise moment, that will be the tipping point
ganeshie8
  • ganeshie8
If \(\theta\) is anything more than that, the torque changes its direction, and the crate would tip to the right side
ganeshie8
  • ganeshie8
I think I get you :)
baru
  • baru
ok so we agree :)
ganeshie8
  • ganeshie8
In below configuration, the \(mg\) force vector is to the right of \(P\). So, the crate will tip to the right because the torque about \(P\) produces clockwise angular momentum now. |dw:1446720267771:dw|
ganeshie8
  • ganeshie8
|dw:1446720428974:dw|
ganeshie8
  • ganeshie8
then the minimum angle for tipping should be \(\arctan \frac{0.6}{0.9}\)
baru
  • baru
i agree. I think breaking mg into components along T and N (from my diagram) is equivalent, if T>N, then there is a net clockwise moment. also we need not say that component along T will cause tipping, we can break mg into any two 2 components which always cause clockwise and anti-clock moments. but T and N are the most convenient. {also neglect 45 deg i mentioned somewhere :p}
ganeshie8
  • ganeshie8
completely got it, thanks :)
baru
  • baru
also, we need to include friction, tangent force should not exceed friction force

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