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Incline Plane

  • Thread starter physicsss
  • Start date
Knowing that the incline plane has an angle of 30 degrees and the coefficient of friction between a 60 lb block and the incline is 0.25, determine the smallest force P for which motion of the block up the incline is impending and the corresponding angle the force makes with the incline plane.

Is P= 0.25*cos(30)*60+sin(30)*60? But the book gives the answer of 41.7 while my answer is 42.99...any ideas?
 
3,761
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First, where is your gravitaional constant g ?

Isn't lb just pounds ?

Secondly, the gravity component (ie the sine-part) is bigger then the friction component (the cosine part). This means that the block is going down with a force equal to the difference between the two parts in your equation. You need to apply a force , equal to that difference but opposite in direction, to make sure that the block does not go down anymore

marlon
 
Doesn't the sine part points at the same direction as the friction part...?
 
And pound is a force.
 

Fermat

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P is at an angle to the incline. If you keep P acting along the plane then you will get P = 42.99...
But if you angle P upwards from the plane. then it will have a lifting effect on the block which will reduce the friction, hence reduce the slope component required from P.

Edit: looks like a minimisation problem, perhaps ???
 
Can you get me started?
 

Fermat

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Hold on a moment.
 

Fermat

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Use the sketch below.
Do the same as before, but adjust the friction value and the force normal to the plane to take the new position of P into account.
You will get a function involving beta, the angle of P.

Now differentiate P wrt beta in order to minimise P.
http://img508.imageshack.us/img508/1593/physicss0fb.th.jpg" [Broken]

Edit: changed maximise to minimise.
 

Attachments

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Fermat

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I'll post my attachment onto image shack.
 
Is this right?
sum of forces alont the incline:
P*cos(B)-0.25*cos(30)*60-sin(30)*60=0

P*sin(B)+cos(30)*60-cos(30)*60=0
 

Fermat

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Just worked through it myself. I got 41.7 deg :biggrin:
 

Fermat

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P.sinB reduces the normal force on the plane and so reduces the friction component. You haven't included that in your calculations yet.
 
I don't quite get it...
 

Fermat

Homework Helper
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The normal force on the plane is mg.cos@ less the normal component of P, which is P.sinB.
So, the normal force on the plane is NR = mg.cos@ - P.sinB.
 
P*cos(B)-0.25( cos(30)*60-P*sin(B) ) - sin(30)(60)=0?
 
Last edited:

Fermat

Homework Helper
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Yas!

Now get P as a function of B and "minimise" it.
 
Thank you!
 

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