Incline Plane

  • Thread starter physicsss
  • Start date
  • #1
319
0
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?
 

Answers and Replies

  • #2
3,763
8
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
 
  • #3
319
0
Doesn't the sine part points at the same direction as the friction part...?
 
  • #4
319
0
And pound is a force.
 
  • #5
Fermat
Homework Helper
872
1
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 ???
 
  • #6
319
0
Can you get me started?
 
  • #7
Fermat
Homework Helper
872
1
Hold on a moment.
 
  • #8
Fermat
Homework Helper
872
1
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

Last edited by a moderator:
  • #9
Fermat
Homework Helper
872
1
I'll post my attachment onto image shack.
 
  • #10
319
0
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
 
  • #11
Fermat
Homework Helper
872
1
Just worked through it myself. I got 41.7 deg :biggrin:
 
  • #12
Fermat
Homework Helper
872
1
P.sinB reduces the normal force on the plane and so reduces the friction component. You haven't included that in your calculations yet.
 
  • #13
319
0
I don't quite get it...
 
  • #14
Fermat
Homework Helper
872
1
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.
 
  • #15
319
0
P*cos(B)-0.25( cos(30)*60-P*sin(B) ) - sin(30)(60)=0?
 
Last edited:
  • #16
Fermat
Homework Helper
872
1
Yas!

Now get P as a function of B and "minimise" it.
 
  • #17
319
0
Thank you!
 

Related Threads on Incline Plane

  • Last Post
Replies
19
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
4
Views
4K
Top