Calculating Force and Angle on Incline Plane with Friction

[physicsss]

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?

 

[marlon]

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

 

[physicsss]

Doesn't the sine part points at the same direction as the friction part...?

 

[physicsss]

And pound is a force.

 

[Fermat]

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 ?

 

[physicsss]

Can you get me started?

 

[Fermat]

Hold on a moment.

 

[Fermat]

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"

Edit: changed maximise to minimise.

 

[Fermat]

I'll post my attachment onto image shack.

 

[physicsss]

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]

Just worked through it myself. I got 41.7 deg

 

[Fermat]

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

 

[physicsss]

I don't quite get it...

 

[Fermat]

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.

 

[physicsss]

P*cos(B)-0.25( cos(30)*60-P*sin(B) ) - sin(30)(60)=0?

 

[Fermat]

Yas!

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

 

[physicsss]

Thank you!