EQUILIBRIUM CASE V: At rest on an incline

EQUILIBRIUM1 CASE 5

The inclined plane is a simple machine that has a number of excellent Physics teaching concepts associated with it.

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Perhaps the best way to investigate theh r

forces acting on a body on an inclined angle of inclinationt

plane is to draw an x-y reference frameor angle of tilt y

thru the cg of the body, making the x- base x

axis parallel to the plane.

Using the cg as the origin, and inserting the forces at work on the object, we can break down everything into components:

  

Some of the Forces are familiar: N is the Normal force, perpendicular to the surface of the plane. The -N is the anti-Normal because it points in the opposite direction to the Normal. It is equal to the y-component of the weight (W) vector.

F' is effective weight and is equal to the x-component of the weight vector. It is the vector that tends to pull the object down the plane. The angle between W and -N is , the same as the tilt angle of the plane. Can you see why this is so?

N and -N are equal in magnitude so the normal force on an incline can always be determined by: All bodies on an incline have an effective weight F'. The geometry shows it is equal to the x-component N = W cos of W, so:

F' = W sin

NOTE: The association of sine is with x, not y. Cosine also follows the opposite alliance.

Between the surface of the plane and the under surface of the object, the force of friction will exist. We found in Case III that fs = sN and since we have shown that N = W cos, we can substitute for N and come up with an equation that helps us on the incline (only):

fs= sW cosONLY ON INCLINES!!

For any set of surfaces the static frictional Force (fs) will have a maximum value. As an example, let’s look at a plane tilted at 10. The body on the plane weighs 100 lbs and the coefficient of friction (s) is 0.27.

 = 10Investigate the maximum fsfirst:

W = 100 lbs

s = .27fs = sW cos = (.27)(100lbs)(cos10) = 26.59 lbs

That is the maximum frictional force!

The effective weight is found by:

F' = W sin = (100lbs)(sin10) = 17.36 lbs

17.36 lbs wants to drag the body down the

plane, but friction is too strong. The body can’t

move and static Equilibrium exists.

Since all forces up the plane equal all forces down the plane:

F' = fs and F' = 17.36 lbs

then fs = 17.36 lbs

Now set the angle of inclination to 20 - the effective weight will increase and the maximum frictional force will decrease:

fs = sW cos = (.27)(100lbs)(cos20) = 25.37 lbs

F' = W sin = (100lbs)(sin20) = 34.20 lbs

Friction will no longer hold the body back.

There will be motion down the plane.

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Somewhere between 10 and 20 - for this example - there is a critical angle - an angle where the effective weight equals the maximum frictional force. Below this angle friction is too strong - above this angle friction cannot hold the body. The critical angle is called the SLIP ANGLE. The slip angle is unique for each given system.

We can determine the slip angle easily by working a little algebraic hocus-pocus:

At the slip angle F' = fsand W sin = sW cos so,

W sin = sW cos  s cos = sin s = sin / cos

s = tan

The coefficient of static friction on an inclined plane is the tangent of the slip angle! This can be stated other ways:

s = tan (slip ) or slip  = tan-1s

If the slip  > no motion down the plane - friction is too great (fs)maxF'

If the slip  =  equilibrium - impending motion(fs)max = F'

If the slip  <  motion down the plane - friction is too weak(fs)maxF'

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An inclined plane is tilted at 8 and the coefficient of static friction for the surfaces is 0.3:

slip  = tan-1s = tan-1 (.3) = 16.7 slip  > 8 No motion!

------An inclined plane is tilted at 25 and the coefficient of static friction for the surfaces is 0.21:

slip  = tan-1s = tan-1 (.21) = 11.8 slip  < 25 Motion down the plane - no

equilibrium (Fx= ma)

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A 200lb crate rests on an inclined plane in a state of impending motion. If s = .4

find (a) slip  , (b) F' , (c) fs , and (d) N

In impending motion  = slip  = tan-1s = tan-1 (.4) = 21.8

On an inclined plane:F' = W sin = (200lbs)(sin21.8)

= 74.27 lbs

fs = sW cos = (.4)(200lbs)(cos21.8)

= 74.27 lbs

N = W cos = (200lbs)(cos21.8)

= 185.70 lbs

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10/15/2018