RAMP FRICTION

Even the simplest of ramps require a friction analysis to ensure proper applicationSolvit Half-Ramp II

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Ramp Friction There are quite a number of ways to measure the coefficient of

static friction. These include: Leonardo da Vinci's weight ratio method Spring balance method Clamping Pendulum Motorized tribometer Tilted plane

Nanovea Tribometer With Pin-On-Disk and Linear Reciprocating Modules

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Ramp Friction One of the simpliest and most

common methods is the tilted plane method. The device used to perform this test is a called a friction table

Such a device allows a mass or a certain material to rest on a table. The table surface may be covered or coated with a second material

The table is simply tilted until the mass begins to move

The angle at which motion begins is recorded and a specified analysis procedure is used to calculate the coefficient of friction

Testing Machines, Inc. Model 32-25 Coefficient of Friction Tester

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Ramp Friction The use of a friction table requires the analysis of what is called

ramp friction. Such an analysis requires the development of the Free Body Diagram

Assume mass 'm' with weight 'W' rests on an inclined plane. Impending motion occurs when the plane is tilted to angle θ A FBD of the mass is shown on the following page. Of course, the

weight vector is vertical. But the normal force will be perpendicular to the inclined plane and equal to the component of weight perpendicular to the inclined plane

Of course, the component of weight parallel to the inclined plane is what tends to drive the mass down the plane

The developed force of friction will be parallel to the inclined plane, be equal to the component of weight parallel to the inclined plane, and always oppose the direction of impending motion

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Ramp Friction The FBD explained on the previous page is shown below. Note

the use of a rotated X-Y coordinate axis system. By similar triangles, the y-component of W is at an angle θ with the y-axis

The determination of the coefficient of friction is a simple sum of forces analysis

W y =W⋅cos and W x =W⋅sin

F x = F −W⋅sin . . . F =W⋅sin

F y= N −W⋅cos . . . N =W⋅cos

Since: =FN

Then: =W⋅sinW⋅cos

= tan

θ W

Motion

NF

X

Y

Wx

Wy

θ

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Ramp Friction From the discussion on plain friction, we know

Resultant 'R' of the normal and friction forces is at an angle φ with the normal force

The tangent of angle φ also defines the coefficient of friction

It then follows:When motion is impending for a mass of weight 'W' when resting on an inclined plane at angle θ, then the angle of incline is equal to the friction angle φ

But what happens when we apply some force 'P' to: Push the mass up the ramp? Pull the mass down the ramp? Hold the mass in place to prevent it

from sliding down the ramp?

θ W

Motion

NF

X

Y

Wx

Wy

θ φR

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Ramp Friction Assume force 'P', at some angle α with the horizontal, pushing

a mass of weight 'W' up a ramp inclined to angle θ. Motion is impending. The value of µs is given

As with all statics problems, first develop a proper free body diagram. This includes an x-y axis rotate to angle θ

Break all forces into their x- and y-components and apply equilibrium.

Solve for any three of five possible variables: P, W, F, N, or µ

P x = P⋅cos and P y = P⋅sin

W x =W⋅sin and W y =W⋅cos--------------------------------

F x = P⋅cos −W⋅sin −F F y= P⋅sin −W⋅cos N

=FN N

Y

θ W

Motion

N

F

XWx

θ

θ Rφ

αP

WyPy

Px

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Example 7 - Ramp Friction What force, 'P', acting at angle α = 10o, must be applied to a

mass of 1000 lbm for impending motion up a ramp at angle 30o

µs = 0.268

Y

θ W

Motion

N

F

XWx

θ

θ Rφ

αP

WyPy

Px

A mass of 1000 lbmhas a weight of 1000 lb fBreaking all forces into their x and y components, we know:

P x = P⋅cos 40P y= P⋅sin 40W x = 1000⋅sin 30 = 500 lb fW y= 1000⋅cos 30 = 866 lb f

Notice that since weight is a vertical forceand the normal force is not, N≠W

N = unknown y-componentF = unknown x-component

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Example 7a - Ramp Friction Writing the equations of equilibrium, we have:

F x= 0.766⋅P−500−F F y =−0.643⋅P−866N

This results in two equations with three unknowns!We need another relationship. We do know:

=FN

. . . F = 0.268⋅N

Substituting F into F x , we have: F x= 0.766⋅P−500−0.268⋅N

Now divide through by and add to F y.

F x= 0 = 2.86⋅P−1866−N F y = 0 =−0.643⋅P−866N

Adding to eliminate N and solve for P:

0 = 2.22⋅P − 2732 . . . P = 1230 lb f

Y

θ W

Motion

N

F

XWx

θ

θ Rφ

αP

WyPy

Px

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Example 7b - Ramp Friction Alternatively, we can solve this problem trigonometrically. Use the

information known about P, W and R to build a force triangle and solve using the law of sines.

W = 1000 lbf

θ + φ

R

P

β

γα

= tan−1 = 15o

Resultant 'R' is at = 45o to a vertical. = 90 = 100o

= 180−−= 35o

Applying the law of sines:

P

sin 45o=

1000

sin 35o=

R

sin 100o

P = 1230 lb fR= 1720 lb f

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Example 8a - Ramp Friction Now assume we wish to apply a force 'P' to pull the same mass

down the same ramp with impending motion In such a scenario, we first need to check to ensure the block is

resting in place, not already sliding down the ramp. To do so, the angle θ cannot exceed the tan-1(µ)

= 30o

= tan−1 = 15o

Since ≪ , this mass will move withoutapplication of force 'P'. In fact, a force 'P'must be applied to prevent slidingLet's determine that force

Y

θ W

Motion

NF

XWx

θ

φR θ

αP

Wy

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Example 8b - Ramp Friction For clarity, let's modify the FBD to show the proper directions of

force 'P' and resultant force 'R' The components of 'P' and 'W' are:

Y

θ W

Motion

NF

XWx

Wy

θφR

θ

αP

F x=0.766⋅P−500F F y =−0.643⋅P−866N

=FN

. . . F = 0.268⋅N

Substituting: F x=0.766⋅P−5000.268⋅N

Divide through by − and add to F y.

F x= 0 =−2.86⋅P1866−N F y = 0 =−0.643⋅P−866N

0 =−3.50⋅P 1000 . . . P = 286 lb f

Py

Px

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Example 8c - Ramp Friction Let's solve the previous problem using trigonometic methods.

Notice that the angle between the resultant and a vertical is calculated as θ−φ; rather than θ+φ for a mass moving up a ramp

W = 1000 lbf

θ − φ

R

P

βγα

= tan−1 = 15o

Resultant 'R' is at −= 15o to a vertical.= 90 = 100o

= 180−−−= 65o

Applying the law of sines:

P

sin 15o=

1000

sin 65o=

R

sin 100o

P = 286 lb fR= 1086 lb f

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Example 9a - Ramp Friction What would happen if angle θ was reduced to 10o, an angle less

than φ? The angle α remains at 10o

The components of 'P' and 'W' will change as follows

Y

θ W

Motion

NF

XWx

Wy

θ

φR θ

αP

P x = P⋅cos 20 and P y= P⋅sin 20

W x= 1000⋅sin 10 = 174 lb fW y= 1000⋅cos 10 = 985 lb f

F x=−0.940⋅P−174F F y =0.342⋅P−985NF = 0.268⋅N

Substituting 'F' and dividing through by − :

F x= 0 = 3.51⋅P649−N F y = 0 = 0.342⋅P−985N

0 = 3.85⋅P − 336 . . . P = 87.3 lb f

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Example 9b - Ramp Friction Let's solve the previous problem using trigonometic methods.

Notice that the angle between the resultant and a vertical is calculated as θ−φ; rather than θ+φ for a mass moving up a ramp

W = 1000 lbf

φ−θR

Pβ

γ

α = tan−1 = 15o

Resultant 'R' is at − = 5o to a vertical. = 90− = 80o

= 180−−− = 95o

Applying the law of sines:

P

sin 5o=

1000

sin 95o=

R

sin 80o

P = 87.5 lb fR= 989 lb f

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Example 10 - Ramp Friction A ramp is being designed to load equipment into a pickup truck.

The tailgate of the pickup is 1.2 meters above the ground. The ramp has a horizontal dimension of 3.6 meters. The equipment being loaded has a mass of 1000 kg. What is the minimum required coefficient of friction?

To build a FBD for this problem, weneed to know the angle of the rampand the weight of the mass.

= tan−1 = 1.23.6 = 18.43o

W = m⋅g= 1000 kg⋅9.807m/ s2

= 9807N

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θ W

Motion

NF

X

Y

Wx

Wy

θ φR

Example 10 - Ramp Friction The minimum coefficient of friction is that value that causes

impending motion if the mass is at rest

Since =FN

, then we need to find F and N.

From equilibrium: N =W⋅cos = 9807⋅cos 18.43o= 9304N

F =W⋅sin = 9807⋅sin 18.43o= 3100 N

Then =FN

=31009304

= 0.333

Notice this is also the tangent of angle θ of the ramp. In actuality, we would increase µ based upon some factor of safety

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Example 11 - Ramp Friction In the previous problem, for the calculated coefficient of friction,

what is the maximum weight that can be driven up the ramp without slipping?

Since =FN

From equilibrium: N =W⋅cos F =W⋅sin

Then:

=FN

=W⋅sin W⋅cos

= tan

In other words, weight is not a factor when determining when sliding occurs. This is determined strictly by ramp angle and coefficients of friction