Statics

Lesson 7

(Static) Friction

🟦 7.1 Introduction to friction

🟦 7.2 Review of key friction concepts from Physics

🟦 7.3 Graphic symbols for different types of planes

🟦 7.4 When do engineers neglect or include friction?

🟦 7.5 When and how to use a half arrow for friction force

🟦 7.6 Showing friction on a FBD (using N3L pairs)

🟦 7.7 Normal force can be expressed as a line or area load

🟦 7.8 Using friction to prevent translation (slipping or sliding)

🟦 7.9 Using friction to prevent rotation (tipping or overturning)

🟦 7.10 Introduction to limit states

🟦 7.11 Trigonometry tips

➜ Practice Problems

🟦 7.1 Introduction to friction

So far, we have focused on three categories of forces: normal force (N), internal tension force (T), and body force or self-weight (W=mg).

In this unit, we will add shear force to our studies.

A shear force is defined as a force that is parallel (or coplanar) to a plane or surface.

The symbol for shear force is V.

❏ Categories of force (contact and non-contact forces)

You saw this diagram earlier this semester, but take a closer look. I bet your understanding of force has deepened a bit!

While normal force (N) is a push or pull contact force, a shear force (V) is a contact force that lies in the plane of a surface.

Shear forces are neither pushes nor pulls. Shear is a planar force.

❏ Normal and shear forces

Friction is a contact force, but it's neither a push nor a pull. It's a type of shear force, because it lies in the surface between two objects.

For a system in static equilibrium, the friction vector points in the direction that opposes impending motion.

We will use either Ff or Ffr to indicate the force of friction. Both variants are commonly used.

❏ Symbols for friction force

🟦 7.2 Review of key friction concepts from Physics

The type of friction force we study in Statics is called dry friction (or Coulomb friction).

(Dry) friction force is a function of two parameters:

the roughness of the interface between the two surfaces in contact (more roughness = more friction capacity)
the magnitude of compressive force transferred over the contact surface (more compression = more friction capacity)

The roughness of the surface is quantified in the static and kinetic coefficients of friction. These are determined experimentally.

❏ Friction force at the microscale

The rough interface between blocks 1 and 2 creates horizontal force transfer when the little hills and valleys interlock. Note that in order to develop friction, the surface must also transfer compression. In other words, blocks 1 and 2 are pushing against each other.

We use μ (pronounced "mu") for the coefficient, with subscripts to differentiate between static equilibrium and motion:

μs = static coefficient of friction
μk = kinetic coefficient of friction

Since our focus in this class is static equilibrium, we won't be using the kinetic coefficient of friction. You do need to know conceptually that it exists, and that it's smaller than the static coefficient. That means that it's harder to initiate motion than to maintain motion.

Here is the formula for the limiting (maximum) friction force:

❏ Static and kinetic friction

This plot shows that until we reach the maximum possible static friction force, the friction force matches the shear demand.Once the shear force exceeds the maximum frictional resistance (or limiting friction force), the system begins to move. While in motion, the frictional resistance slightly drops, because it is easier to maintain motion than initiate it.

The equation tells us that the shear resistance of a friction plane is proportional to the amount of compressive force that is perpendicular (or normal) to the plane.

By the way, in Statics, we only focus on on dry (Coulomb) friction. If you want to minimize the effects of friction, you may decide to lubricate the surfaces. If you do that, then you're in the realm of fluid friction. We don't cover that in Statics.

🟦 7.3 Graphic symbols for different types of planes

In this text, a frictionless plane is shown as a thick black line. Sometimes, for emphasis, the surface is colored blue, which is meant to evoke an icy (slippery) surface.

This could mean that the coefficient of friction is nearly zero.
Alternatively, it could mean that we are choosing to neglect friction.

❏ Frictionless planes

In this text, a frictioned plane is signified by black dots adjacent to a black line.

The (static) coefficient of friction will be provided on the graphic or in the problem statement.
If the mu doesn't have a subscript, assume it is the static coefficient, as we do not study dynamic friction in Statics.

❏ Frictioned plane

As a reminder, a solid black line adjacent to black hatchmarks signifies a fixed connection. At this type of engineering connection, the body can neither translate nor rotate. We do not develop friction at a fixed plane. It's a mechanical connection. Instead, you'd solve for Ax , Ay, and MA (the reactions required for static equilibrium).

❏ Fixed plane (plane of fixity)

🟦 7.4 When do engineers neglect or include friction?

In engineering practice, we generally neglect friction when we engineer systems to resist motion.

This is because we generally do not rely on friction for static equilibrium.

For example, this image shows two ways that engineers might prevent a block from sliding down a ramp using mechanical connections (a cable in the middle image, and anchor bolts in the bottom image).

We would engineer the cable or bolt system to transfer 100% of the load. Friction would not affect the system behavior unless the cable or bolts were to give way.

When you study dynamics, you'll take friction into account more often. When engineers try to set a system into motion, the design must overcome the force of friction. So, in that scenario, you will need to take friction into account.

In this class we will learn and apply the basics of static friction, even though they are rarely used in most engineering applications.

❏ Friction vs. Engineering Connections

If you want to keep a heavy block from sliding down a ramp, you probably should attach it to a cable or bolt it down. These are both examples of mechanical connections.

🟦 7.5 When and how to use a half arrow for friction force

This is the second time in this text that you've seen this image of a coffee mug subjected to a horizontal force. You may recall the use of the half arrow symbol for the friction force in the 2D view of the coffee cup.

You already know that a shear force lies in a planar surface.

In the 3D view of the transparent coffee mug, you can see that surface, and therefore you would draw the shear force with the regular style of arrow (← or →).

In the 2D view of the coffee mug, the friction force is coincident with the bottom surface of the coffee mug. It would be hard to see the vector it if you were to draw the friction force on the proper line of action. That impedes our ability to communicate effectively.

Past engineers have come up with a solution to this problem. In order to easily communicate the presence of a shear force in 2D, we simply offset the vector slightly away from the body. Then, we draw a half arrow (↽ or ⇁ or ↼ or ⇀ or ↿ or ⇃ or ⇂ or ↾).

Also note that the flag on the half arrow is always away from the body, as illustrated.

❏ Half arrow example

This is not a FBD in static equilibrium. It is just an infographic to help you learn how to use half arrows to illustrate shear (and friction) in a 2D drawing.

🟦 7.6 Showing friction on a FBD (using N3L pairs)

Newton's Third Law applies to friction forces.

Unfortunately, we can't use the push/pull logic that was so helpful for normal forces (arrows toward the body for push pairs vs. arrows away from the body for pull pairs).

Instead, we describe the N3L pairs for friction force in terms of their direction. If one is rightwards, the other is leftwards.

You only draw friction forces on a FBD when you disassemble the parts at the friction surface, like this:

❏ N3L pairs for the force of friction

Two blocks (1 and 2) of negligible weight are supported by (frictionless) rollers in tracks. Forces (P and P) are applied to the blocks. This generates a friction force at surface a-a. If we neglected friction on Block 1, it would translate rightwards. That's how we know the force of friction required for static equilibrium must be leftward. Similarly, if you look at Block 2, without friction, it would move leftwards. Therefore, the force of friction needed for static equilibrium must be rightwards.The compressive force (which is required for developing a friction force) has been omitted from these drawings to focus on the N3L friction pairs. These are not FBDs; they are just educational diagrams.

The key take-away is that the friction vector is always drawn in the direction that opposes impending motion of the free-body.

🟦 7.7 Normal force can be expressed as a line or area load

We recently learned that a resultant force can be expressed as a line load (w) or an area load (p).

If we consider the simple case of a body supported by a surface, we get the FBDs at right. Both are correct and equivalent; they are interchangeable.

The left diagram is what you have been trained to draw so far.

The right diagram is a more realistic depiction of the transfer of force. In Mechanics of Materials, this idea is called bearing stress.

❏ Normal force expressed as a constant line load

Note that in this problem, the normal force, N, is located at the midpoint of the bottom surface.

Now, consider a trapezoidal body. It looks like a triangle connected to a rectangle.

In the left FBD, the normal force, N, is no longer at the midpoint of the bottom surface. Due to moment equilibrium, we know that N must occur at the third point between W1 and W2 , since W2 is twice W1 .

The two FBDs are statically equivalent. The force, N, can be expressed as a linear line load function. To verify that the two systems are equivalent, we would use the equations of static equivalency.

❏ Normal force expressed as a linear line load

You can use the Lever Law to deduce that the location of the normal force (N) is at the third-point between W1 and W2.

Things get more complex when we add a horizontal load to a body. Let's return to the rectangular body.

This is when many of the "rules" for making FBDs we learned earlier become critically important.

In the left image, notice that the normal and shear forces are shown as components of a resultant force. There are 3 forces in the system, and they are not parallel, so this must be a concurrent force problem. All 3 vectors must intersect at the same point. Therefore, we can use a moment equilibrium equation to solve for x̄, and force equilibrium equations to solve for V and N.

In the right image, we see an equivalent FBD. The resultant force has been replaced with a linear loading function that produces the same tendency to rotate and the same tendency to translate.

❏ Self-weight plus a horizontal load

Key takeaways:

Each vector in the system has a precise point of application, line of action, direction, and magnitude.
Principles of concurrent force systems are in effect for friction problems.
The equations of static equilibrium must be satisfied if we want the friction to prevent motion (translation or rotation).

🟦 7.8 Using friction to prevent translation (slipping or sliding)

The most basic type of friction problem is to evaluate translation (sometimes called slipping or sliding) on a single friction plane.

For symbols, we'll use:

V = the force component that is parallel to the friction surface
N = the force component that is normal to the a surface
FR = the resultant force of components N and V: due to the Pythagorean Theorem, FR = sqrt(V2 + N2)
Ffr,max = the limiting friction force (the threshold between equilibrium and motion)

The following examples will walk you through the basic process for solving slipping/sliding problems. Work each in parallel. In slipping/sliding problems, you only need to use a force equilibrium equation. We'll use a moment equilibrium equation when we get to the tipping/overturning type of problems.

Question 1.

A 50N block lays on surface d-d. The coefficient of static friction at the surface is 0.5.

A horizontal 40N force is applied. Does the block translate to the left or can the friction keep the block in static equilibrium?

Answer 1.

In order to maintain static equilibrium, a rightward shear force of 40N is required at d-d. We compare that to the limiting friction force of (0.5)*(50N) = 25N.

Since the required shear force (40N) exceeds the maximum possible friction force (25N), the box is in motion. It translates to the left.

❏ Flipbook: Example 1

Question 2.

This is a continuation of the prior problem. In order to arrest the sliding, you find another block (W=30N) and pile it on top. Is this enough weight to maintain static equilibrium?

Answer 2.

There is no tendency to slip at plane e-e because there is no shear force at that location.

At d-d, the normal force increased to 80N. That makes the limiting friction force equal to (0.5)*(80N) = (40N).

Since the 40N of shear force (demand) is the same as the maximum friction force (40N), this system is in static equilibrium. In fact, it's right at the threshold between equilibrium and motion.

❏ Flipbook: Example 2

Question 3.

Blocks 1 and 2 are the same as above, but a 10N horizontal force is applied as shown. Is this system in static equilibrium or in motion?

Answer 3.

First, check plane e-e. Since the demand of 10N does not exceed the capacity of 12N, the top plane doesn't slip.

Next, investigate plane d-d. Since the demand of 10N does not exceed the capacity of 40N, the bottom plane doesn't slip.

Since we have checked all possible slip planes, we conclude that the block is in static equilibrium with respect to slipping/sliding.

❏ Flipbook: Example 3

Question 4.

We have the same assembly of blocks, but this time, there's a 10N force applied to the top block and a 40N force applied to the bottom block. Is the system in static equilibrium, or does either plane slip?

Answer 4.

First, check friction plane e-e. We notice that this set-up is identical to the prior example. That plane does not slip.

Second, we check plane d-d. The shear force required for equilibrium is 30N leftwards. Since the demand (30N) is less than the maximum friction force (40N), the system is in static equilibrium with respect to slipping/sliding.

❏ Flipbook: Example 4

🟦 7.9 Using friction to prevent rotation (tipping or overturning)

If asked to determine whether or not a system tips or overturns, you'll need to use a moment equilibrium equation.

This flipbook shows one approach to solving these types of problems.

We can see that the self-weight of the body creates a stabilizing moment while the horizontal push creates an overturning moment.

❏ Flipbook: where not to push a bookcase

Sometimes you may be asked to solve for the force that initiates overturning, Ptip.

To do this, place the resultant force at the pivot point, and then solve a moment equilibrium equation.

Let's call the pivot point O (for "overturning").

By using an E.o.E., you're solving for the threshold between equilibrium and rotation. If the push is equal to Ptip or less than Ptip , then you're in static equilibrium. If the push exceeds Ptip, you're in motion.

❏ How much of a push tips the bookcase?

🟦 7.10 Introduction to limit states

In friction problems, we need to use an engineering philosophy called limit states.

What this means is that if you are relying on friction force for static equilibrium, then you need to identify all possible types of motion and set up calculations to check each and every motion scenario. This could include slipping (with respect to multiple surfaces), and tipping (with respect to multiple pivot points). If every limit state is evaluated, and if no limit state results in motion, then you can state that the system is in static equilibrium.

❏ A box on a ramp is pushed

The first limit state is depicted here.

If P is not large enough to keep the box in static equilibrium, it slides down the ramp.

Note that the shear force is rightward (in the opposite direction of the leftward impending motion).

The exact location of N is not important. To analyze this limit state, sum forces. You'll calculate a minimum requirement for the magnitude of P.

❏ Sliding down the ramp

The second limit state is depicted here.

If P is too large, the box will slide up the ramp.

Note that the shear force is leftward (in the opposite direction of the rightward impending motion).

The exact location of N is not important. To analyze this limit state, sum forces. You'll calculate a maximum requirement for the magnitude of P.

❏ Sliding up the ramp

The third limit state is depicted here.

If P is not large enough to impede tipping about A, it will rotate counterclockwise.

The direction of the shear force is not important. To analyze this limit state, place the normal force at A, and sum moments. You'll calculate another minimum requirement for the magnitude of P.

❏ Tipping counterclockwise

The fourth limit state is depicted here.

If P is too large, the box will tip clockwise about B.

The direction of the shear force is not important. To analyze this limit state, place the normal force at B, and sum moments. You'll calculate another maximum requirement for the magnitude of P.

❏ Tipping clockwise

After you solve all four limit states, compare the results of 1 and 3 to determine P_min (the minimum force required for static equilibrium). Then, compare the results of 2 and 4 to determine Pmax (the maximum force you can apply and still attain static equilibrium).

🟦 7.11 Trigonometry tips

Some classic friction problems involve pushing heavy objects up and down ramps (inclined planes).

It's common for students to mix up sines and cosines on these types of problems.

This flipbook provides a method for constructing the geometry thoughtfully.

➜ Practice Problems

Problem 1.

Do these blocks translate or remain in static equilibrium?

Part (a). Check slipping at surface e-e by constructing a FBD of the top block. Solve for V and then compare it to Ffr,max.

Part (b). Check slipping at surface d-d by constructing a FBD of both blocks together. Solve for V and then compare it to Ffr,max.

Access the partial answer by highlighting the invisible text immediately below:The top plane does not slip, but the bottom plane does.

Problem 2.

Block A weighs 90 pounds and Block B weighs 240 pounds.

The coefficient of friction for surface c-c is 0.7. The coefficient of friction for surface d-d is 0.3.

Rope E is tensioned to 50 pounds and Rope F is tensioned to 150 pounds.

Is the system in static equilibrium or in motion?

You have to check 2 limit states.

Check slipping at c-c by drawing a FBD of Block A.

Checking slipping at d-d by drawing a FBD of Blocks A and B together.

Access the partial answer by highlighting the invisible text immediately below:The system is in motion; it slips at the ground plane.

Problem 3.

Work through the Geogebra interactive in 9.1.3 of Engineering Statics. There are three parameters: magnitude of W, magnitude of P, and the location of P.

From the interactive, sketch FBDs for the following scenarios:

While holding the magnitudes of W and P constant, move P close to the bottom and compare it to when it is close to the top.
While holding the magnitude and location of P constant, make W light and then heavy.
While holding the location of P constant along with the magnitude of W, make the magnitude of P high and then low.

Problem 4.

First, confirm that this system is in static equilibrium.

Then, determine the location of the resultant force on surface a-a. (Measure x bar from the left bottom corner of the box).

The weight of the box is 4 kN. The push force, P, is equal to 2 kN.

Access the partial answer by highlighting the invisible text immediately below: xbar is equal to 466 mm (measured rightwards from the bottom left corner of the box)

Problem 5.

This is the same box as in the previous problem (weight of 4 kN). This time, it's on a 30 degree ramp with a 0.7 coefficient of static friction.

Solve for Pslip, the force that puts the box into translational motion down the ramp.

Then, solve for Ptip, the force that overturns the box about the pivot point.

Access the partial answer by highlighting the invisible text immediately below:Pslip = 0.425NPtip = 0.479NThe smaller governs, so slipping is the limit state that tells us the maximum force that can be applied to maintain static equilibrium.

Problem 6.

Four identical bodies are supported by a vertical and a horizontal surface.

You can think of these as ladders leaning on a building.

The black dots illustrate planes with friction. The smooth blue color is meant to evoke ice, and is a frictionless surface.

Determine whether or not each of the four ladders is in motion or static equilibrium by sketching a FBD and determining whether or not the equations of equilibrium can be satisfied.

You do not need to solve these numerically; this is a conceptual problem. Use the qualitative equations of equilibrium to aid your logical deductions.

Access the partial answer by highlighting the invisible text immediately below:The ladders in static equilibrium are number 2 and number 4. Ladders 1 and 3 cannot satisfy the equations of equilibrium and are therefore set into motion.

Problem 7.

A block (W=600 pounds) is to be pulled rightward with the use of a frictionless pulley by pulling with force P.

What force, P, initiates motion of the block up the ramp?

Include a FBD of block A in your solution.

The dimensions of block A are not known.

Access the partial answer by highlighting the invisible text immediately below:A force, P, of 698 pounds sets the block in motion up the ramp.

Problem 8.

A heavy triangular-shaped box (ABC) is supported on ramp d-d.

Solve for the force that causes sliding, Pslip.

Solve for the force that causes overturning, Ptip.

The weight of the triangular-shaped box is 120kN.

Be sure to include a FBD with your solution.

Access the partial answer by highlighting the invisible text immediately below:Pslip = 36.9 kNPtip = 12.3 kNthe smaller of the two is the governing limit state, so Pmax is equal to the 12.3 kN force that puts the body into rotational motion.

Problem 9.

There is a heavy box (W = 3 kip) on a ramp (surface g-g). A force, P, is applied as shown.

Part (a). Solve for Pmin, the minimum pull required to prevent the box from sliding down the ramp.

Part (b). Then, solve for Pmax, the maximum pull that can be applied before the box begins to slide up the ramp.

Hint: remember that the force of friction must be drawn in the opposite direction of the impending motion.

Answer intentionally not supplied.

Problem 10.

This is a CHALLENGE PROBLEM (not because of the Statics, but because of the algebra).

You're cooking dinner and want to video chat with your Mom.

You prop up your phone on the ceramic tile and ceramic counter in the kitchen, but the coefficient of friction is low, and the phone slides down the wall until it's laying on the counter.

You do several experiments until you discover that a 60 degree angle is sufficient for static equilibrium. In other words, a 59 degree angle would not be in static equilibrium.

You decide it would be fun to calculate the coefficient of static friction between your phone and the polished ceramic tile. What is it? Note that both planes are at the verge of slipping (set the shear force at both planes equal to the maximum friction force).

Answer: 0.268 (this value is high enough that we can deduce that the phone is likely enclosed in a case of some sort)

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