Statics

Friction is a force that lies in the surface between two objects. For systems in static equilibrium, the friction vector points in the direction that opposes impending motion.

We will use either Ff or Ffr as a symbol for friction. Both are commonly used.

❏ Common symbols for the force of friction

🟦 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
the magnitude of compressive force transferred over the contact surface

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.

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 just means that it's harder to initiate motion than maintain motion, as far as friction is concerned.

The formula for the limiting (maximum) friction force shows that friction resistance is a function of the normal (compressive) force transferred over a surface:

❏ Static and kinetic friction

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

Be aware that there are other types of friction (drag friction, fluid friction, etc.) that are not in the scope of Statics.

🟦 7.3 When do engineers neglect or include friction?

In engineering practice -- at least in Statics problems -- friction is generally neglected. This is because it (mostly) tends to aid static equilibrium; it's a stabilizing mechanism.

Engineers generally design systems (in static equilibrium) that do not rely on friction for static equilibrium. Whatever friction is there just makes the system a little more redundant (more safe).

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

Please keep in mind that friction is often a critically important parameter in dynamics problems. If you are engineering a system that needs to be set in motion, then your design must overcome the force of friction. So, in that type of scenario, you would take it into account. That's not in the scope of Statics, though.

❏ 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.

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

This is the second time you've seen the image of a coffee mug subjected to a horizontal force. This section is all about the half arrow.

A shear force, such as the force of friction, lies in a plane or a surface.

In a 3D view, when you can see the surface, you would want to draw the regular style of arrow (← or →, etc.). This is depicted in the main part of the image.

We generally prefer to draw in 2D, projecting the geometry onto a 2D plane. This is shown in the top right corner of the image. Since the friction force is coincident with the bottom surface of the coffee mug, it's hard to see it if you draw on top of the body geometry.

In order to easily communicate a shear force (including friction forces), we can offset the vector slightly away from the body. Then, we draw a half arrow (↽ or ⇁ or ↼ or ⇀ or ↿ or ⇃ or ⇂ or ↾). In FBDs, the flag on the half arrow is always on the opposite side of the body, as illustrated here.

Throughout this tutorial, I have decided to use the double-arrow head as a learning tool for you, even on the 2D drawings. It would be equally accurate to depict them as half arrows.

❏ Half arrow example image

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.

🟦 7.5 FBDs that include friction and 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, the N3L pairs for friction force are described in terms of direction. If one is rightwards, the other is leftwards.

You can 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.

In friction problems, you must always remember to draw the friction vector in the direction that opposes impending motion (of that free-body, relative to everything else).

🟦 7.6 Normal force can be expressed as a line load 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 will be called a bearing stress.

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

❏ Scenario for a constant line load

Now, consider a body that looks like a connected rectangle and triangle.

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 right FBD is equivalent to the left FBD. The force, N, has been converted into a linear line load function. To make the two systems equivalent, we must use the equations of static equivalency.

❏ Scenario for a linear line load

You can use the Lever Law to deduce that the location of the normal force is at the thirdpoint between W1 and W2. You could also do a moment summation to make the same conclusion.

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 FBDs we learned earlier become critically important.

In the left image, notice that the normal and shear forces are shown in relation to their resultant force reaction. If there are 3 forces in the system, it is a concurrent force problem. All 3 vectors must intersect at the same point. Therefore, we can use moment equilibrium to solve for xbar, and force equilibrium to solve for the shear force and normal force required for static equilibrium.

In the right image, we see another valid FBD. The resultant force has been replaced with a line load 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.
Static equilibrium (all E.o.E.) must be satisfied if we want the friction to prevent motion (translation or rotation)

🟦 7.7 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, sqrt (V^2 + N^2)
Ffr,max = the limiting friction force

The following examples will walk you through the basic process for solving slip/slide problems. In these types of problems, you will only use a force equilibrium equation. When we get to the rotation (tipping/overturning) type of problems, you'll use a moment equilibrium equation.

Question 1.

A 50N block lays on rough 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 (slip or slide), or does the friction keep the block in static equilibrium?

Answer 1.

In order to maintain static equilibrium, a shear force of 40N is required at surface d-d. The limiting friction force is equal to (0.5)*(50N) = 25N.

Since the shear force (V=40N) required for equilibrium exceeds the maximum possible friction force (Ffr,max = 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 and pile it on top. It weighs 30N. Is this enough weight to maintain static equilibrium?

Answer 2.

Now, the total normal force at surface d-d is 80N. The limiting friction force is (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 (at least theoretically). But it's at the verge of slipping and translating leftward.

❏ Flipbook: Example 2

Question 3.

A new block (30N) has been placed on top of the 50N block. A 10N horizontal force is applied as shown. Is this system in static equilibrium or in motion?

Answer 3.

We have to check both slip planes. First, we draw a FBD of the top block. We see that a shear force of 10N must be transferred to the lower block if we are to maintain static equilibrium. The normal force transferred at surface e-e is 30N. Therefore, the maximum friction resistance is (30N)*(0.4)=(12N). Since the demand of 10N does not exceed the capacity of 12N, the top plane doesn't slip.

Next, we investigate plane d-d. Here, N is equal to 80N, which makes Ffrmax 40N. We compare the demand (10N of shear force needed) to the resistance (up to 40N of friction force) and conclude that this plane doesn't slip either.

Since we have checked all possible slip planes, we conclude that the block is in static equilibrium.

❏ 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 equilbrium, or does either plane slip?

Answer 4.

First, we check friction plane e-e. We notice that this set-up is identical to the prior example. Since the shear demand is 10N and friction resistance is 12N, that plane does not slip.

Second, we check plane d-d. The net shear force in that plane is (40N) - (10N) = 30N leftwards. The normal force at the plane is (30N) + (50N) = (80N). Therefore the frictional resistance is 0.5*(80N) = 40N. Since the demand is less than the maximum friction force, the system is in static equilibrium.

❏ Example 4

🟦 7.8 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, all you have to do is place the resultant force right at the pivot point, and then solve a moment equilibrium equation.

I like to call the pivot point Point O (for "overturning").

By using an equilibrium equation, you're solving for the exact amount of force that tips the structure. If the push is any less, you're in static equilibrium. If the push is any more, it will tip.

❏ How much of a push tips the bookcase?

🟦 7.9 Thinking about limit states

In friction problems, we need to use a kind of engineering design 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 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 just one limit state reveals motion, then the system is not in static equilibrium.

➜ Practice Problems

Problem 1.

Do these blocks translate or remain in static equilibrium?

You have to check two limit states.

Check slipping at surface e-e by constructing a FBD of the top block.

Check slipping at surface d-d by constructing a FBD of both blocks together.

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

Problem 2.

Mass A weighs 90 pounds and mass 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 equilbrium 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.

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 fully solve these numerically; these are a conceptual problem only. Do use the qualitative equations of equilibrium and definition of friction force to aid your logic.

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 applying force P.

What force, P, initiates motion of the block?

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 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.

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