15. Dry Static Friction
• Friction is a force that exists between any two object in contact. This can sometimes work against the engineer, other times it can be of great advantage.
• This natural phenomenon explains the resistance of one object to slide across another when they have common surfaces in contact.
• It is primarily the result of surface roughness, material properties, and if the object is moving.
• The graph of applied load versus friction helps illustrate the nature of friction. Notice that while the force is static, the force increases linearly up to the limit. After the object begins moving the force can be approximated with a constant value, using the dynamic coefficient of friction. Note that dynamic friction is shown to be lower that the maximum static friction.
• The basic assumptions that we will use are,
1. the maximum friction force is proportional to the normal force
2. the maximum friction force is not proportional to the area of contact
3. the static friction force is always higher than the dynamic friction force
4. the kinetic friction force is independent of velocity
• A couple of the major applications for friction calculations is the determination if an object will slip or tip. The following problem shows a typical application, ([Hibbeler, 1992], prob 8-8, pg. )
• Consider the simple tip/slip problem below,
[woking model file]
• The general approach to slip-tip problem is,
1. Find the center of gravity for the object.
2. Determine which corner the object is most likely to tip (as if the corner is a pin joint). Sum the moments about the corner. If the sum of moments is equal to zero to block is about to tip. If not equal to zero look at the resulting moment to see if it will cause motion about the corner.
3. Find the component of the gravity and any other non-friction forces acting perpendicular to the surface of contact. Find the components of applied forces acting parallel to the plane of contact.
4. Compare the actual parallel component to the maximum friction force possible. The the resultant is larger than the maximum the block will slip.
Problem 15.1 We conduct an experiment using the 10 kg. block below on a slope that is being slowly tilted. The block tips over at 20°, and then stops moving, but then it starts to slip at 40°. What is the height ‘h’, and the coefficient of friction?
Answer 15.1 coeff. = 0.84, h = 137 cm
Beer, F.P., Johnson, E.R., Statics & Mechanics of Materials, McGraw-Hill, 1992.
Hibbeler, R.C., Engineering Mechanics: Statics and Dynamics, 6th edition, MacMillan Publishing Co., New York, USA, 1992.
15.1 Applications of Friction
• When dealing with these problems the direction of friction forces must be assigned with care. If the directions are selected backwards, the solutions will be incorrect.
• Before assigning friction forces the impending motion should be analyzed. To do this think of the possible cases that might cause the bodies to start moving: do not assume that all friction surfaces must go into motion. At times this approach may mean that multiple solutions will have to be done to solve a problem.
• The general methods to be followed in these problems involves,
1. Examine the problem to determine impending motion for each individual object, and the overall system. There may be one or more possible cases, each will require a separate solution.
2. Based upon the assumed motion at the points of contact, drawn on friction forces that oppose the motion. Also draw on normal forces.
3. Solve the problem using normal statics (but avoid using sums of moments for friction forces when they don’t act on a clear point).
4. Examine the solution (and compare to others) for anomalies such as normal forces that separate friction surfaces. This will help determine problems, and to eliminate unreasonable solutions.
• Wedges are a useful engineering tool, and the approach used for wedges also finds its way into other engineering applications.
• A good rule to stick to is that when a wedge is in use, the forces on the faces will both be in the same direction. That is either towards, or away from the point of the wedge.
• When solving friction problems we look for friction that is about to let go and start slipping. Keep in mind that not all surfaces will slip, this should be verified after the solution. For all surfaces that slip the friction force will be at the maximum value.
• The example below shows how to deal with a multiple wedge problem.
Problem 15.2 ([Hibbeler, 1992], prob 8-55, pg. ) The two wedges are stacked as shown, and a load is applied. What is the minimum force P required to pull the bottom wedge out?
[working model file]
Problem 15.3 A vertical force, F, of 500N acts at one end of a bracket while a wedge is pushed against the other end, as depicted in the diagram below. Given that the weight of the wedge and the bracket can be neglected and that the coefficient of static friction for the contacting surfaces of the wedge is 0.2, determine the horizontal force P required to push the wedge.
Answer 15.3 216N
188.8.131.52 - Problems
Problem 15.4 The angled bar below has two forces applied that are tending to push it in a counterclockwise direction. These forces are resisted by a wedge that is kept in place by friction (the coefficient of friction is 0.20). determine the force P that is required to pull the block out.
[working model file]
Problem 15.5 We are designing a firing mechanism for a new gun that uses two identical rails that are pressed together to accelerate a projectile. In the figure below we see the two rails at an angle before firing begins. When firing begins, the force ‘F’ will be applied, overcoming the coefficient of friction of 0.05. The length of the projectile is negligible, but it has a wedge shape that matches the rails before firing. What is the initial force ‘F’ that must be applied to the rails before the shot begins to move?
Problem 15.6 What force P will have to be applied to the wedge A to force the blocks B and C apart? You can assume that the coefficient of friction is 0.4 at all points of contact.
Answer 15.6 P = 0.854 MN
Problem 15.7 Find the force required to pull out the wedge below.
Problem 15.8 Find the force required to pull out the wedge below.
184.108.40.206 - References
15.1 Hibbeler, R.C., Engineering Mechanics: Statics and Dynamics, 6th edition, MacMillan Publishing Co., New York, USA, 1992.
15.1.2 Belt Friction
• Belts are a common tool for transmission of forces, motions and velocities.
• If we have a flat belt, it primarily depends on friction to hold it in place.
• The basic rules of static friction still apply for local friction between the belt and the drum, but over the length of the belt the effective normal force changes.
• If we consider one element of the belt we can see an element of friction and a differential of tension.
• An example of belt friction is given below (8.123, pg. 40, Beer and Johnston). In this problem the upper drum is moving slowly (this means the belt sticks with static friction), the lower drum allows the belt to slide (the belt slides with dynamic friction). We need to find the force W that will balance the 150lb load on the other side.
Problem 15.9 A belt is wrapped about drums A and B, both have a radius of 3”. How much force F will have to be applied to lift a weight of 1 lb. if the drums are not moving, and the coefficient of static friction is 0.2?
Answer 15.9 2.57 lb
• V-belts use the same principle as flat belts, except the friction is increased by the angle of the sides.
220.127.116.11 - References
15.2 Beer, F.P., Johnson, E.R., Statics & Mechanics of Materials, McGraw-Hill, 1992.
15.3 Soustas-Little, R.W. and Inman, D.J., Engineering Mechanics Statics, Prentice-Hall, 1997.