Imagine repeating the experiment, only now with a second brick on top of the first. This doubles the normal force and it also doubles the maximum force of static friction. Thus, the maximum force is proportional to the magnitude of the normal force; that is, 6-3 The constant of proportionality is called a, (pronounced "mew sub s"), the coeffi- clent of static friction. Notice that a, is dimensionless, just like a. Typical values are given in Table 6-1. In most cases, a, is greater than a, indicating that the force of static friction is greater than the force of kinetic friction. In some cases, like rubber in contact with dry concrete, a, is greater than 1. Two additional experimental results regarding static friction are important: (1) Static friction, like kinetic friction, is independent of the area of contact. (li) The force of static friction is parallel to the surface of contact, and opposite to the direction the object would move if there were no friction. All of these observations are summarized in the following rules of thumb: Rules of Thumb for Static Friction The force of static friction between two surfaces has the following properties: 1. It takes on any value between zero and the maximum possible force of static friction, OSASAN 2. It is independent of the area of contact between the surfaces. 3. It is parallel to the surface of contact, and in the direction that opposes relative motion. The next Example presents a practical method of determining the coefficient of static friction in a real-world system.Calculate the normal force using Newton's Second Law: what pulls up equals what pulls down. The normal force should be greater than the weight of the crate because of the additional vertical component pushing down on the crate. Calculate the maximum force of static friction using Eq. 6-3. Your Answer: Question 8 (1 point] To move a large crate across a rough floor, you push on it with a force F at an angle of 210 below the horizontal, as shown in the figure. Find the force necessary to start the crate moving, in newtons, given that the mass of the crate is 50.0 kg and the coefficient of static friction between the crate and the floor is 0.584. Decompose the applied force into the hon'zontal and vertical components like we learned earlier in the course (Chapter 3}. Calculate the normal force using Ne wton's Second Law: what pulls up equals what pulls down. The normal force should be greater than the weight of the crate because of the additional vertical component pushing down on the crate