Friction

PHYSICS 1 MIDTERM EXAM REVIEW #2 1. A 6.0-N force and an 8.0-N force act concurrently on a point. As the angle between these forces increases from 0° to 90°, the magnitude of their resultant (1) decreases (2) increases (3) remains the same 2. A car increases its speed from 9.6 m/s to 11.2 m/s in 4.0 s. The average acceleration of the car during this 4.0-second interval is (1) 0.40 m/s2 (3) 2.8 m/s2 (2) 2.4 m/s2 (4) 5.2 m/s2 3. What is the speed of a 2.5-kilogram mass after

Theoretical Background General This chapter summarizes all the theoretical background that is necessary in order to understand the pipeline walking phenomenon. This includes the theory of pipe-soil interaction, the parameters of thermal expansion design as well as detailed definition of the pipeline walking, description of its mechanism and common mitigation methods. The theoretical background behind pipeline walking phenomenon is quite vast as it includes aspects of structural, geotechnical

be the smoothest surface in the experiment - only had a friction coefficient value of 0.2401, whilst surface C (wood) - considered to be the roughest surface - indicated a much higher coefficient value of 0.432. These results were found to correspond with almost all surfaces tested, with surface D (carpet) found to have the second lowest coefficient value of 0.2477, and surface E (bitumen), being found to have a much higher static friction coefficient value - similar to that of wood - of 0.4233

A block of mass m = 5.8 kg is pulled up a ( = 25° incline as in figure with a force of magnitude F = 32 N. (a) Find the acceleration of the block if the incline is frictionless. (b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.10. [(a) 1.4 m/s2, (b) 0.49 m/s2] 11. A 276-kg glider is being pulled by a 1950-kg airplane along a horizontal runway with an acceleration of a = 2.20 m/s2 to the right as in figure. Find (a) the thrust

coefficient of friction, or the μ, is a scalar value which is equivalent to the ratio of the force of friction (FF) between two surfaces to the normal force (FN) which pushes two surfaces together. Each set of surfaces has a different μ, as it depends on the particular materials. Basically, the μ increases as it becomes more difficult for objects to move against each other. When a large amount of force is required to move two surfaces against each other (and therefore a large amount of friction is produced)

the velocity of the parachutist. Water drag is acted upon swimmers at all times. Hydrodynamics show that materials of the swimmers closely affect drag. Swimmers wear sleek swim caps, shave their body, and wear tightly fitted swimsuits all to reduce friction. Skill level and technique of a swimmer also affect drag resistance. “Water is 800 times denser than air, drag forces

of the string and g represents the acceleration due to gravity. There are some limitations to this mathematical model. The model assumes that the stopper is spinning at a constant velocity and does not take into account the force of friction or air resistance. Friction and air resistance can affect the experimental data because it can create a drag which slows down the string and can change the orbital period. Procedure: To begin the experiment, we measured the masses of the two stoppers and the

Microelectromechanical Systems (MEMS) are systems that are designed on a micro metre scale and have become more popular as the demand for devices to get smaller has increased. The main uses of these systems are for sensors, such as accelerometers and gyroscopes and other such devices like microscopy and inkjet nozzles for example. There are many materials that can be used for MEMS as the cost of the material is almost eradicated due to the micro size of the systems being produced. This brings materials

on the body must be equalized somehow. The main force is the one caused by extension of the back leg into the ground (a). This is held in check by dropping the torso into the stance and allowing the friction on the front foot from the ground, μb + ~(1/3)μmg where μ is the coefficient of static friction, and b is the force of the body attempting to move forward due to a, and the tensing of the front leg quads to keep the stance in place. If the stance is correct, this force is easily released to throw

F; double fn, ft, vt; double friction; // compute total spring and gravity force XToxv(x, p, l, q, X); F = pl.m * g - sp.k * ((x - x0).norm() - sp.l0) * (x - x0).normalize(); // if in resting contact compute the friction force and support force fn = F * pl.n; // force component normal to the surface if(resting && fn < 0){ if(stopped){ // static friction ft = F * pl.tangent; // force component tangent to the surface friction = -Min(Abs(pl.mus * fn), Abs(ft))