Using a model of the sliding process, derive that the static friction force F, between two touching objects does not depend on the touching area of the two objects and can be appropriately expressed as F5P/2, where P is the force pressing the two objects together.
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- A 5.00-kg box is initially at rest on a level surface. The coefficient of static friction between the box and the surface is 0.400. Calculate the maximum force you can apply before the box starts to move, if the force is directed at an angle of 30 degrees from the horizontal, with the vertical component downward. Calculate the maximum force you can apply before the box starts to move, if the force is directed at an angle of 30 degrees from the horizontal, with the vertical component upward.Continuing from problem 8, the force F is increased to 50 N. The data are as follows: a) The object's mass is 4 kg, b) g = 10 m/s². c) Coefficient of the kinetic friction = 0.5 / sqrt(3) and coefficient of the static friction = 1 / sqrt(3). d) Magnitude of the force F = 50 N. 30⁰ F What is the acceleration of the object in m/s²? Round to the nearest decimal.Consider an object at rest on a rough horizontal surface. Which of the following is necessarily correct if we use the model of friction described in the videos and book? The force of friction on the object is smaller than or equal to μg FN, where μs is the coefficient of static friction and FN is the normal force exerted on the object by the surface. The force of friction on the object is proportional to the normal force exerted by the surface on the object. The force of friction on any object opposes motion; since the object is at rest, the force of friction is zero. The force of friction on the object is proportional to the object's weight, i.e. Ff = p.mg, where μs is the coefficient of static friction, m is the mass of the object, and g is acceleration due to Earth's gravity.
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