The glue on a piece of tape can exert forces. Can these forces be a type of simple friction? Explain, considering especially that tape can stick to vertical walls and even to ceilings.

How do you calculate the friction coefficient of a surface with various areas of different frictions?

A 20-kg box of oranges slides from rest down a frictionless incline from a height of 5.0 m. A constant frictional force, introduced at point A, brings the block to rest at point B, 19 m to the right of point A. What is the coefficient of kinetic friction, Hk, of the surface from A to B? 5.0 m A 19 m 0.11 0.52 0.26 0.33 0.47

A ski that is placed on snow will stick to the snow. However, when the ski is moved along the snow, the rubbing warms and partially melts the snow, reducing the coefficient of friction and promoting sliding. Waxing the ski makes it water repellent and reduces friction with the resulting layer of water. A magazine reports that a new type of plastic ski is especially water repellent and that, on a gentle 230 m slope in the Alps, a skier reduced his top-to-bottom time from 69 s with standard skis to 42 s with the new skis. Old skis acceleration = 0.097 m/s2 New skis acceleration = 0.26 m/s2 Assuming a 3.0° slope, compute the coefficient of kinetic friction for each case. Old skis _______ New skis  _______

Many birds can attain very high speeds when diving. Using radar, scientists measured the altitude of a barn swallow in a vertical dive; it dropped 208 m in 3.0 s. The mass of the swallow was estimated to be 0.018 kg, and its cross-section area as 5.6 x 10-4 m2. What was the drag coefficient for this swallow as it dove?

Write Newton's laws for a static system (1) EF, = mg sin 0 – µ̟n = 0 in component form. The gravity force has two components. (2) EF, = n - mg cos 0 = 0 Rearrange Equation (2) to get an n = mg cos e expression for the normal force n. Substitute the expression for n into EF, = mg sin 0 – µ̟mgcos 0 = 0 → tan 0 =µ, Equation (1) and solve for tan 0. Apply the inverse tangent function to tan 0 = 0.350 → 0 = tan 1 (0.350) = 19.3° get the answer. LEARN MORE REMARKS It's interesting that the final result depends only on the coefficient of static friction. Notice also how similar Equations (1) and (2) are to the equations developed in previous problems. Recognizing such patterns is key to solving problems successfully. QUESTION A larger static friction constant would result in a: (Select all that apply.) O larger component of normal force at the maximum angle. O larger component of gravitational force along the ramp at the maximum angle. smaller component of gravitational force along the ramp…

A rope one meter long rests on a table where the coefficient of friction is 0.50. What maximum length of rope can hang over the side before the whole rope will slip off?

Suppose there is a box of books with a mass of 10 kg resting on a flat horizontal table, and that the coefficient of static friction between the table and the box is 0.5. At a minimum, how much force do you have to apply to lift the box off the table vertically? At a minimum, how much force do you have to apply to shift the box horizontally across the table? Explain your answers.

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.

i'm doing chapter 4 problem 12p in the college physics 11th edition text book. On bartleby it says that "total force along the x axis is  fx= f1sin30degrees - f2sin10degrees" my question is that wouldnt it be the cosine since it is the x axis and not sin?? i put a image below