#  Friction opposes motion  Friction is dependent on the texture of the surfaces  Friction is dependent on normal force motionfriction.

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 Friction opposes motion  Friction is dependent on the texture of the surfaces  Friction is dependent on normal force motionfriction

 Friction opposes motion  Friction is dependent on the texture of the surfaces  Friction is dependent on normal force F fr =  F n μ is called coefficient of friction  has no units  depends on characteristics of both surfaces  Higher   Higher  rougher surface / more frictionmotionfriction Note: Friction does NOT depend on the surface area of contact

The coefficient of friction is different for when an object is at rest and when it is moving. μ s = coefficient of static friction (object at rest) μ k = coefficient of kinetic friction (object moving) Static friction is greater than kinetic friction - its harder to start an object moving than it is to keep it moving.

A 28 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor. What do we do first?

A 28 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor. Remember our strategy: 1) Draw a free body diagram 2) Identify all variables 3) Identify relevant equations 4) Solve! 5) Check your work!

A 28 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor. μ s = 0.27

A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s 2 along a horizontal surface. a. How large is the frictional force? b. What is the coefficient of friction?

A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s 2 along a horizontal surface. a. How large is the frictional force? b. What is the coefficient of friction? F f = 10 N μ k = 0.2

2) A 12 kg suitcase is pushed with a force of 38 N to the left. If the coefficient of kinetic friction between the suitcase and the floor is 0.3, how far will the suitcase move after 5 sec? 2.8 m

FOLLOW THE STEPS! Make a plan before you plug in numbers! Be able to explain your reasoning! 1) A 30 kg crate requires a 53 N force to keep it moving at 1 m/s. Find the coefficient of kinetic friction. 2) You need to move a 105-kg sofa to a different location in the room. It takes a 403-N force to start the sofa moving. What is the coefficient of static friction between the sofa and the carpet?  k = 0.18

Question: How does the weight of a person in an elevator depend on the motion of that elevator? What will the scale show if the elevator is 1.at rest or moving with constant speed 2.speeding up 3.slowing down Newton’s 3. law: Force with which the person acts on the scale (reading of the scale) is equal to the normal force on the person. So, if we find normal force we know the reading of the scale, so called APPARENT WEIGHT

Let’s assume that elevator is moving upward, and let this be positive direction. 1. draw free body diagram 2. apply Newton’s 2. law : F net = ma F n – mg = ma = 0 → F n = mg apparent weight = weight apparent weight = weight+ 1. elevator is at rest or moving with constant speed 2. elevator is speeding up: a is positive F n – mg = ma → F n = mg + ma apparent weight > weight apparent weight > weight the scale would show more, and you would feel heavier 3. elevator is slowing down: a is negative F n – mg = - ma → F n = mg - ma apparent weight < weight apparent weight < weight the scale would show less, and you would feel lighter FnFnFnFn FnFnFnFn FnFnFnFn

You are riding in an elevator holding a spring scale with a 1kg mass suspended from it. You look at the scale and see that it reads 9.3 N. What, if anything, can you conclude about the elevator’s motion at that time?

1) Solve the elevator problem for an elevator traveling downward. YOU MUST EXPLAIN YOUR REASONING USING NEWTON’S 2 nd LAW 2) Write down 2 things you learned friction problems.

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