Assessment

Friction

Friction resists motion. If an object is stationary, frictiontries to keep it from beginning to move. If an object is moving,friction slows it down and tries to stop it.

In all cases all you need to do is add an extra arrow (vector)to your free-body diagram. This new arrow always points oppositethe direction of motion. When you use Newton's law to sum theforces, there will be one more term in the equation.

The magnitude of this new arrow/term is always given by Ff = μFN where

μ is the \"coefficient of friction\", a number (theat yougenerally look up on a table) that tells how hard it is to slidetwo objects in contact.FN is the \"Normal Force\", or how hard thesurface pushes up on the object. Generally you find the Normalforce by summing all the forces in the y-direction and solving forFN. Most often however, if there aren't any forces acting in they-direction other than gravity and the normal force, then forhorizontal surfaces, FN = mg. Therefore, Ff = μmgfor inclined planesurfaces, FN = mg cos θ. Therefore, Ff = μmg cos θ

Question 1 (1 point)

Match the following formulas about calculating friction:

Question 1 options:

123

Always works. FN can be found by summing all the y-dir forces,recognizing that the acceleration in the y-dir is (probably) zero,and solve for FN

123

Works whenever the surface is an inclined plane and there are noy-direction forces except for gravity and the normal force

123

Works whenever the surface is horizontal and there are noy-direction forces except for gravity and the normal force

1.

Ff = μmg

2.

Ff = μmg cos θ

3.

Ff = μ FN

Question 6 (1 point)

A 3 kg box sits on a ramp of 6 degrees where the coefficient offriction is 0.2. A 24 N force pulls the box uphill. Find theacceleration.

Your Answer:

Question 6 options:

Answer

Question 7 (1 point)

A 2 kg box sits on a ramp where the coefficient of friction is0.4. Find the angle that will cause the box to slide downhill atconstant velocity.

Hints:

constant velocity means a = 0 sum the forces (downhill pull andfriction) and solve for θsin θ / cos θ = tan θtake the arctan

Your Answer:

Question 7 options:

Answer

Question 8 (1 point)

A 2 kg box sits on a ramp of 14 degrees where the coefficient offriction is 0.4. A string runs uphill over a pulley and back downto a hanging mass of 7 kg. Assuming the box on the ramp is pulleduphill by the weight of the hanging mass, find theacceleration.

Your Answer:

Question 8 options:

Answer

Question 9 (1 point)

A 2 kg box sits on a ramp of 14 degrees where the coefficient offriction is 0.2. A string runs uphill over a pulley and back downto a hanging mass of 9 kg. Assuming the box on the ramp is pulleduphill by the weight of the hanging mass, find theacceleration.

Your Answer:

Question 9 options:

Answer

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