Question

Assessment

Friction

Friction resists motion. If an object is stationary, friction tries 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 opposite the direction of motion. When you use Newton's law to sum the forces, there will be one more term in the equation.

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

μ is the "coefficient of friction", a number (theat you generally look up on a table) that tells how hard it is to slide two objects in contact.FN is the "Normal Force", or how hard the surface pushes up on the object. Generally you find the Normal force by summing all the forces in the y-direction and solving for FN. Most often however, if there aren't any forces acting in the y-direction other than gravity and the normal force, then for horizontal surfaces, FN = mg. Therefore, Ff = μmgfor inclined plane surfaces, 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 no y-direction forces except for gravity and the normal force

123

Works whenever the surface is horizontal and there are no y-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 of friction is 0.2. A 24 N force pulls the box uphill. Find the acceleration.

Your Answer:

Question 6 options:

Answer

Question 7 (1 point)

A 2 kg box sits on a ramp where the coefficient of friction is 0.4. Find the angle that will cause the box to slide downhill at constant velocity.

Hints:

constant velocity means a = 0 sum the forces (downhill pull and friction) 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 of friction is 0.4. A string runs uphill over a pulley and back down to a hanging mass of 7 kg. Assuming the box on the ramp is pulled uphill by the weight of the hanging mass, find the acceleration.

Your Answer:

Question 8 options:

Answer

Question 9 (1 point)

A 2 kg box sits on a ramp of 14 degrees where the coefficient of friction is 0.2. A string runs uphill over a pulley and back down to a hanging mass of 9 kg. Assuming the box on the ramp is pulled uphill by the weight of the hanging mass, find the acceleration.

Your Answer:

Question 9 options:

Answer

Submit Assessment0 of 9 questions saved

Answer 1

Works whenever the surface is horizontal and there are no y-direction forces except for gravity and the normal force

umg

________________

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

u * F_N

___________________

Works whenever the surface is an inclined plane and there are no y-direction forces except for gravity and the normal force

u * mg * cos

_____________________________

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

Fnet = ma

mgsin + umgcos - F = ma

3 * 9.8 * sin 6 + 0.2* 3 * 9.8 * cos 6 - 24 = 3 * a

so,

a = 5.026 m/s²

___________________

A 2 kg box sits on a ramp where the coefficient of friction is 0.4. Find the angle that will cause the box to slide downhill at constant velocity.

= arctan (u)

= arctan ( 0.4)

= 21.8 degree

_____________________

A 2 kg box sits on a ramp of 14 degrees where the coefficient of friction is 0.4. A string runs uphill over a pulley and back down to a hanging mass of 7 kg

a = m_hg - mgsin - umgcos / m_h + m

a = 7 * 9.8 - 2 * 9.8 * sin 14 - 0.4 * 2 * 9.8 * cos 14 / ( 7 + 2)

a = 6.25 m/s²

_____________________

A 2 kg box sits on a ramp of 14 degrees where the coefficient of friction is 0.2. A string runs uphill over a pulley and back down to a hanging mass of 9 kg. Assuming the box on the ramp is pulled uphill by the weight of the hanging mass

a = m_hg - mgsin - umgcos / m_h + m

a = 7.24 m/s²