Friction

This problem set deals with friction.

What always results from two moving surfaces moving against each other?

Surface wear and heat generation.

Dry friction depends on the force that presses two surfaces together and on (c.) the type of material of which the two surfaces are made.

A wooden box weighing 150 lb slides along a horizontal surface of wood. The coefficient of sliding friction for wood on wood is 0.4. What force is needed to keep the box sliding at a constant speed?

Knowns	Unknowns	Equations/Remarks
N = w = 150 lb m_k = 0.4	F_k = ?	F_k = m_kN
F = m_sN = (0.4)(150 lb) F = 60 lbs		Substitute in the known values and calculate the answer.

Kinetic friction is always less than static friction between two surfaces.

A large metal plate weighs 300 N. The coefficient of kinetic friction for this plate moving on wood is 0.5. Find the force needed to start the plate moving on the wood surface.

Knowns	Unknowns	Equations/Remarks
N = w = 300 N m_k = 0.5	F_s = ?	F_k = m_kN
F_k = m_kN = (0.5)(300 N) = 150 N F_s must be greater than F_k since m_s is always greater than m_k.		Find the force needed to keep the plate moving.

Dry friction between two surfaces is reduced in mechanical systems by lubrication and the use of roller surfaces.

List three specific examples where friction is an undesirable form of resistance. (These are from the book. There are many more examples)

overheated piston rings
electric motors
damage to furniture dragging it across the floor

List three specific examples where friction is desirable and its presence is used to make a device work. (These are from the book. There are many more examples)

clutches
drive belts
friction drives

A wooden box weighing 50 lb rests on a flat table made of wood. Find the horizontal force to overcome static friction and just cause the box to slide (m_s = 0.7). How much force is required to just keep it moving at constant speed?

Knowns	Unknowns	Equations/Remarks
N = w = 50 lb m_s = 0.7 m_k = 0.3	F_s = ? F_k = ?	F_s = m_sN F_k = m_kN
F_s = m_sN = (0.7)(50) F_s = 35 lb F_k = m_kN = (0.3)(50 lb) F_k = 15 lb		Substitute in the known values and calculate the answers.

10.

A 45 lb leather covered block rests on a horizontal surface of wood.

How much force is required to start the block moving if m_s = 0.5?
How much force is required to keep the block moving at constant speed?

Knowns	Unknowns	Equations/Remarks
N = w = 45 lb m_s = 0.5 m_k = 0.4	a. F_s = ? b. F_k = ?	F_s = m_sN F_k = m_kN
a. F_s = m_sN = (0.5)(45 lb) = 22.5 lb b. F_k = m_kN = (0.4)(45 lb) = 18 lb		Substitute in the known values and calculate the answers.

11.

A steel block with a mass of 13.5 kg sits on a horizontal steel surface.

What is the normal force on the block?
How much force is required to start the block moving if m_s = 0.15?
How much force is required to keep the block moving at constant speed?

Knowns

Unknowns

Equations/Remarks

m = 13.5 kg
m_s = 0.15
m_k = 0.09
g = 9.8 N/kg (another way of writing m/s² that makes canceling units easier)
N = w

a. N = ?
b. F_s = ?
c. F_k = ?

w = mg
F_s = m_sN
F_k = m_kN

a. N = w = mg
N = (13.5 kg)(9.8 N/kg) = 132.3 N

b. F_s = m_sN = (0.15)(132.3 N) = 19.8 N

c. F_k = m_kN = (0.09)(132.3 N) = 11.9 N

Find the normal force, then substitute in the known values and calculate the answers.

12.

A wooden chest is placed on a horizontal wooden floor. The coefficient of static friction is 0.7 and the coefficient of sliding friction is 0.4. The chest weighs 100 lb. In each case below, find the force of friction (magnitude and direction), the net force on the block and describe the motion of the block. (Forces are positive to the right and negative to the left)

A force of 50 lb to the right is applied to the chest.
A force of 80 lb to the left is applied to the chest.
The chest is already moving to the right at constant speed when a force of 50 lb is applied to the right.

Knowns

Unknowns

Equations/Remarks

N = w = 100 lb
m_s = 0.7
m_k = 0.4
a. F = 50 lb to the right
b. F = 80 lb the the left
c. F = 50 lb to the right (already moving)

F_f = ?
F_net = ?
motion = ?

F_s = m_sN
F_k = m_kN
F_net = F + F_f

a. F_s = m_sN = (0.7)(100 lb) = 70 lb; F<F_f therefore F_f = –F
Ff = 50 lb to the left.
Fnet = F + F_f = (50 lb) + (–50 lb) = 0
Since the applied force is not greater than the static force of friction, the block does not move.

b. F>F_f
F_f = m_kN = (0.4)(100 lb) = 40 lb to the right
Fnet = F + F_f = (–80 lb) + (40 lb) = –40 lb (negative means to the left)
The block accelerates to the left since the net force is non-zero and to the left.

c. The block is already moving to the right, so F_f = 40 lb to the left (–40 lb).
Fnet = F + F_f = (50 lb) + (–40 lb) = 10 lb (positive means to the right.)
The block accelerates to the right since the net force is non-zero and to the right.

First check to see if the applied force will start the block moving if it is not already doing so. In the first case the force is less than that necessary to start the block moving and the force of friction is equal to the applied force, but in the opposite direction. In the second case the applied force is large enought to start the block moving. The kinetic force of friction is 40 lbs to the right, leaving a net force of 40 lbs to the left to accelerate the block to the left. In the third case the block is already moving, so the force of friction is 40 lbs to the left. The applied force is 50 lbs to the right, resulting in a net force of 10 lbs to the right that accelerates the block to the right.

13.

A force of 150 N to the right is applied to a 20 kg block of concrete resting on a level rubber surface.

If the block is originally at rest will it start to move (m_s = 0.9)?
Find the force of friction on the block.
If the block is already moving, find the magnitude and direction of its acceleration (note: a = F_net/m)

Knowns

Unknowns

Equations/Remarks

F = 150 N to the right
m = 20 kg
m_s = 0.9
m_k = 0.7
g = 9.8 N/kg
N = w
Forces to the right are positive, those to the left are negative

a. will it move?
b. F_f = ?
c. a = ?

F_s = m_sN
w = mg
F_k = m_kN
F_net = F + F_f
a = F_net/m

a. N = w = mg = (20 kg)(9.8 N/kg) = 196 N
F_s = m_sN = (0.9)(196 N) = 176.4 N

The force is not great enough to start the block moving

b. Therefore F_f = 150 N to the left.

c. F_f = F_k = m_kN = (0.7)(196 N) = 137.2 N
F_net = F + F_f = (150 N) + (–137.2 N) = 12.8 N

a = F_net/m = (12.8 kg·m/s²)/(20 kg)
a = 0.64 N to the right

First find the normal force, then find the amount of force needed to start the block moving. The applied force is less than the needed force, so the block does not move and the force of friction is equal in magnitude but in the opposite direction. For part (c.) find the kinetic force of friction since the block is moving, then find the net force acting on the block. The net force, along with the mass of the block, is used to find the acceleration of the block.

14.

Describe the force of friction and the net force in each case below:

A 200 lb block of wood sliding at constant speed along a level wood floor.
An 80 lb force is applied to a 100 lb block of concrete resting on a level steel surface.
A 7.0 kg block of steel is pushed at a constant speed along a level steel surface.
An additional force of 8.0 N is applied to a 9.0 kg block of steel which is already moving at constant velocity along a level steel floor. The force is applied in the direction of motion.

Knowns

Unknowns

Equations/Remarks

a. moving at constant speed: N = 200 lb; m_k = 0.3 (wood on wood)
b. not initially moving: F = 80 lb; N = 100 lb; m_s = 0.9 (concrete on steel)
c. moving at constant speed: m = 7.0 kg; m_k = 0.09 (steel on steel)
d. moving at constant speed: F_add = 8.0 N; m = 9.0 kg; m_k = 0.09 (steel on steel)

g = 9.8 N/kg
N = w

F_f = ?
F_net = ?

F_s = m_sN
F_k = m_kN
F_net = F + F_f
w = mg

a. F_f = F_k = m_kN = (0.3)(200 lb) = 60 lb opposite the direction of motion
F_net = 0 since the speed is constant

b. F_s = m_sN = (0.9)(100 lb) = 90 lb
The force applied is less than that, so the block doesn't move, F_f is 80 in the opposite direction of the applied force, and F_net is zero.

c. N = w = mg
F_f = F_k = m_kN = m_kmg (0.09)(7.0 kg)(9.8 N/kg) = 6.17 N opposite to the direction of motion
F_net = 0 since the speed is constant

d. N = w = mg
F_f = F_k = m_kN = m_kmg (0.09)(9.0 kg)(9.8 N/kg) = 7.94 N opposite to the direction of motion
The additional force is over and above the force that is needed to keep the block moving at a constant speed, so F_net = 8.0 N in the direction of motion.

First found out if the applied force is enough to start the block moving if initially at rest. If the block is at rest or moving at a constant speed then the net force is zero.