© Nuffield Foundation 2011

Nuffield Free-Standing Mathematics Activity

Solve friction problems

What forces are acting on the sledge?

What force is making the suitcases accelerate?

The friction modelBefore sliding occurs …

Friction is just sufficient to maintain equilibrium and prevent motion

F < FMAX

On the point of sliding and when sliding occurs …

F = mR where F is the friction, R is the normal contact force

and m is a constant called the coefficient of friction

Friction problems

F 15N

15NIf = 0.4, will the box move?

Solution

R

5g

Vertical forces: R = 5g

where g = 9.8 ms–2

Maximum possible friction

FMAX = R

The pushing force is less than 19.6 N

Example

= 0.4 5g = 19.6 N

The box will not move

5 kg

Think about What is the smallest force that will make the box slide along the table?

smoothIf the package is on the point of moving, find . 400 grams

Friction problemsExample

200 grams

F T

Solution

R

0.4g

Vertical forces: R = 0.4g

where g = 9.8 m s–2

As the pulley is smooth

F = R

0.4g = 0.2g

=

On the point of moving

T = 0.2g

F = T

0.2g0.4g

12=

= 0.4g

Think about What forces are acting on the package?

More difficult friction problems

Resultant force

Newton’s Second Law

= mass accelerationwhere the force is in newtons, mass in kg, and acceleration in m s–2

Equations of motion in a straight line with constant acceleration

v = u + at s = ut + at212 v2 = u2 + 2as

(u + v)t2

s =

where u is the initial velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacement

may require the use of … Think about Why does the friction model allow the use of these equations?

More difficult friction problems

F

20 m s–1

The car brakes sharply then skids.

SolutionR

1200g

Vertical forces: R = 1200g

where g = 9.8 m s–2

Newton’s Second Law gives:

Example

= 9408 N

–9408 = 1200a

1.2 tonnesIf = 0.8, find a the decelerationb the distance travelled in coming to rest

F = R= 0.8 1200g

a = –7.84 m s–2

v2 = u2 + 2as 02 = 202 - 2 7.84s

15.68s = 400 s = 25.5 metres

a

b

Think about What is the friction when the car is skidding?

Think about Which equation can be used to find the distance the car travels as it comes to a halt?

• When can you use F = R?• How does the friction model allow you to use F = ma

and the constant acceleration equations to solve problems?

• Can you think of other situations when friction prevents an object from moving?

• Can you think of other situations when friction causes an object to accelerate?

Reflect on your work

Solve friction problems