1. Solve friction problems
Nuffield Free-Standing Mathematics Activity
Solve friction problems

2. What forces are acting on the sledge?
What force is making the suitcases accelerate?

3. The friction model Before 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

4. Friction problems Example If m = 0.4, will the box move?
5 kg
15N
If m = 0.4, will the box move?
Think about What is the smallest force that will make the box slide along the table?
Solution R 5g
Vertical forces:
R = 5g F
15N
Maximum possible friction
FMAX = m R
= 0.4  5g
= 19.6 N
The pushing force is less than 19.6 N
where g = 9.8 ms–2
The box will not move

5. Friction problems Example
If the package is on the point of moving, find m.
smooth
400 grams
Think about What forces are acting on the package?
Vertical forces:
R = 0.4g
200 grams
Solution
On the point of moving
F = m R
= m  0.4g R
0.4g
F = T
As the pulley is smooth
T = 0.2g F T
m  0.4g = 0.2g
0.2g
0.4g 1 2 =
m =
where g = 9.8 m s–2

6. More difficult friction problems
may require the use of …
Think about Why does the friction model allow the use of these equations?
Newton’s Second Law
Resultant force
= mass  acceleration
where the force is in newtons, mass in kg, and acceleration in m s–2
(u + v)t 2
s =
Equations of motion in a straight line with constant acceleration
s = ut at2 1 2
v = u + at
v2 = u2 + 2as
where u is the initial velocity, v is the final velocity, a is the acceleration, t is the time and s is the displacement

7. More difficult friction problems
Example
The car brakes sharply then skids.
If m = 0.8, find
a the deceleration
1.2 tonnes
b the distance travelled in coming to rest
Solution
Think about What is the friction when the car is skidding? R
1200g a
Vertical forces:
R = 1200g
F = m R
= 0.8  1200g
= 9408 N
Newton’s Second Law gives: F
–9408 = 1200a
a = –7.84 m s–2
Think about Which equation can be used to find the distance the car travels as it comes to a halt? b
v2 = u2 + 2as
02 =  7.84s
where g = 9.8 m s–2
15.68s = 400
s = 25.5 metres

8. Solve friction problems
Reflect on your work
When can you use F = mR?
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?