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Nuffield Free-Standing Mathematics Activity
Solve friction problems
© Nuffield Foundation 2011
What forces are acting on
the sledge?
What force is making the
suitcases accelerate?
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
Friction problems
Example
If m = 0.4, will the box move?
5 kg
15N
Think about
What is the smallest force that will
make the box slide along the table?
Solution
R
Vertical forces: R = 5g
15N
F
Maximum possible friction
FMAX = m R = 0.4  5g = 19.6 N
5g
where g = 9.8 ms–2
The pushing force is less than 19.6 N
The box will not move
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:
Solution
R = 0.4g
On the point of moving
R
F = m R= m  0.4g
F=T
T
F
200 grams
As the pulley is smooth T = 0.2g
m  0.4g = 0.2g
0.4g
where g = 9.8 m s–2
m = 0.2g = 1
0.4g
2
More difficult friction problems
may require the use of …
Newton’s Second Law
Think about Why does the
friction model allow the use
of these equations?
Resultant force = mass  acceleration
where 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 + 1 at2
2
s=
(u + v)t
2
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
More difficult friction problems
Example
The car brakes sharply then skids.
20 m s–1
1.2 tonnes
If m = 0.8, find a the deceleration
b the distance travelled in coming to rest
Solution
Think about What is the friction when
R
a Vertical
R = 1200g
the carforces:
is skidding?
F = m R = 0.8  1200g = 9408 N
Newton’s Second Law gives:
–9408 = 1200a a = –7.84 m s–2
F
1200g
where g = 9.8 m s–2
Think about Which equation can be
2 = distance
find0the
car
b v2used
202 - 2 the
7.84s
= u2 +to2as
travels as it comes to a halt?
15.68s = 400 s = 25.5 metres
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?