© Nuffield Foundation 2011 Nuffield Free-Standing Mathematics Activity Parking permits.
© Nuffield Foundation 2011 Nuffield Free-Standing Mathematics Activity Solve friction problems.
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Transcript of © Nuffield Foundation 2011 Nuffield Free-Standing Mathematics Activity Solve friction problems.
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