Lab #5: FRICTION
only 7 more to go
Friction is related to the weight of an object, the heavier an object the more
friction it has.
We talk about static friction and kinetic friction:
static friction is friction an object experiences before it moves
the static friction force is given by:
Ff = s FN
where FN is the normal force and s is called the coefficient of static
friction
kinetic friction is friction an object experiences while it is moving
the kinetic friction force is given by:
Ff = k FN
where FN is the normal force and k is called the coefficient of kinetic
friction
What is the frictional force?
FN
In this situation when we write Newton’s second
law in the y-direction we get: FN – mg = 0
So,
Fn = mg,
50 kg
s = 0.7
Ff = S FN = S mg = (0.7)(50 kg)(9.8 m/s2)= 343 N
mg
What is the frictional force now?
FN
F = 200 N
50 kg
s = 0.7
mg
It’s the same!! Because if we write Newton’s
2nd law in the y-direction we see that once
again FN = mg
So Ff = S FN = S mg = (0.7)(50 kg) (9.8 m/s2)
= 343 N
Is the box going to move?
Let’ s look at this situation: A box slides on the surface with a constant velocity.
+y
+x
F
FN

Ff
mg
What is the relationship between F
and Ff?
1st step: Draw FBD
2nd step: write Newton’s 2nd law for the x
and y directions.
x: -Ff + Fx = 0, so then Ff = Fx= F cos()
Y: FN – mg + Fy = 0 so then, FN = mg – Fy,
FN = mg – F sin()
FN
What are you going to do today?
Ff
m
mg
If we write Newton’s second law in the x direction:
FSpring – Ff = 0 so then FSpring = Ff = S FN
Recall the equation for a straight line:
y=mx+b
Fspring = S FN
FSpring
y FSpring
slope 

 S
x
FN
FN
FSpring
FN
Ff
mb
+
+
m W:
m B:
T
+
T
m bg
mw
m wg
Using this equation we solve for K:
mWg – T = mwa
T – Ff = mBa
Remember that: Ff = K FN and for the
Block FN = mg, so Ff becomes K mg
+ So we can rewrite the equation for mB,
as:
T – K mg = mBa
So we have:
mW:
mWg – T = mwa
mB:
T – K mg = mBa
If we add these equations we get:
mWg - K mg = (mW + mB) a
mw g  (mw  mB )a
k 
mB g