Tue Sept 22
•
•
•
Assign 5 - Friday
Midterm – Monday 10/05
7:15-9:15 PM | Morton 235
– Email if class conflict
• Help Room Wed/Thur 6-9 PM
SI sessions:
Monday- 8:10-9:10 PM in Morton 126
Tuesday- 8:10-9:10 PM in Morton 227
Thursday- 7:05-8:05 PM in Morton 227
Today:
• Tension
• Incline Plane
Isolate object or group of objects
Draw all EXTERNAL forces on object
Choose a coordinate system
Write down F=ma (in each dimension)
FFriction,KINETIC = µ KINETIC Fnormal
FFriction,STATIC ≤ µSTATIC Fnormal
Static Force of Friction "Reaction Force“ – only as big as it needs to be
A box weighing 10 N is moving to the right under the influence of a 10
N horizontal force. µs=0.4 and µk=0.3. What is the magnitude and
direction of the frictional force?
(1) 1N to the left
(2) 1N to the right
(3) 3N to the left
(4) 3N to the right
(5) 4N to the left
(6) 4N to the right
ΣFY = may
FN – FG = 0
FN – 10 N = 0
FN = 10 N
Moving, so use µk.
Ff = µk * FN
Ff = µk * FN
Ff = 0.3 * 10N
Ff = 3N
The 5 N force is now applied at an angle downwards. The box is still
moving. What happens to the frictional force?
5N
1.
2.
3.
4.
increases
stays the same
decreases
Cannot determine from information given
Normal force increases, same materials, friction force increases.
A box with a weight of 10 N starts at rest. A 2 N force is exerted on the
box. µs=0.4 and µk=0.3. What is the magnitude and direction of the
frictional force?
2N
(1) 2N to the left
(2) 2N to the right
(3) 3N to the left
(4) 3N to the right
(5) 4N to the left
Ff,kinetic = µk * FN = 3N
(6) 4N to the right
Ff,static MAX = µs * FN = 4N
2N is not enough to move it!!!!
ΣFX = 0
2N – FFRICT = 0
FFRICT = 2N
"How do I figure out the force due to friction?"
• If know µ, can try to find FN – then calculate directly
– Remember: FN is not always just the weight.
• Or if you know all the other forces and the acceleration, can find
with forces:
– Example: A 10 kg box is sliding across the floor with an
acceleration of 1m/s2. You are pushing horizontally with a force
of 15 N. What is the magnitude and direction of the frictional
force?
ΣFx = max
FN
15N – FFRIC =(10kg)(1m/s2)
15N
FFRIC
FFRIC = 15N – 10N
FFRIC = 5N
FG
µk = ?
When friction is the only force in a direction …
• Sometimes friction is the only thing causing changes in motion
• Crate on back of a truck:
– If acceleration too great, static friction force not strong enough
to cause needed acceleration
– This holds for both speeding up and slowing down
Car stopping: "…force of friction or force of brakes?"
What actually stops a car is the friction between the wheels and
the road. The brakes use friction to slow down wheels.
A box with a mass of 5 kg is just sitting on the flat bed of a truck (it is
not tied down in any way). The truck accelerates at 2 m/s2, as does the
box (so it’s not slipping). The coefficient of frictions between the box
and the bed of the truck are µs=0.6 and µk=0.4. What is the magnitude
of the frictional force acting on the box?
(1) 2.0 N (2) 3.0 N (3) 10. N (4) 16 N (5) 20N (6) 29N (7) 49N
Friction is causing box to accelerate
Is it slipping? No – so find
maximum static force of friction
ΣFY = may
FN – FG = 0
FN – 5kg*(9.8 m/s2) = 0
FN = 49 N
Ff ≤ µs * FN
Ff ≤ 0.6 * 49N
Ff ≤ 29.4 N
If friction only force, what
force required for a=2 m/s2?
F = ma
F = (5kg)(2m/s2) = 10 N
Suppose friction is NOT negligible between the bottom of the sled
and the snow. You can either push forward and down at an angle θ or
pull up and forward at the same angle. If F1=F2 and the angles are the
same, which situation has the greater acceleration?
(1) The push
(2) The pull
(3) Both are equal
Horizontal component of force same.
Normal force, and therefore frictional force, less in second case.
Coordinate Axes
• Must be perpendicular to each other
• Can be oriented any way want
• Often one orientation most convenient
Perfectly Valid
More Convenient
The block on the incline has a
weight of 2.0 N. The angle θ is
30º. What is the component of
the force due to gravity in the x
direction?
(1) 0.5 N
(3) 1.7 N
(5) 2.3 N
(2) 1.0 N
(4) 2.0 N
(6) 4 N
For a Weight of 2.0 N, the component
parallel to the incline would be:
FN
+y
θ
Fg
Fg cosθ
+x
θ
Fg sinθ = 2.0N sin30º
Fg
= 1.0 N
Fg sinθ
Incline
• Gravity downward - has parallel and
perpendicular components
• Normal force still perpendicular to
surface
• Friction still parallel
• Often convenient to tilt coordinate axes
Components of weight
shown in blue
Leap of faith – if don't know
mass, put in m and look for it
to possibly cancel
FN
+y
θ
Fg
Fg cosθ
+x
θ
Fg
Fg sinθ
Suppose you increase the angle θ. What happens to the x component
of Fg and the normal force FN?
FN
(1) F increases; F increases
g,x
N
(2) Fg,x increases; FN same
(3) Fg,x increases; FN decreases
(4) Fg,x decreases; FN increases
(5) Fg,x decreases; FN same
+y
θ
Fg
(6) Fg,x decreases; FN decreases
(7) Fg,x same; FN increases
(8) Fg,x same; FN same
(9) Fg,x same; FN decreases
Parallel:
ΣFx = max
Fg sinθ = max
Perpendicular:
ΣFy = may
FN – Fgcos θ = 0
FN = Fgcos θ
+x
A 30 kg (294 N) crate is sliding down an incline at an angle 30° below
the horizontal. The kinetic coefficient of friction is 0.3 between the
crate and the ramp. What is the acceleration of the crate?
• Draw picture and define axes
Normal
• Isolate objects – draw Free Body Diagrams
Friction
(FBD) for each object
– gravity?
– anything else touching object?
• Write down sum of forces in each direction
Weight
• Solve
Parallel:
ΣFx = max
Perpendicular:
Fg sin30° - FFRICTION = ma
ΣF = ma
y
y
FN – Fgcos 30° = 0
294 sin(30°) – µFN = 30 a
FN = 294 cos30° = 254.6N
147 – 0.3(254.6N) = 30 a
a = 2.35 m/s2
Let's take the previous incline problem and see how the acceleration
depends on the mass. A block is sliding down an incline with friction.
Let's solve it algebraically:
+y
Friction
Normal
Parallel:
+x
Perpendicular:
ΣFx = max
Weight
Fg sinθ - FFRICTION = ma
ΣFy = may
mg sinθ – µFN = m a
FN – Fgcos θ = 0
mg sinθ – µ mg cosθ= m a
FN = mg cos θ
Mass cancels out!
g sinθ - µg cosθ = a
Note: this only works because all the forces involved are
proportional to the mass.
A block with a weight of 10 N is sitting at rest on an incline which is
tilted at an angle of 30º. The force of friction is 5.0 N. What is the net
force acting on the block?
(1) 0 N
(2) 5N down the incline
(3) 5N up the incline
(4) 10N straight down
(5) 5N straight up
(6) 15N straight down
(7) 15N straight up
"At rest" – acceleration is zero (no change in velocity)
If acceleration is zero, the net force is zero.
Tension – Ropes/Pulleys
•
•
•
•
•
Tension – Force rope exerts on other objects
Rope can only pull – force in direction of rope
Tension same in all parts of rope (massless rope)
Mag of v and a same for all objects on rope
Pulley changes direction of force, but if massless and frictionless
axle, no affect on T
• For now, massless pulley and rope
Two people pull on opposite ends of a massless rope. Each pulls with
a force of 40 N. What is the tension of the rope?
(1) 0N
(2) 20N
(3) 40N
(4) 80N
Look at point where one person is holding onto rope.
Not accelerating.
- FPerson on Rope + Frope on person = 0
FPerson on Rope = Frope on person
Force of rope on person is the tension.
A person of weight 800 N is sitting on a chair of weight 10 N. The
chair is supported by a rope over a pulley. The person pulls down on
the rope with a force of F to support themself. What force, F, is
required to hold themselves stationary?
(1) 400N
(2) 405N
(3) 800N
(4) 810N
(5) 1600N
(6) 1620N
ΣF = ma
2F – FG = 0
2F = (800N+10N)
F
Consider seat as part of
person. Two ropes pulling up
on person and seat, each
with same force
F
FG
A person of weight 800 N is sitting on a chair of weight 10 N. The
chair is supported by a rope over a pulley. The person pulls down on
the rope with a force of F to support themself. What force is required
to accelerate upward at 0.5 m/s2?
(1) 362N
(2) 384N
(3) 400N
(4) 405N
(5) 426N
(6) 483N
Fg = mg
m = Fg/g
m = 810N/9.8m/s2
m = 82.6 kg
F
F
ΣF = ma
2F – FG = ma
2F = (800N+10N) + ((82.6kg)0.5m/s2)
FG
A 200 N box is hanging from a rope. Two ropes attach the box to the
ceiling at the angles given. What is the tension in each rope?
T1
T2
θ1=30°
θ2 = 60°
T3
200N
ΣFx = max
-T1cosθ1 + T2cosθ2 = 0
T1 = T2 (cos θ2/cos θ1)
T1 = T2 (0.577)
ΣFy = may
T1sinθ1 + T2sinθ2 – 200N= 0
T1 (0.5) + T2 (0.866) = 200 N
T2(0.577) (0.5) + T2 (0.866) = 200 N
Solve: T2 = 173 N
T1 = T2 (0.577) = 100. N
A 200 N box is hanging from a rope. Two ropes attach the box to the
ceiling at the angles given. What is the tension Rope 3?
(1) 50 N
(4) 136 N
(2) 86 N
(5) 173 N
(3) 100
(6) 200 N
T1
T2
θ1=30°
θ2 = 60°
T3
Point where T3 attaches to box, a=0
ΣF = ma
T3 – 200N = 0
T3 = 200 N
200N
Three boxes are accelerating to the right at a rate of 2.0 m/s2.
All 3 have non-zero mass.
How do T1, T2, and T3 relate?
(1) T1 = T2 = T3
(2) T1 < T2 < T3
(3) T3 < T2 < T1
(4) T1 < T2 = T3
(5) T3 = T2 < T1
(6) T1 = T2 < T3
all 3 boxes same acceleration
ΣFX = ma
T1 = mAa
T2 = (mA+mB)a
T3 = (mA+mB+mC)a
Example: Three Boxes
Three boxes are accelerating to the right at a rate of 2.0 m/s2.
The tension in rope 2 (T2) is 6 N. The mass of block B is 1 kg. The
mass of block C is 2 kg.
Find the tension in rope 1 (T1).
ΣFX = ma
Find T1:
T1
B
T2
T2 - T1 = ma
T1 = T2 - ma
T1 = 6N – (1kg)*(2.0m/s2)
T1 = 4N
Three Boxes
Three boxes are accelerating to the right at a rate of
2.0 m/s2.
The tension in rope 2 (T2) is 6 N. The tension in rope
1 (T1) is 4 N. The mass of block B is 1 kg. The
mass of block C is 2 kg.
What is the mass of block A?
1.
2.
3.
4.
5.
6.
ΣFX = ma
A
T1
T1 = ma
4N = m*(2 m/s2)
m = 2 kg
0.5 kg
1.0 kg
2.0 kg
4.0 kg
8.0 kg
16 kg
Example: Three Boxes
Three boxes are accelerating to the right at a rate of 2.0 m/s2.
The tension in rope 2 (T2) is 6 N. The mass of block B is 1 kg. The
mass of block C is 2 kg.
Find the tension in rope 3 (T3).
ΣFX = ma
T3 – T2 = ma
Find T3:
T2
C
T3
T3 = ma + T2
T3 = (2kg)*(2.0m/s2) +6N
T3 = 10N
Can also use all 3 blocks:
T3 = (2kg+1kg+2kg) * 2m/s2
Example: Three Boxes
Three boxes are accelerating to the right at a rate of 2.0 m/s2.
The mass of Block A is 2 kg. The mass of block B is 1 kg. The mass
of block C is 2 kg.
Suppose friction were involved. µS = 0.45 and µK=0.35. What is the
magnitude of the tension T1?
+y
ΣFX = max
In this case:
FN
+x
T1
–
F
=
ma
ΣFy = may
FRICT
x
Ff
T1 = FFRICT + max
FN – Fg = 0
T1
A
2
FN=Fg=19.6N T1 = 6.9N + 2kg*2m/s
T1 = 10.9 N
Fg
FFRICT = µKFN
FFRICT = µKFg
FFRICT = 0.35*2kg*9.8m/s2