Download Solving force problems

Solving force problems
•  CAPA due Tuesday night.
•  Exam solutions posted on D2L
•  Exam scores will be posted
either tonight or Monday night.
•  Today will be multiple body
problems and the dynamics of
circular motion.
1
Pulleys
Until we get to Chapter 10, all of our pulleys will be
ideal pulleys which are massless and frictionless.
An ideal pulley simply changes the direction of the
tension force in the rope which goes around the pulley.
2
Statics problem
A massless rope is looped over a pulley with
one end connected to a 30 kg box and the
other end held by a person. What force must
the person exert to keep the box stationary?
30
kg
.
Free body
diagram
for the box
T
FG = mg
Fnet,y = T − mg = 0
T = mg
In a massless rope the tension
is the same everywhere!
Free body
diagram
for the hand
T
Fhand
Fnet,y = T − Fhand = 0
Fhand = T = mg
Fhand = 30 kg ⋅ 9.8 m/s2
= 290 N
3
Statics problem
What is the magnitude of the force exerted
by the rod holding the pulley to the ceiling?
30
kg
.
Free body
Frod
diagram
for the
T T
pulley
Fnet,y = Frod − 2T = 0
from the last problem
we know T = mg
Frod = 2T = 2mg = 2 ⋅ 30 kg ⋅ 9.8 m/s2 = 590 N
This is the force on the ceiling
4
Clicker question 1
Set frequency to BA
The person holding the rope steps back a
few feet. What is the magnitude of the
force exerted by the rod on the ceiling now
(note the angle)?
30
kg
.
Pulley free body diagram
A.  Frod = 2mg
B.  mg ≤ Frod < 2mg
C.  Frod < mg
D.  Frod > 2mg
E.  Impossible to tell
Frod
T T
Frod < 2T and T=mg.
Given the angles, we
could calculate Frod.
5
Clicker question 2
Set frequency to BA
An Atwood's machine is a pulley with two masses
connected by a string as shown. The mass of object
A, mA, is twice the mass of object B, mB. The tension
T in the string is...
A. T = mA g B. T = mB g C. Neither of these
A
TA
A
net,y
A
A y
F = T − mA g = m a
FG = mA g
B
TB
A
B
B
Fnet,y
= T − mB g = mB ayB
FG = mB g
Since the weights are unequal, it should be clear that there
will be some non-zero acceleration. Thus, since mAa yA and
mB a yB are not zero, neither A nor B can be correct.
6
Full solution to Atwood’s machine
A
TA
T − mA g = m a
B
A
A y
TB
T − mB g = mB a yB
FG = mB g
FG = mA g
A
B
We defined up as positive for both masses which means a yA = −a yB
T = mA g − mAa yB & T = mB g + mB a yB give mA g − mAa yB = mB g + mB a yB
mA − mB
g
Solving for acceleration: a =
mA + mB
B
y
B
B y
T = mB g + m a
so the tension is
m A + mB
m A − mB
2mA mB
=
mB g +
mB g =
g
m A + mB
m A + mB
mA + m7B
Remember circular motion
What is the acceleration for circular motion with
varying speed?
Can divide acceleration vector into two parts
!
v2
arad
! !
a v1
atan
Tangential acceleration is related to
change in speed and is parallel to the d v!
velocity vector with magnitude atan = dt
Radial acceleration is perpendicular to the velocity vector 2
v
and points to the inside of the curve with magnitude arad = r
!
2
2
Since they are perpendicular: a = atan
+ arad
We can consider the coordinate system as r,t rather than
the more usual x,y.
8
What is the force?
We know we need forces to cause acceleration.
For each of these scenarios, what is the force
which causes the radial acceleration?
A car going around a corner on a level surface
Friction between tires and pavement
A car going around a banked curve
Friction between tires and pavement and
a component of the normal force
Satellite orbiting the Earth
Earth’s gravity
9
Vertical circular motion
Consider a motorcycle doing a vertical loop.
10
Vertical loops
What does the free body diagram
look like at the bottom of the loop?
n
Let’s apply Newton’s 2nd law
mg
Adding the forces: Fnet,r = n − mg
2
2
v
mv
We will have radial acceleration arad = r so mar =
r
2
mv
2
Therefore n − mg =
mv
Can also write n = mg +
r
r
making it clear that n > mg
The normal force is what you feel as weight which is why you
feel heavier at the bottom of the roller coaster.
11
Clicker question 3
What is the free body diagram at the top?
A.
B.
C.
D.
Let’s apply Newton’s 2nd law:
2
mv
Therefore n + mg =
r
mv2 < mg
What happens if r
?
n mg
E.
Fnet,r = n + mg = marad
2
mv
Can also write n =
− mg
r
Can the normal force be negative?
No. In this case, n=0 and there is more than radial acceleration;
it is in free fall and follows a parabola. It falls off the loop.
mv2 ≤ mg
For r
, n = 0 so you feel weightless.
12