Download F 1 F 2 F 3 Everyday Forces

UNIT 3
Forces and
the Laws of Motion
Tuesday October 18th
FORCES & THE LAWS OF MOTION
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ConcepTest 4.9b Collision Course II
In the collision between
1) the car
the car and the truck,
2) the truck
which has the greater
3) both the same
acceleration?
4) it depends on the velocity of each
5) it depends on the mass of each
ConcepTest 4.9b Collision Course II
In the collision between
1) the car
the car and the truck,
2) the truck
which has the greater
3) both the same
acceleration?
4) it depends on the velocity of each
5) it depends on the mass of each
We have seen that both
vehicles experience the
same magnitude of force.
But the acceleration is
given by F/m so the car
has the larger acceleration,
since it has the smaller
mass.
TODAY’S AGENDA
Tuesday, October 18
 Laws of Motion
 Mini-Lesson: Everyday Forces (1st Law Problems)
 Hw: Practice D (All) p139
UPCOMING…
 Wed: Even More Everyday Forces
 Quiz #1 1st Law Problem
 Thurs: NO SCHOOL (Teacher In-Service Day)
 Fri:
NO SCHOOL (Teacher In-Service Day)
 Mon: Lab Review: LAWS OF FORCE Lab
Everyday Forces
q
Find Tensions T1 and T2
 Fy  0
T1 sin(60)
mg  T1 sinq 
T1
mg
20(9.8)
T1 

 226 N
sinq 
sin60 
 Fx  0
T2  T1 cosq 
T1 cos(60)
T2
mg
m
q = 60o
m = 20 kg
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Everyday Forces
F3
A lantern of mass m is suspended
by a string that is tied to two
other strings, as shown to the
right. The free-body diagram
shows the forces exerted by the
three strings on the knot.
F2
q1
In terms of F1, F2, and
F3, what is the net force
acting on the knot?
q2
F1
F3
Fnet = F1+F2+F3 = 0
F1
F2
Everyday Forces
A lantern of mass m is suspended
by a string that is tied to two
other strings, as shown to the
right. The free-body diagram
shows the forces exerted by the
three strings on the knot.
F3
F2
Find the magnitudes of the x and y components
for each force acting on the knot.
q1
q2
F1
Everyday Forces
F3
Find the magnitudes of the x and y components
for each force acting on the knot.
X component
q1
q2
y component
F1
0.0
F2
-F2cosӨ1
+F2sinӨ1
F3
+F3cosӨ2
+F3sinӨ2
Fx net = F3cosӨ2 – F2cosӨ1 = 0
F2
-mg
Fy net = F2sinӨ1 + F3sinӨ2 - mg = 0
F1
Everyday Forces
F3
F2
Assume that Ө1= 30°, Ө2= 60°, and the mass of
the lantern is 2.1 kg. Find F1, F2, and F3.
Fx net = F3cosӨ2 – F2cosӨ1 = 0
q1
Fy net = F2sinӨ1 + F3sinӨ2 - mg = 0
F1
Find F3 in terms of F2.
F3cosӨ2 = F2cosӨ1
q2
F3 = F2
cos 𝜃1
cos 𝜃2
F3 = F2
cos 30
cos 60
F3 = F2(1.73)
Substitute for F3 with F2(1.73).
mg = 20.6N
F2sinӨ1 + F3sinӨ2 - mg =0
F2sin(30) + F2(1.73)sin(60) – 20.6 N =0
F2(.5) + F2(1.73)(.866) = 20.6 N
F2(.5) + F2(1.50) = 20.6 N
F2(.5 + 1.50) = 20.6 N
F2(2.00) = 20.6 N
F2 = 10.3 N
Everyday Forces
F3
Assume that Ө1= 30°, Ө2= 60°, and the mass of
the lantern is 2.1 kg. Find F1, F2, and F3.
F1 = -mg = -20.6 N
F3 = F2(1.73)
F3 = 17.8 N
F2 = 10.3 N
F3 = 10.3N(1.73)
F2
q1
q2
F1
END
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