UNIT 3
Forces and the Laws of Motion

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FORCES & THE LAWS OF MOTION
2

ConcepTest 4.9b
Collision Course II
In the collision between the car and the truck, which has the greater acceleration?
1) the car
2) the truck
3) both the same
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 the car and the truck, which has the greater acceleration?
1) the car
2) the truck
3) both the same
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 (1 st Law Problems)
 Hw: Practice D (All) p139
UPCOMING…
 Wed: Even More Everyday Forces
 Quiz #1 1 st 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
Find Tensions T

F y

0
1 and T
2 mg

T
1 sin
T
1
 mg sin
  
20 ( sin
9 .
8
 
)

226 N

F x

0 T
2

T
1 cos
  q
T
1
T
1 sin(60)
T
2
T
1 cos(60) m mg q
= 60 o m = 20 kg
24

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.
In terms of F
1
, F
2
F
3
, and
, what is the net force acting on the knot?
F
3
F net
= F
1
+F
2
+F
3
= 0
F
1
F
2
F
2 q
1 q
2
F
1
F
3

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.
F
2
Find the magnitudes of the x and y components for each force acting on the knot.
q
1 q
2
F
1
F
3

Everyday Forces
Find the magnitudes of the x and y components for each force acting on the knot.
F
1
F
2
F
3
X component
0.0
-F
2 cos
Ө
1
+F
3 cos
Ө
2 y component
-mg
+F
2 sin
Ө
1
+F
3 sin
Ө
2
F x net
= F
3 cos
Ө
2
–
F
2 cos
Ө
1
= 0
F y net
= F
2 sin
Ө
1
+
F
3 sin
Ө
2
- mg
= 0
F
2 q
1 q
2
F
3
F
1

Everyday Forces
Assume that
Ө
1
= 30
°
,
Ө
2
= 60
°
, and the mass of the lantern is 2.1 kg. Find F
1
, F
2
, and F
3
.
F x net
= F
3 cos
Ө
2
–
F
2 cos
Ө
1
= 0
F y net
= F
2 sin
Ө
1
+
F
3 sin
Ө
2
- mg
= 0
F
2 q
1 q
2
F
Find F
3 in terms of F
2
.
F
3 cos Ө
2
= F
2 cos Ө
1 F
3
= F
2 cos 𝜃1 cos 𝜃2
F
3
= F
2 cos 30 cos 60
F
3
= F
2
(1.73)
Substitute for F
3 with F
2
(1.73).
F
2 sin
Ө
1
+
F
3 sin
Ө
2
- mg =0
F
2 sin(30)
+
F
F
2
(.5)
+
F
2
2
(1.73)
(1.73) mg = 20.6N
sin(60) – 20.6 N =0
(.866) = 20.6 N
F
2
(.5)
+
F
2
(1.50) = 20.6 N
F
2
(.5
+
1.50) = 20.6 N
F
2
(2.00) = 20.6 N
F
2
= 10.3 N
1
F
3

Everyday Forces
Assume that
Ө
1
= 30
°
,
Ө
2
= 60
°
, and the mass of the lantern is 2.1 kg. Find F
1
, F
2
, and F
3
.
F
1
= -mg = -20.6 N
F
3
= F
2
(1.73)
F
2
= 10.3 N
F
3
= 10.3N(1.73)
F
3
= 17.8 N
F
2 q
1 q
2
F
3
F
1

END
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