chapter 7 forces in two dimensions
DESCRIPTION
Chapter 7 Forces in Two Dimensions. A 50 kg sign is supported in a motionless position supported by two ropes that make a 40 º angle with the horizontal What is the Tension in each rope?. F A. F B. 40 °. 40 °. F g. F B. 50 °. 490 N. 80 °. 50 °. F A. - PowerPoint PPT PresentationTRANSCRIPT
Chapter 7Forces in Two Dimensions
A 50 kg sign is supported
in a motionless position
supported by two ropes that
make a 40º angle with the
horizontal What is the
Tension in each rope?
Chapter 7Forces in Two Dimensions
Draw a free body diagram FBFA
Fg
40°40°
The force of gravity is F=maor F= (50kg)(9.8 m/s2)=490 N
Since the system is in equilibrium,the sum of the vectors is zero and by the head tail method
= 490 N
490 N50°
50°
FA
FB
80°
50sin80sin
490 AFN FA = FB = 381 N
Chapter 7Forces in Two Dimensions
Motion along inclined planesFy
Fx
Fg
By vector addition and similartriangles Fg
Fy
Fx
Chapter 7Forces in Two Dimensions
A trunk weighing 300 N
is placed on a plane and
starts to slide when inclined
at 30°.Find the parallel and
perpendicular components
of the force and the
coefficient of friction.
Fy
Fx
300 N 30
300 N
Fy
Fx
30
Chapter 7Forces in Two Dimensions
Fx=F sin 30°=300 N *.5
Fx = 150 N
300 N
Fy
Fx
30
Check (150 N)2+(260 N)2 = (300 N)2
Fy=F cos 30°=300 N *.866
Fy = 260 N
Fy
Fx
300 N 30
.577260N
150N
F
F
N
f or µ = tan ø
Chapter 7Forces in Two Dimensions
Suppose the crate is moving down the plane
and the coefficient of kinetic friction between
the surfaces is .10. How fast is the crate going
2 seconds after starting from rest?
Chapter 7Forces in Two Dimensions
y direction
FN= mg cos Ø
x direction
Fnet= Fx-kFN
Fnet= 150 N- .10(260 N) = 124 N
F = ma
a = 4.0 m/s2 vf = vi + at vf = 0 + 4.0 m/s2*2 s = 8 m/s
Fy
Fx
300 N 30
Ff
Fnet = Fx - Ff
Applet
124 = (300N/9.8 m/s2) a
Projectiles Launched Horizontally
A projectile is thrown horizontally at 20 m/sfrom the top of a cliff 490 m high.
a. How long is the object in the air?b. How far from the base of the cliff does it land?c. How fast is it moving the instant it hits the ground?
x0 = 0vx0 = 20 m/sy0 = -490 may = -9.8 m/s2
20 m/s
490 m
a. d = vit + ½ at2
solve for t
t = 29.8m/s
490m)2(
a
2d
t = 10 s
b. d = vt d = (20m/s)(10s) d = 200 m
c. vf = vi+at vf = 0 + (-9.8m/s2)(10s) vf = -98 m/s
A projectile is thrown horizontally at 5 m/sfrom the top of a cliff 19.6 m high.
a. How long is the object in the air?b. How far from the base of the cliff does it land?c. How fast is it moving the instant it hits the ground?
x0 = 0vx0 = 5 m/sy0 = -19.6 may = -9.8 m/s2
a. d = vit + ½ at2
solve for t
t = 29.8m/s
19.6m)2(
a
2d
t = 2 s
b. d = vt d = (5m/s)(2s) d = 10 m
5 m/s
19.6 m
c. vf = vi+at vf = 0 + (-9.8m/s2)(2s) vf = -19.6 m/s
Projectiles Launched at an AngleA ball is thrown with an initial velocityof 5 m/s at an angle of 60º with the horizonat ground level.
a. How long is the ball in the air?b. How high does it go?c. How far does it go?
x0=0y0 = 0v = 5 m/s = 60ºa = -9.8 m/s2
60º
60º
5m/svx = v cos Ø = 5 cos 60º= 2.5 m/svy = v sin Ø = 5 sin 60º = 4.33 m/s
vx
vy
.88s9.8m/s
(4.33m/s)4.33m/s
a
vvt a.
2if
.956m)(.44s)9.8m/s(2
1.44s)(4.33m/s)(at
2
1tvd b. 222
i
2.2mm/s)(.88s) (2.5 vt d c.
Projectiles Launched at an AngleA projectile is thrown with an initial velocityof 98 m/s at an angle of 53º with the horizonat ground level.
a. How long is the ball in the air?b. How high does it go?c. How far does it go?
x0=0y0 = 0v = 98 m/s= 53ºa = -9.8 m/s2
53º
53º
98 m/svx = v cos Ø = 98 cos 53º= 58.98 m/svy = v sin Ø = 98 sin 53º = 78.27 m/s
vx
vy
15.97s9.8m/s
(78.27m/s)78.27m/s
a
vvt a.
2if
m 312)(8s)9.8m/s(2
1s) (8(78.27m/s)at
2
1tvd b. 222
i
941.88ms)m/s)(15.97 (58.98 vt d c.
Projectiles Launched at an AngleA projectile is thrown with an initial velocityof v m/s at an angle of ø with the horizonat ground level.
a. How long is the ball in the air?b. How high does it go?c. How far does it go?
x0=0y0 = 0v = v= a = -9.8 m/s2
ø
ø
vvx = v cos Øvy = v sin Ø
vx
vy
22if
m/s9.8
2vsin
9.8m/s
m/s)(vsinm/sv
a
vvt a.
)sin(
2
22
22
i 9.8m/s-
2vsin-)9.8m/s(
2
1
9.8m/s-
2vsin-m/s)(vsinat
2
1tvd b.
2
2
2 9.8m/s
cossin2v
m/s9.8
sin2vm/s) cos(v vt d c.
Projectile Motion
What goes faster, the inside of a record or the outside?
Animation
Circular Motion
Fc
ac
As a car makes a turn, the force of friction acting upon the turned wheels of the car provide the centripetal forcerequired for circular motion.
As a bucket of water is tied to a string and spun in a circle, the force of tension acting upon the bucket provides the centripetal force required for circular motion.
As the moon orbits the Earth, the force of gravity acting upon the moon provides the centripetal force required for circular motion.
T
R2
t
dv
rr
AB
v
Δv
r
tv
v
Δv
ar
v
t
Δv 2
2
22
cc T
R4πm
R
mvmaF
A 900-kg car moving at 10 m/s takes a turn around a circle with a radius of 25.0 m. Determine the acceleration and the net force acting upon the car.
Known Information:
m = 900 kg v = 10.0 m/sR = 25.0 m
Requested Information:
a = ???? Fnet = ????
To determine the acceleration of the car, use the equation a = (v2)/R.The solution is as follows: a = (v2)/R
a = ((10.0 m/s)2)/(25.0 m)a = (100 m2/s2)/(25.0 m)
a = 4 m/s2
To determine the net force acting upon the car, use the equation F net = m*a. The solution is as follows.Fnet = m*a
Fnet = (900 kg)*(4 m/s2)
Fnet = 3600 N
Joe Brick, the new 95-kg halfback makes a turn on the football field. He sweeps out a path which is a portion of a circlewith a radius of 12-meters. The halfback makes a quarter of a turn around the circle in 2.1 seconds. Determine the speed, acceleration and net force acting on Joe.
Known Information:m = 95.0 kg R = 12.0 mTraveled 1/4-th of the circumference in 2.1 s
Requested Information:v = ???? a = ????Fnet = ????
To determine the speed of the halfback, use the equation v = d/t where the d is one-fourth of the circumference and the time is 2.1 s. The solution is as follows: v = d/t v = (0.25 * 2 π * R)/t v = (0.25 * 2 * 3.14 * 12.0 m)/(2.1 s) v = 8.97 m/s
To determine the acceleration of the halfback, use the equation a = (v2)/R. The solution is as follows:•a = (v2)/R •a = ((8.97 m/s)2)/(12.0 m)
•a = (80.5 m2/s2)/(12.0 m)
•a = 6.71 m/s2
To determine the net force acting upon the halfback, use the equation Fnet = m*a. The solution is as follows.•Fnet = m*a
•Fnet = (95.0 kg)*(6.71 m/s2)
•Fnet = 637 N
Simple Harmonic Motion Definitions
•Period (s) The time required to complete one cycle.•Frequency (hz) The number of cycles per second. •Amplitude (m) The point of maximum displacement from rest.
AB
CD
E
+-
Simple Harmonic Motion
A
B
C
D
E
D
C
B
A
Time (sec)
+
- Dis
tan
ce
(mete
rs)
AB
CD
E
+-Simple Harmonic Motion
Time (sec)
+
- Velo
city
(m
/s)
B
D
ED
C
A
B
C
A
AB
CD
E
+-Simple Harmonic Motion
Time (sec)
+
- Acc
ele
rati
on
(m
/s
2)
A
B
D
ED
C
A
B
C
