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Forces and Newton’s Laws
Forces and newton’s laws
Newton’s Laws of Motion
Newton’s laws of motion
Newton’s Second Law of Motion
Newton’s
nd
2
law of motion 2
Newton’s Second Law of Motion
Newton’s
nd
2
law of motion Ex1b
x component of the net force: Fnet = 275+395-560 = 110 N (+)
Acceleration (x direction): a = Fnet / m = 110/1850 m/s2
a = 0.059 m/s2 in the + x direction.
Newton’s Second Law of Motion: Example
Newton’s
nd
2
law of motion Ex3
Newton’s Second Law of Motion: Example
Newton’s
nd
2
law of motion Ex3
v = v0 + at
v0 = 220 m/s
v=0
t = 0.0065 s  a = -33,850 m/s2
Fnet = ma = - 1.39*106 N
Newton’s Second Law of Motion: Example
nd law of motion
Newton’s
2
Find the acceleration
(vector) of the raft.
Ex4a0
Newton’s Second Law of Motion: Example
nd
2
Find the acceleration
(vector) of the raft.
Newton’s
law of motion Ex4a
Newton’s Second Law of Motion: Example
Newton’s
Fx = 22.9 N
Fy = 13.8 N
nd
2
law of motion Ex4b
Newton’s Second Law of Motion: Example
Newton’s
nd
2
law of motion Ex4c
Prob 4-11a
The component of the Net Force (F) in
the horizontal (x) direction:
Fx = F1x + F2x
F1x = F1 cos70o = 59.0 cos70o = 20.2 N
F2x = - 33.0 N
Fx = -12.8 N
The component of the acceleration
in the horizontal (x) direction:
ax = Fx / m = -12.8 / 7.0 m/s2
ax = -1.83 m/s2
Example 4.17 (7e) Only two forces act on an object (mass = 3.00 kg),
as in the drawing. Find the magnitude and direction (relative to the x
axis) of the acceleration of the object.
Prob 4-11
The x,y components of the forces:
F1x = F1 cos45o = 60.0 cos45o = 42.4 N
F1y = F1 sin45o = 60.0 sin45o = 42.4 N
F2x = F2 cos0o = F2 = 40.0 N
F2y = F2 sin0o = 0
The x,y components of the net force:
Fx = F1x + F2x = 82.4 N
Fy = F1y + F2y = 42.4 N
Example 4.17 (7e) Only two forces act on an object (mass = 3.00 kg),
as in the drawing. Find the magnitude and direction (relative to the x
axis) of the acceleration of the object.
Prob 4-11
The x,y components of the net force:
Fx = F1x + F2x = 82.4 N
Fy = F1y + F2y = 42.4 N
The x,y components of the acceleration:
ax = Fx / m = 82.4/ 3.00 = 27.5 m/s2
ay = Fy / m = 42.4/ 3.00 = 14.1 m/s2
Example 4.17 (7e) Only two forces act on an object (mass = 3.00 kg),
as in the drawing. Find the magnitude and direction (relative to the x
axis) of the acceleration of the object.
Prob 4-11
The x,y components of the acceleration:
ax = Fx / m = 82.4/ 3.00 = 27.5 m/s2
ay = Fy / m = 42.4/ 3.00 = 14.1 m/s2
θ
Magnitude and direction of the
acceleration:
a = (ax2 + ay2)½ = 30.9 m/s2
tanθ = ay / ax → θ = tan-1 (ay / ax )
θ = 27.2o
Newton’s Third Law of Motion
Newton’s
rd
3
law 1
Newton’s Third Law of Motion
Newton’s
rd
3
law 2
Newton’s Third Law of Motion
Newton’s
rd
3
law 3
Newton’s Third Law of Motion
Newton’s
rd
3
law Ex b
Example: Two boxes are being pushed across a (frictionless) surface
by a pushing force in the x direction.
Prob 4-11
a) What is the acceleration of the boxes?
b) What is the force exerted by box 1 on box 2 ?
c) What is the force exerted by box 2 on box 1 ?
Example: Two boxes are being pushed across a (frictionless) surface
by a pushing force in the x direction.
Prob 4-11
a) What is the acceleration of the boxes?
b) What is the force exerted by box 1 on box 2 ?
c) What is the force exerted by box 2 on box 1 ?
Free body diagram for m1 + m2 .
ax = Fx / M
→ a = F / ( m1 + m2 )
Example: Two boxes are being pushed across a (frictionless) surface
by a pushing force in the x direction.
Prob 4-11
a) What is the acceleration of the boxes?
b) What is the force exerted by box 1 on box 2 ?
c) What is the force exerted by box 2 on box 1 ?
Free body diagram for m2 .
a = F / ( m1 + m2 )
a = F21 / m2
→ F21 = { m2 / ( m1 + m2 ) } F
Example: Two boxes are being pushed across a (frictionless) surface
by a pushing force in the x direction.
Prob 4-11
a) What is the acceleration of the boxes?
b) What is the force exerted by box 1 on box 2 ?
c) What is the force exerted by box 2 on box 1 ?
Free body diagram for m1 .
a = F / ( m1 + m2 )
a = F1, NET / m1 = (F - F12 ) / m1
→ F12 = { m1 / ( m1 + m2 ) } F