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Chapter 4
Forces and Newton’sLaws of Motion
4.11 Equilibrium Application of Newton’s Laws of Motion
Definition of EquilibriumAn object is in equilibrium when it has zero acceleration.
! = 0xF
! = 0yF
e.g. brick at rest on a table
4.11 Equilibrium Application of Newton’s Laws of Motion
Reasoning Strategy to solve equilibriumproblems
• Select an object(s) to which the equations of equilibrium areto be applied.
• Draw a free-body diagram for each object chosen above.Include only forces acting on the object, not forces the objectexerts on its environment.
• Choose a set of x, y axes for each object and resolve all forcesin the free-body diagram into components that point along theseaxes.
• Apply the equations and solve for the unknown quantities.
4.11 Equilibrium Application of Newton’s Laws of Motion
035sin35sin 21 =!+ oo TT
035cos35cos 21 =!++ FTT oo
Example. Find the magnitude of F if the system is in equilibrium.
! = 0xF
! = 0yF
4.11 Equilibrium Application of Newton’s Laws of Motion
The y-equation tells us that T1 = T2 = T
Using this in the x-equation and solving for F,
F = 2T cos 35o
But what is T ?
T
mg
T
T
! = 0yF
Free body diagram for thehanging mass
= T - mg
T = mg
F = 2T cos 35o = 2mg cos 35o = 2(2.2)(9.8)cos 35o = 35 N
4.11 Equilibrium Application of Newton’s Laws of Motion
4.11 Equilibrium Application of Newton’s Laws of Motion
Example. An automobile engine has aweight which is W = 3150 N. The engineis being positioned as in the figure. Find the tensions T1 and T2 in equilibrium.
4.11 Equilibrium Application of Newton’s Laws of Motion
N 3150=W
y componentx componentForce
1Tr
2Tr
Wr
o0.10sin1T!
o0.80sin2T+
0
o0.10cos1T+
o0.80cos2T!
W!
4.11 Equilibrium Application of Newton’s Laws of Motion
00.80sin0.10sin 21 =+!=" oo TTFx
00.80cos0.10cos 21 =!!+=" WTTFyoo
The first equation gives 21 0.10sin0.80sin TT !!"
#$$%
&=
o
o
Substitution into the second gives
00.80cos0.10cos0.10sin0.80sin
22 =!!""#
$%%&
'WTT oo
o
o
4.11 Equilibrium Application of Newton’s Laws of Motion
oo
o
o
0.80cos0.10cos0.10sin0.80sin2
!""#
$%%&
'=
WT
N 5822 =T N 1030.3 31 !=T
4.12 Nonequilibrium Application of Newton’s Laws of Motion
! = xx maF
! = yy maF
When an object is accelerating, it is not in equilibrium.
Example. A supertanker of mass m = 1.50 x 108 kg is being towed bytwo tugboats. The tensions in the towing cables apply the forces T1 and T2 at equal angles of 30.0o with respect to the tanker’s axis. Inaddition, the tanker’s engines produce a forward drive force D, whosemagnitude is D = 75.0 x 103 N. Moreover, the water applies an opposingforce R, whose magnitude is R = 40.0 x 103 N. The tanker movesforward with an acceleration that points along the tanker’s axis andhas a magnitude of 2.00 x 10-3 m/s2. Find the magnitudes of the tensions T1 and T2.
4.12 Nonequilibrium Application of Newton’s Laws of Motion
4.12 Nonequilibrium Application of Newton’s Laws of Motion
m = 1.50 x 108 kg R = 40.0 x 103 N D = 75.0 x 103 N
ax = 2.00 x 10-3 m/s2 ay = 0
4.12 Nonequilibrium Application of Newton’s Laws of Motion
y componentx componentForce
1Tr
2Tr
Dr
Rr
o0.30cos1T+
o0.30cos2T+
0
0
D+
R!
o0.30sin1T+
o0.30sin2T!
4.12 Nonequilibrium Application of Newton’s Laws of Motion
00.30sin0.30sin 21 =!+=" TTFyo
x
x
ma
RDTTF
=
!+++=" 0.30cos0.30cos 21o
TTT == 21
N 1053.10.30cos2
5!="+
=o
DRmaT x
= max
fk
Example. Block 1 (mass m1 = 8.00 kg) is moving on a 30.0o incline witha coefficient of kinetic friction between the block and incline of 0.300.This block is connected to block 2 (mass m2 = 22.0 kg) by a masslesscord that passes over a massless and frictionless pulley. Find theacceleration of each block and the tension in the cord.
fkfk
m1 = 8.00 kg m2 = 22.0 kg µk = 0.300
Find a, T and T’
Block 1:
Σ Fx = -fk - W1 sin 30.0o + T = m1a
Σ Fy = FN - W1 cos 30.0o = 0 FN = W1 cos 30.0o
Block 2:
Σ Fy = T’ - W2 = m2(-a)
We also know:
T = T’ since the pulley and cord are massless
fk = µkFN = µkm1g cos 30.0o = (0.300)(8.00)(9.80)(0.866) = 20.4 N
fk
Equations we’re left with to solve for a and T:
-fk - W1 sin 30.0o + T = m1a
T - W2 = -m2a T = W2 - m2a
Substituting for T in the first equation:
-fk - W1 sin 30.0o + W2 - m2a = m1a
a = (-fk - W1 sin 30.0o + W2)/(m1 + m2) = (-20.4 - (8.00)(9.80)(0.500) + (22.0)(9.80))/(8.00 + 22.0) = 5.20 m/s2
T = W2 - m2a = (22.0)(9.80) - (22.0)(5.20) = 101 N
11.2 Pressure
AFP =
SI Unit of Pressure: 1 N/m2 = 1Pa
Pascal
Pressure, P, is defined as theforce, F, per unit area, A.
11.2 Pressure
Example 2 The Force on a Swimmer
Suppose the pressure acting on the backof a swimmer’s hand is 1.2x105 Pa. Thesurface area of the back of the hand is 8.4x10-3m2.
(a)Determine the magnitude of the forcethat acts on it.(b) Discuss the direction of the force.
11.2 Pressure
AFP =
( )( )N 100.1
m104.8mN102.13
2325
!=
!!== "PAF
Since the water pushes perpendicularly against the back of the hand, the forceis directed downward in the drawing.
11.2 Pressure
Atmospheric Pressure at Sea Level: 1.013x105 Pa = 1 atmosphere