chapter 4humanic/p111_lecture9.pdf · chapter 4 forces and newton’s ... 4.11 equilibrium...

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


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.


035sin35sin 21 =!+ oo TT

035cos35cos 21 =!++ FTT oo

Example. Find the magnitude of F if the system is in equilibrium.

! = 0xF

! = 0yF


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



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.


N 3150=W

y componentx componentForce

1Tr

2Tr

Wr

o0.10sin1T!

o0.80sin2T+

0

o0.10cos1T+

o0.80cos2T!

W!


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


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.



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


y componentx componentForce

1Tr

2Tr

Dr

Rr

o0.30cos1T+

o0.30cos2T+

0

0

D+

R!

o0.30sin1T+

o0.30sin2T!


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

chapter 4humanic/p111_lecture9.pdf · chapter 4 forces and newton’s ... 4.11 equilibrium...

Documents