PROBLEM 2-44 (page 41, 13th edition)

The magnitude of the resultant force acting on the bracket

is 400 N. Determine the magnitude of

!F1.

Take ! = 30o . Disregard the u axis.

2-44 (page 41) 13th edition

COMPLETION OF PROBLEM

Determine F1 so that F

R= 400 N :

!FR= !!F

Cartesian Vector Method (suppressing units)

!FR= F

Rx

!i + F

Ry

!j

FRx

= ! Fx= "650

3

5

#$%

&'(+F

1cos30! +500cos45!

FRy= ! F

y= 650

4

5

"#$

%&'+F

1sin30! (500sin45!

FR= FRx

2+ FRy

2

400 = (!390+F1 cos30o+500cos45

o)2+(520+F1 sin 30

o!500sin 45

o)2

Solution

Solve using Solver on a TI-83 Plus calculator:

F1= 314 N or F

1= ! 417 N

The negative sign indicates for that answer that

!F1

must act in the

opposite direction to that shown in the diagram.

PROBLEM 2-51 (page 41, 13th edition)

PROBLEM 2-40 (page 40, 12th edition)

Determine the magnitude and direction measured

counterclockwise from the positive x-axis of the resultant

force of the three forces acting on the ring A.

Take F1= 500 N and ! = 20o.

2-51 (page 41, 13th edition) 2-40 (page 40, 12th edition)

COMPLETION OF PROBLEM

Determine the magnitude and direction of

!FR= !!F .

Cartesian Vector Method (suppressing units):

!FR= F

Rx

!i + F

Ry

!j

= F

R(cos!

!i + sin!

!j )

FRx

= ! Fx= 400cos30! +500cos70! "600(

4

5) = 37.42 = A

FRy= ! F

y= 400sin30! +500sin70! +600(

3

5) =1030 = B

Answers

FR= FRx

2+ FRy

2= A

2+ B

2= 1.03 kN

! = tan"1(FRy/F

Rx) = tan"1(B /A) = 87.9!

PROBLEM 2-53 (page 42, 13th edition)

PROBLEM 2-57 (page 42, 12th edition)

Determine the magnitude of force !F so that the resultant of

the three forces is a small as possible.

Determine the magnitude of the resultant force.

2-53 (page 42) 13th edition 2-57 (page 42) 12th edition

COMPLETION OF PROBLEM

Determine F so that FR

is a small as possible:

!FR= !!F .

Cartesian Vector Method (suppressing units):

!FR= F

Rx

!i + F

Ry

!j

FRx

= ! Fx= 8"Fcos45! "14cos30!

FRy= ! F

y= "Fsin45! +14sin30!

FR= FRx

2+ FRy

2

FR(F) = (8!F cos45

!!14 cos30

!)2+(!F sin 45

!+14sin 30

!)2

We can determine the minimum of FR(F) using calculus.

Alternatively, since we want numerical answers, we determine the

minimum of FR(F) using the minimum function on a graphing

calculator. This yields:

F = 2.03 kN FR= 7.87 kN

PROBLEM 2-59 (page 42, 13th edition)

Determine the magnitude of force

!F1 so that the resultant of

the three forces is a small as possible when ! = 30!.

(Disregard the u axis.)

Determine the magnitude of the resultant force.

2-59 (page 42) 13th edition

COMPLETION OF PROBLEM

Determine F1 so that F

R is a small as possible:

!FR= !!F

Cartesian Vector Method (suppressing units):

!FR= F

Rx

!i + F

Ry

!j

FRx

= ! Fx= F

1cos60! + 200 + 260

5

13

"#$

%&'

FRy= ! F

y= F

1sin60! " 260

12

13

#$%

&'(

FR= FRx

2+ FRy

2

FR(F1) = F1 cos60

o+300( )

2

+ F1 sin60o!240( )

2

We can determine the minimum of FR(F1) using calculus.

Alternatively, since we want numerical answers, we determine the

minimum of FR(F1) using the minimum function on a graphing

calculator. This yields:

F1= 57.8 N F

R= 380 N

PROBLEM 2-78 (page 55, 13th edition)

Three forces act on the ring. The resultant force

!FR

has

magnitude and direction as shown.

Determine the magnitude and coordinate direction angles

of

!F3.

2-78 (page 55, 13th edition)

COMPLETION OF PROBLEM (page 1)

Forces (suppressed units)

!F1= 80

4

5

!i +

3

5

!k

!"#

$%&

!F2= !110

!k

!F3= F

x

!i +F

y

!j +F

z

!k

!FR= 120(cos45"sin30"

!i + cos45"cos30"

!j +sin45"

!k)

Equations for the resultant force

!FR= F

Rx

!i +F

Ry

!j +F

Rz

!k

FRx

= ! Fx:

120cos45!sin30! = 804

5

!"#

$%&+ F

x

FRy= ! F

y:

120cos45!cos30! = Fy

FRz= ! F

z:

120sin45! = 803

5

!"#

$%&' 110 + F

z

2-78 (page 55, 13th edition)

COMPLETION OF PROBLEM (page 2):

SOLVING THE EQUATIONS

Solve for Fx, F

y, F

z:

Fx= 120cos45!sin30! ! 80

4

5

"#$

%&'= A

Fy= 120cos45!cos30! = B

Fz= 120sin45! ! 80

3

5

"#$

%&'+ 110 = C

Magnitude and coordinate direction angles:

F3= Fx

2+Fy

2+Fz

2 = A

2+B

2+C

2= F = 166 N

!3= cos"1(F

x/F3)

= cos!1(A /F) = 97.5!

!3= cos"1(F

y/F3)

= cos!1(B /F) = 63.7!

!3= cos"1(F

z/F3)

= cos!1(C /F) = 27.5!

PROBLEM 2-79 (page 54, 12th edition)

Specify the magnitude of

!F3 and its coordinate direction

angles !3, "

3, #

3 so that the resultant force is

9!j kN .

2-79 (page 54, 12th edition)

COMPLETION OF PROBLEM (page 1)

Forces (suppressed units)

!F1= 12(cos30"

!j !sin30"

!k)

!F2= 10 !

12

13

!i +

5

13

!k

"#$

%&'

!F3= F

x

!i +F

y

!j +F

z

!k

!FR= 9!j

Equations for the resultant force

!FR= F

Rx

!i +F

Ry

!j +F

Rz

!k

FRx

= ! Fx: 0 = !

120

13+ F

x

FRy= ! F

y:

9 = 12cos30! + Fy

FRz= ! F

z:

0 = !12sin30! +50

13+ F

z

2-79 (page 54, 12th edition)

COMPLETION OF PROBLEM (page 2):

SOLVING THE EQUATIONS

Solve for Fx, F

y, F

z:

Fx=120

13= A

Fy= 9 ! 12cos30! = B

Fz= 12sin30! !

50

13= C

Magnitude and coordinate direction angles:

F3= Fx

2+Fy

2+Fz

2 = A

2+B

2+C

2= F = 9.58 kN

!3= cos"1(F

x/F3)

= cos!1(A /F) = 15.5!

!3= cos"1(F

y/F3)

= cos!1(B /F) = 98.4!

!3= cos"1(F

z/F3)

= cos!1(C /F) = 77.0!