Beer11e Chapter 3 ISM PDF

Preview

Summary

Download Beer11e Chapter 3 ISM PDF

Description

CHAPTER 3

PROBLEM 3.1 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at B that creates a moment of equal magnitude and opposite sense about E.

SOLUTION

(a)

By definition, We have

W  mg  80 kg(9.81 m/s 2)  784.8 N M E : M E  (784.8 N)(0.25 m) M E  196.2 N  m

(b)



For the force at B to be the smallest, resulting in a moment (ME ) about E, the line of action of force F B must be perpendicular to the line connecting E to B. The sense of FB must be such that the force produces a counterclockwise moment about E. Note: We have

d  (0.85 m)2  (0.5 m)2  0.98615 m

M E : 196.2 N  m  FB (0.98615 m) FB  198.954 N

and or

 0.85 m    59.534  0.5 m 

  tan1 

FB  199.0 N

59.5° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.2 A crate of mass 80 kg is held in the position shown. Determine (a) the moment produced by the weight W of the crate about E, (b) the smallest force applied at A that creates a moment of equal magnitude and opposite sense about E, (c) the magnitude, sense, and point of application on the bottom of the crate of the smallest vertical force that creates a moment of equal magnitude and opposite sense about E.

SOLUTION First note. . .

W mg (80 kg)(9.81 m/s2 )  784.8 N

(a)

We have

M E  rH /EW  (0.25 m)(784.8 N)  196.2 N m

For F A to be minimum, it must be perpendicular to the line joining Points A and E. Then with FA directed as shown, we have (  M E )  rA/ E ( FA )min .

(b)

(c)

or M E  196.2 N m 

Where

rA /E  (0.35 m)2  (0.5 m)2  0.61033 m

then

196.2 N  m  (0.61033 m)(F A )min

or

( FA ) min  321 N

Also

tan  

0.35 m 0.5 m

or

  35.0 

( FA )min  321 N

35.0° 

For Fvertical to be minimum, the perpendicular distance from its line of action to Point E must be maximum. Thus, apply (Fvertical)min at Point D, and then ( M E )  rD / E ( Fvertical ) min 196.2 N  m  (0.85 m)(Fvertical )min

or ( Fvertical )min  231 N

at Point D 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.3 It is known that a vertical force of 200 lb is required to remove the nail at C from the board. As the nail first starts moving, determine (a) the moment about B of the force exerted on the nail, (b) the magnitude of the force P that creates the same moment about B if α  10°, (c) the smallest force P that creates the same moment about B.

SOLUTION (a)

M B  rC /B FN

We have

 (4 in.)(200 lb)  800 lb  in.

or M B  800 lb  in.  (b)

By definition,

M B  rA /B P sin    10  (180  70)  120

Then

800 lb  in.  (18 in.)  P sin120

or P  51.3 lb  (c)

For P to be minimum, it must be perpendicular to the line joining Points A and B. Thus, P must be directed as shown. Thus

or or

M B  dPmin d  rA /B 800 lb  in.  (18 in.)Pmin Pmin  44.4 lb

Pmin  44.4 lb

20° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.4 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the smallest force applied at B that creates the same moment about D.

SOLUTION (a)

Fx  (300 N) cos 25  271.89 N F y  (300 N) sin 25  126.785 N F  (271.89 N) i  (126.785 N) j  r  DA   (0.1 m)i  (0.2 m) j

M D  r F M D  [ (0.1 m)i  (0.2 m)j ]  [(271.89 N)i  (126.785 N)j ]   (12.6785 N  m)k  (54.378 N  m)k  (41.700 N  m) k M D  41.7 N  m

(b)



The smallest force Q at B must be perpendicular to  DB at 45°  M D  Q ( DB) 41.700 N  m  Q(0.28284 m) Q  147.4 N

45.0° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.5 A 300-N force is applied at A as shown. Determine (a) the moment of the 300-N force about D, (b) the magnitude and sense of the horizontal force applied at C that creates the same moment about D, (c) the smallest force applied at C that creates the same moment about D.

SOLUTION (a)

See Problem 3.3 for the figure and analysis leading to the determination of MD M D  41.7 N  m

(b)



Since C is horizontal C  C i  r  DC  (0.2 m)i  (0.125 m)j MD  r  C i  C (0.125 m)k 41.7 N  m  (0.125 m)(C ) C  333.60 N

(c)

C  334 N 

The smallest force C must be perpendicular to DC; thus, it forms with the vertical

0.125 m 0.2 m   32.0 

tan  

M D  C (DC ); DC  (0.2 m)2  (0.125 m)2  0.23585 m 41.70 N  m  C (0.23585 m)

C  176.8 N

58.0° 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.6 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α  25°, determine the moment of the force about Point B by resolving the force into horizontal and vertical components.

SOLUTION Free-Body Diagram of Rod AB:

x  (9 in.)cos 65  3.8036 in. y  (9 in.)sin 65  8.1568 in.

F  Fx i  Fy j

rA /B

 (20 lb cos 25 ) i  ( 20 lb sin 25) j  (18.1262 lb)i  (8.4524 lb) j   BA  (3.8036 in.)i  (8.1568 in.)j

M B  rA /B  F  ( 3.8036i  8.1568j ) (18.1262i  8.4524j )  32.150k  147.852k  115.702 lb-in.

M B  115.7 lb-in. 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.7 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that α  25°, determine the moment of the force about Point B by resolving the force into components along AB and in a direction perpendicular to AB.

SOLUTION Free-Body Diagram of Rod AB:

  90   (65   25 )  50 

Q  (20 lb) cos50  12.8558 lb M B  Q(9 in.)  (12.8558 lb)(9 in.)  115.702 lb-in.

M B  115.7 lb-in. 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.8 A 20-lb force is applied to the control rod AB as shown. Knowing that the length of the rod is 9 in. and that the moment of the force about B is 120 lb · in. clockwise, determine the value of α.

SOLUTION Free-Body Diagram of Rod AB:

    25

Q  (20 lb) cos

and Therefore,

M B  (Q )(9 in.)

120 lb-in. (20 lb)(cos )(9 in.) 120 lb-in. cos   180 lb-in.

or

  48.190

Therefore,

  48.190  25

  23.2 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.9 Rod AB is held in place by the cord AC. Knowing that the tension in the cord is 1350 N and that c = 360 mm, determine the moment about B of the force exerted by the cord at point A by resolving that force into horizontal and vertical components applied (a) at point A, (b) at point C.

SOLUTION Free-Body Diagram of Rod AB: (a)

F 1350 N

cos  

AC  (450)2  (600)2  750 mm

450  0.6 750

sin  

600  0.8 750

 F  F cos i + F sin  j  (1350 N)0.6i  (1350 N)0.8j  (810 N)i  (1080N)j  rA /B  (0.45 m)i  (0.24 m)j

   MB  rA/ B  F  ( 0.45 i  0.24 j)  (810i  1080j)  486k  194.4k

 (291.6 N  m)k

(b)

MB 

292 N m

F  (810 N)i  (1080 N) j

rC /B  (0.36 m)j

MB  rC / B  F  0.36 j  (810 i  1080 j)  (291.6 N  m)k

M B  292 N m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.10 Rod AB is held in place by the cord AC. Knowing that c = 840 mm and that the moment about B of the force exerted by the cord at point A is 756 N·m, determine the tension in the cord.

SOLUTION Free-Body Diagram of Rod AB: MB  (756 N  m ) k

cos  

450 1170

AC  (450)2  (1080)2  1170 mm

sin  

1080 1170

 F  F cos i + F sin j 450 1080 Fi  Fj  1170 1170  rA /B  (0.45 m)i  (0.24 m)j    F M B  r A/ B  F  ( 0.45 i  0.24 j)  (450i  1080 j) 1170  ( 486k  108k )  378  F k    1170 

F 1170

Substituting for M B : 756  

378 F 1170

F  2340 N 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.11 The tailgate of a car is supported by the hydraulic lift BC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B, determine the moment of the force about A.

SOLUTION First note

d CB  (12.0 in.) 2  (2.33 in.)2  12.2241 in.

12.0 in. 12.2241 in. 2.33 in. sin   12.2241 in.

cos  

Then

FCB  FCB cos  i  FCB sin  j

and



125 lb [(12.0 in.)i  (2.33 in.)j ] 12.2241 in.

Now

M A  rB/A  FCB

where

r B /A  (15.3 in.) i  (12.0 in.  2.33 in.) j  (15.3 in.) i  (14.33 in.) j

Then

M A  [(15.3 in.)i  (14.33 in.)j ]

125 lb (12.0i  2.33j ) 12.2241 in.

 (1393.87 lb  in.)k

 (116.156 lb  ft)k

or M A  116.2 lb  ft

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.12 The tailgate of a car is supported by the hydraulic liftBC. If the lift exerts a 125-lb force directed along its centerline on the ball and socket at B , determine the moment of the force about A .

SOLUTION First note

d CB  (17.2 in.) 2  (7.62 in.)2  18.8123 in.

17.2 in. 18.8123 in. 7.62 in. sin   18.8123 in.

cos  

Then

FCB  (FCB cos  )i  (FCB sin  )j

and



125 lb [(17.2 in.)i  (7.62 in.)j ] 18.8123 in.

Now

M A  rB/A  FCB

where

rB /A  (20.5 in.) i  (4.38 in.) j

Then

MA  [(20.5 in.)i  (4.38 in.)j ]   (1538.53 lb in.)k  (128.2 lb ft)k

125 lb (17.2i  7.62j ) 18.8123 in. or M A  128.2 lb  ft

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.13 It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

SOLUTION Using (a): MC  y1 ( FAB )x  x1 ( FAB )y  7   24   (0.056 m)   2500 N   (0.042 m)   2500 N   25   25   140.0 N  m

(a) M C  140.0 N  m



Using (b): M C  y2 ( FAB ) x  7   (0.2 m)   2500 N  25    140.0 N  m

(b) M C  140.0 N  m

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.14 It is known that the connecting rod AB exerts on the crank BC a 2.5-kN force directed down and to the left along the centerline of AB. Determine the moment of the force about C.

SOLUTION Using (a): M C   y1 ( FAB ) x  x1 ( FAB ) y 7   24   (0.056 m)   2500 N   (0.042 m)   2500 N   25   25   61.6 N  m

(a) MC  61.6 N  m



Using (b): MC  y2 ( FAB )x 7   (0.088 m)   2500 N  25    61.6 N  m

(b)

M C  61.6 N  m

Copyright © McGraw-Hill Education. Permission required for reproduction or display.



PROBLEM 3.15 Form the vector products B × C and B × C, where B  B, and use the results obtained to prove the identity 1 1 sin  cos   sin (   )  sin (   ). 2 2

SOLUTION Note:

By definition,

Now

B  B (cos i  sin  j) B  B (cos  i  sin  j) C  C(cos  i  sin  j)

|B  C |  BC sin (   )

(1)

|B   C |  BC sin (   )

(2)

B  C  B(cos  i  sin  j)  C (cos  i  sin  j)  BC(cos  sin   sin  cos )k

and

(3)

B  C  B(cos  i  sin  j)  C(cos  i  sin  j)  BC(cos  sin   sin  cos )k

(4)

Equating the magnitudes of B  C from Equations (1) and (3) yields: BC sin(   )  BC(cos  sin   sin  cos )

(5)

Similarly, equating the magnitudes of B   C from Equations (2) and (4) yields: BC sin(   )  BC(cos  sin   sin  cos )

(6)

Adding Equations (5) and (6) gives: sin(   )  sin(   )  2cos  sin  1 1 or sin  cos   sin(   )  sin(   )  2 2

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.16 The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when (a) P = –8i + 4j – 4k and Q = 3i + 3j + 6k, (b) P = 7i – 6j – 3k and Q = –3i + 6j – 2k

SOLUTION (a)

We have

A  |P  Q |

where

P   8i  4j  4k Q  3i  3j  6k

Then

i j k    P Q 8 4 4 3 3 6  [(24  12) i  ( 12  48) j  ( 24  12) k ]  (36)i  (36) j  (36)k A  (36) 2  (36) 2  ( 36) 2

(b)

We have

A  |P  Q|

where

P  7i  6j  3k

or A  62.4 

Q   3i  6 j  2k

Then

i j k PQ  7 6 3 3 6  2  [(12  18) i  (9  14) j  (42  18) k]  (30)i  (23)j  (24)k A  (30) 2  (23) 2  (24)2

or A  44.8 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.17 A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal to, respectively, (a) 2i + 3j – 6k and 5i – 8j – 6k, (b) 4i – 4j + 3k and –3i + 7j – 5k.

SOLUTION (a)

AB |A  B |

We have

λ

where

A  2i  3j  6k B  5i  8j  6k

i Then

j

k 6

AB  2 3 5 8 6

 (18  15)i  (30  12) j  (16  15)k  (33i  18 j  31k ) and

| A  B|  ( 33) 2 ( 18) 2 ( 31) 2  2374

λ

(b)

(33i  18j  31k )

or λ   0.677i  0.369j  0.636k 

2374 A B |A  B|

We have

λ

where

A  4i  4j  3k B   3i  7j  5k

i Then

j

k

AB  4 4 3 3 7  5  (20  21) i  (9  20) j  (28  12) k  (  i  11 j  16 k)

and

2

2

| A  B|  ( 1)  (11)  (16)  378

λ

( i  11j  16k ) 378

or λ  0.0514i  0.566j  0.823k 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.18 A line passes through the points (12 m, 8 m) and (–3 m, –5 m). Determine the perpendicular distance d from the line to the origin O of the system of coordinates.

SOLUTION dAB  [12 m  ( 3 m)]2  [8 m  ( 5 m)]2  394 m

Assume that a force F, or magnitude F(N), acts at Point A and is directed from A to B. F  F  AB

Then where

 AB  

By definition,

rB  rA d AB 1 394

(15i  13j)

MO | rA  F|  dF

where

rA  ( 3 m) i + ( 5 m) j

Then

MO  [( 3 m) i + ( 5 m) j]  

F 394m

[(15 m) i  (13 m) j]

F [ 39k  75k ] N  m 394

 36   F k N  m  394 

Finally,

 36  F  d( F)   394  36 d m 394

d  1.184 m 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.19 Determine the moment about the origin O of the force F  4i  3j  5k that acts at a Point A. Assume that the position vector of A is (a) r  2i  3j  4k, (b) r  8i  6j  10k, (c) r  8i  6j  5k.

SOLUTION MO  r  F

(a)

i j k M O  2 3 4 4 3 5  (15  12) i  ( 16  10) j  ( 6  12)k

(b)

MO  3 i  26 j  18 k 

j k i 8 6 10 MO   4 3 5  (30  30) i  (40  40) j  (24  24)k

MO  0 

j k MO  8  6 5 4 3 5 i

(c)

 ( 30  15) i  (20  40) j  ( 24  24)k

MO   15i  20 j 

Note: The answer to Part (b) could have been anticipated since the elements of the last two rows of the determinant are proportional.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.20 Determine the moment about the origin O of the force F  2i  3j  4k that acts at a Point A. Assume that the position vector of A is (a) r  3i  6j  5k, (b) r  i  4j  2k, (c) r  4i  6j  8k.

SOLUTION MO  r  F

(a)

i j k M O  3 6 5 2 3 4  (24  15)i  (10  12) j  (9  12)k

(b)

i j k M O  1 4 2 2 3 4  (16  6) i  ( 4  4) j  (3  8)k

i

(c)

MO  9i  22j  21k 

M O  22i  11k 

j

k M O  4 6 8 2 3 4  ( 24  24) i  ( 16  16) j  (12  12)k

MO  0 

Note: The answer to Part (c) could have been anticipated since the elements of the last two rows of the determinant are proportional.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 3.21 Before the trunk of a large tree is felled, cables AB and BC are attached as shown. Knowing that the tensions in cables AB and BC are 555 N and 660 N, respectively, determine the moment about O of the resultant force exerted on the tree by the cables at B.

SOLUTION

d BA  ( 0.75 m) 2  (7 m) 2  (6 m) 2  9.25 m d BC  (4.25 m) 2  ( 7 m) 2  (1 m) 2  8.25 m

Have

TB A

 TB A

 BA 555 N  ( 0.75 m i  7 m j  6 m k ) dB A 9.25 m

TBA   (45 N)i  (420 N)j  (360 N)k

TBC  TBC

 660 N BC (4.25 m i  7 m j  k )  8.25 m dBC

TBC  (340 N)i  (540 N)j  (80 N)k R  TBA  TBC

 R  295 N  i  980 N j   440 N k M O  rB / O  R where rB / O  (7 m)j

i MO 

j

k

0 7 0 N m 295  980 440

  3080 N m i  2065 N m k MO   3080 N m i   2070 N m  k 

Copyright © McGraw-Hill Education. Permissio...