Homework 7 Solutions - Basic Mechanics I | CE 29700, Assignments of Civil Engineering

Material Type: Assignment; Class: Basic Mechanics I (Statics); Subject: CE-Civil Engineering; University: Purdue University - Main Campus; Term: Fall 2007;

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2011/2012

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COSMOS: Complete Online Solutions Manual Organization System
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
Homework Solution #7
(Due 10/10/07)
Problem 3.68, 3.74, 3.109
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Vector Mechanics for Engineers: Statics and Dynamics , 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Homework Solution

(Due 10/10/07)

Problem 3.68, 3.74, 3.

Vector Mechanics for Engineers: Statics and Dynamics , 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 3, Problem 68.

A steel plate is acted upon by two couples as shown. Determine ( a ) the

moment of the couple formed by the two 40-N forces. ( b ) the value of

if the resultant of the two couples is 8 N m counterclockwise and

d 820 mm, ( c ) the perpendicular distance between the two 24-N

forces if the resultant of the two couples is zero.

Chapter 3, Solution 68.

( a ) Have

1 1 1

Md F

Where

1 1

d  0.6 m and F 40 N

   1

M  0.6 m 40 N

or

1

M  24.0 N m

( b ) Have

Total 1 2

MMM

   

8 N m  24.0 N m  0.820 m cos  24 N

 cos 0.

or   35.6

( c ) Have

1 2

MM  0

 

2

24 N m  d 24 N  0

or

2

d 1.000 m

Vector Mechanics for Engineers: Statics and Dynamics , 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 3, Solution 74.

Have

1 2

MMM

Where

DE D

M  r  F

1

   

  0.7 m k  80 N j

 

 56.0 N m i

And

GF B

M  r  F

2

Now      

2 2 2

0.300 m 0.540 m 0.350 m

BF

d    

0.710 m

Then

B BF B

F   F

     

 

0.300 m 0.540 m 0.350 m

71 N

0.710 m

i j k

     

  30 N i  54 N j  35 N k

       

2

0.54 m 30 N 54 N 35 N

M j i j k

   

 18.90 N m i  16.20 N m k

Finally      

56.0 N m 18.90 N m 16.20 N m

M i i k

   

 74.9 N m i  16.20 N m k

and    

2 2

M  74.9 N m  16.20 N m

 76.632 N m or M  76.6 N m

cos cos cos

x y z

or 12.20 90.0 77.

x y z

Vector Mechanics for Engineers: Statics and Dynamics , 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,

Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell

Chapter 3, Problem 109.

As four holes are punched simultaneously in a piece of aluminum sheet

metal, the punches exert on the piece the forces shown. Knowing that the

forces are perpendicular to the surfaces of the piece, determine ( a ) the

resultant of the applied forces when   45  and the point of

intersection of the line of action of that resultant with a line drawn

through points A and B , ( b ) the value of  so that the line of action of

the resultant passes through fold EF.