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Assignment 1 Problems - Statistics | ME 212, Assignments of Statics

Material Type: Assignment; Class: Statics; Subject: Mechanical and Materials Engr; University: Wright State University-Main Campus; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/18/2009

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ME 212 HOMEWORK
J. Slater: Assignment 1
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
Solve the following for as many variables as possible.
1)
2)
3)
4)
5)
6)
Find the unit direction vectors between the points defined by the vectors
F
1
and
F
2
(From point 1
to point 2).
1)
2)
3)
4)
1i
ˆ2j
ˆ
+()3j
ˆ2k
ˆ
+()+=
3 2.5i
ˆ4j
ˆ
+()=
1i
ˆ3j
ˆ
+()2j
ˆ3k
ˆ
+()ÿ=
2j
ˆ3k
ˆ
+()1i
ˆ3j
ˆ
+()ÿ=
8i
ˆ
–10k
ˆ
+()4i
ˆ5j
ˆ
()µ=
3i
ˆ2j
ˆ8k
ˆ
++()2i
ˆ8j
ˆ3k
ˆ
++()µ=
2i
ˆ8j
ˆ3k
ˆ
++()3i
ˆ2j
ˆ8k
ˆ
++()µ=
530i
ˆ
sin 30j
ˆ
cos+()=
ai
ˆ10j
ˆ
+()3j
ˆbk
ˆ
+()µ=
8i
ˆbj
ˆ
+()3i
ˆ4k
ˆ
+()ÿ=
a45i
ˆ
sin 45j
ˆ
cos+()b135i
ˆ
sin 135j
ˆ
cos+()+=
ai
ˆbj
ˆ
+()3i
ˆaj
ˆ
+=
ai
ˆbj
ˆ
+3i
ˆ9k
ˆ
+=
3ai
ˆbj
ˆcj
ˆ
++()41i
ˆ3j
ˆ
+()=
ai
ˆbj
ˆ
+()ci
ˆ1k
ˆ
+()µ1i
ˆ2j
ˆ3k
ˆ
++=
2i
ˆbj
ˆ
+()ci
ˆ1k
ˆ
+()ÿ2=
ai
ˆ1i
ˆ2j
ˆ
+()2j
ˆ3k
ˆ
+()µ[]ÿ2j
ˆ3k
ˆ
+()1i
ˆ1i
ˆ2j
ˆ
+()µ[]ÿ=
F10i
ˆ0j
ˆ
+= F2
,1i
ˆ2j
ˆ
+=
F11i
ˆ2j
ˆ
+= F2
,0i
ˆ0j
ˆ
+=
F16i
ˆ14j
ˆ
+= F2
,1i
ˆ2j
ˆ
+=
F11i
ˆ2j
ˆ3k
ˆ
++=F2
,9i
ˆ3j
ˆ2k
ˆ
++=

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ME 212 HOMEWORK

J. Slater: Assignment 1

Solve the following for as many variables as possible.

Find the unit direction vectors between the points defined by the vectors F 1

and F 2

(From point 1

to point 2).

1 i

2 j

( + ) 3 j

2 k

3 2.5 i

4 j

1 i

3 j

( + ) 2 j

3 k

ÿ ( + ) =

2 j

3 k

( + ) 1 i

3 j

ÿ ( + ) =

8 i

  • 10 k

( + ) 4 i

5 j

μ ( – ) =

3 i

2 j

8 k

( + + ) 2 i

8 j

3 k

μ ( + + ) =

2 i

8 j

3 k

( + + ) 3 i

2 j

8 k

μ ( + + ) =

5 30 i

sin 30 j

( +cos ) =

a i

10 j

( + ) 3 j

b k

μ ( + ) =

8 i

b j

( + ) 3 i

4 k

ÿ ( + ) =

a 45 i

sin 45 j

( +cos ) b 135 i

sin 135 j

  • ( +cos ) =

a i

b j

( + ) 3 i

a j

a i

b j

  • 3 i

9 k

3 a i

b j

c j

( + + ) 4 1 i

3 j

a i

b j

( + ) c i

1 k

μ ( + ) 1 i

2 j

3 k

2 i

b j

( + ) c i

1 k

ÿ ( + ) = 2

a i

1 i

2 j

( + ) 2 j

3 k

ÿ [ μ( + )] 2 j

3 k

( + ) 1 i

1 i

2 j

= ÿ[ μ( + )]

F

1

0 i

0 j

= + F

2

, 1 i

2 j

F

1

1 i

2 j

= + F

2

, 0 i

0 j

F

1

6 i

14 j

= + F

2

, 1 i

2 j

F

1

1 i

2 j

3 k

= + + F

2

, 9 i

3 j

2 k