Solution






1
3
0
1
20.1104 Intro to Eng. Analysis, Studio version
Fall 1998 – Test #3 Friday Nov. 6, 1998
Instructions: 60-min time limit. You can use Maple to do the problems and check your answers, but clearly show your setup and solution for each problem on the test pages – read each
problem to see what we want specifically.
1.
(10 points) Using the cofactor method, show us the setup of your method to find the determinant of the following 4 x 4 matrix. Find the numerical value of the determinant; you may do the determinants of any 3 x 3 matrices using Maple.
0
4
5
1
2
0
2
1


1
3
3



2


Using the cofactor method and using row one as the “pivot” row, we have:
4 0 1 3 4 1 3 4 0
1 (

1 )
1

1
5
1
2
1

3
2 
0

2 (

1 )
1

3 
0
1 5
1

3
2 
(

3 )(

1 )
1

4 
0
1 5
1
2
1
= 1(1)(35) + 0 + 2(1)(62) + (-3)(-1)(13)
= 198
2.
(20 points) A person wearing a 5 lb Halloween mask on his 8 lb head is bending over at the waist to pick up some candy. Assume that center of gravity of the total weight of the mask and head is located at a point in space with coordinates (1, 2, 6) ft. The weight can be assumed to act along the z-axis, which is oriented vertically.
(a)
Compute the moment of the mask’s and head’s weight about the point
A (0.5, 1, 3) ft, which is located in the person’s lower back.
F = -13 lb k r = (1,2,6) ft – (0.5,1,3) ft r = 0.5 i + 1 j + 3 k ft
M
A
= r x F = (0.5 i + 1 j + 3 k ft) x (-13 lb k )
M
A
= -13 ft-lb i + 6.5 ft-lb j
(b) Write down the equivalent system that would exist at point A in the person’s lower back if we are just considering the weight of the mask and head.
The equivalent system is comprised of the resultant force and moment.
F = -13 lb k
M
A
= -13 ft-lb i + 6.5 ft-lb j
1

Solution
3.
(20 points) A crowbar is subjected to a vertical force of P = 25 lb at the grip, but it requires a force of F = 155 lb at the claw to pull out the nail. The 14-inch length of the crowbar is from the hand to point B in the diagram. y x
14 ”
20

1.5 “
70

P =25 lb
B
(a) Find the moment about A due to the 25 lb force P .
Using the “Fd” method (see diagram above for help with geometry).
M
A
= Fd = 25 lb {14 in. cos (20

) + 1.5 in. cos (70

)} = 341.7 in-lb k
(b) Find the moment about A of a 155 lb force F .
Using the “Fd” method.
M
A
= Fd = 155 lb cos (30

) (3 in.) = -402.7 in-lb k
(c) Compare your answers to (a) and (b) and state whether the crowbar is pulling out the nail under these conditions.
Since the applied moment found in part (a) is less than the “pull-out” moment found in part
(b), then the crowbar is not pulling out the nail under these conditions.
2

Solution
4.
(10 points) The rod shown is supported by two brackets at points A and B . Determine the moment of the force F about the axis AB . The force F = (-600 i + 200 j – 300 k ) N. r
AF
= 0.6 m i + 0.3 m k
r
AB
= 0.4 m i + 0.2 m j
F = -600 N i + 200 N j – 300 N k e
AB

( 0 .
4 i

0 .
2 j ) m
( 0 .
4 m )
2 
( 0 .
2 m )
2
e
AB
= 0.8944 i + 0.4472 j
M
A
= r
AF
x F = (0.6 m i + 0.3 m k) x (-600 N i + 200 N j – 300 N k )
M
A
= -60 N
 m i + 120 N
 m k
M
AB
= M
A
 e
AB
= (-60 N
 m i + 120 N
 m k )

(0.8944 i + 0.4472 j )
M
AB
= -53.67 N
 m
5.
(40 points) Draw correct free body diagrams for the following four problems; do not solve the problems for the reactions at the supports; just draw correct FBDs!
(a) A force and a moment are being applied to an L - shaped member connected to the support by a built-in (fixed) joint at A . Find the reactions at point A.
3

F
1
=200 lb
Solution
(b) Determine the magnitude of the pin reactions at A for the structure shown being loaded by a force of 200 lb and a couple of 100 ft-lb. L = 2 ft.
(c) An open-end wrench is being used to tighten a bolt at point A. The workman’s hands can be assumed to apply only forces as shown at points B and C .
The wrench has stopped turning when the loads are applied. Find the reactions where the bolt meets the wrench.
(d) The link shown is pin-connected at A and rests against a perfectly smooth support at B .
Computed the reactions at the pin A .
4

Solution
BD
B
D
B
D
First, for convenience, re-draw the fbd with the y-axis parallel to the line which connects points B and D.
Remember that the connections at B and D are pin joints.
Second, apply the conditions of force equilibrium to the linkage.

F x
= 0: B x
– D x
= 0

F y
= 0: B y
– D y
= 0
(1)
(2)
From these we establish that:
B x
= D x
B y
= D y
(3)
(4)
Next apply the condition of moment equilibrium:

M
B
= 0: - D x
(d) = 0 (5)
Thus, D x must be equal to zero and since
B x
= D x
, (3) d from (3), then B x must also be equal to zero .
Finally we are left with just B y
and D y
as non-zero reaction forces, thus member BD must be a
“two-force member.”
5