Eng1010 - Sample Test 4

Transcription

Eng1010 - Sample Test 4
Eng1010 - Sample Test 4
1. An adjustable tow bar connecting the tractor unit H with the landing gear J of a large
aircraft is shown in the …gure. Adjusting the height of the hook F at the end of the tow bar
is accomplished by the hydraulic cylinder CD activated by a small hand pump (not shown).
The triangular linkage ABC pivots on the trailer body at A. For the position of the triangular
linkage ABC shown (AC is vertical), calculate the force required in the hydraulic cylinder
CD to maintain the rig in this position. The tow bar assembly has a total mass of 50 kg with
center of mass at G. Neglect the mass of the hydraulic cylinder, linkage ABC and wheels at
B. Note: the lengths of AB and AC are 500 mm, and \ABC = \ACB = 30 .
FCD
298 N
2. The excavator scoop shown below supports a vertical load of F = 1800 N located at G.
Neglecting the mass of all other elements, determine the axial force (FDH ) in the hydraulic
cylinder required to hold the scoop in static equilibrium. Assume: 1) link HL is rigidly
attached to the excavator; and 2) points C, D and K lie in a vertical line.
FDH
1701 (T ) ( N)
Eng1010 - Sample Test 4
3. In the frame shown below a cable is attached to element AC at B. The cable wraps around
the pulley which is attached to element CE at D. The cable is then pulled vertically down
by a force of 170 N. Determine the magnitude of the reactions at pin connections A, C and
E in the frame. Assume the pulley has radius 60 mm and neglect the mass of all elements in
the frame.
4. The right angle pipe OAB is rigidly attached to a wall at O: The tension in cable
750 N. The force F~ applied at A has magnitude 500 N and is alligned with the y axis.
Cartesian vector notation, determine the reactions at O required for static equilibrium.
section AB of the pipe is in a plane that is parallel to the y; z plane, and is rotated 30
from the x; z plane.
F~R = 158:25^{ 231:25 |^
50:25k^
( N)
~ R = 526:86^{ + 190:07^
M
|
306:7k^
( N m)
BC is
Using
Note:
down
Eng1010 - Sample Test 4
5. Determine the x; y; z components of the reactions exerted by the nut and bolt on the loaded
bracket at O to maintain equilibrium. Write the reactions in Cartesian vector notation in
the table below.
F~R =
1:028^{ 0:852^
| + 1:2k^
( N)
~ R = 0:117^{ + 0:360^
M
| + 0:245k^
( N m)
6. The bent rod lies in the horizontal x; y plane and is supported by a journal bearing at A, and
cables BC and BD. Two forces F~1 and F~2 are applied to the rod. Determine the reactions
~ R at A and the tensions TC and TD in cables BC and BD, respectively. Since
F~R and M
this is a 3D problem you can use the given diagram as the FBD. Notes: (1) Points B, C and
D all lie in a plane parallel to the y; z plane; and (2) a rod can rotate in and slide through a
journal bearing.
Eng1010 - Sample Test 4
7. After a long trip a 100 kg crate, with center of mass at G, is resting on a slope of angle = 10 .
The coe¢ cient of static friction between the slope and the crate is s = 0:6. A horizontal
force P~ is applied to the crate as shown. Determine the minimum required magnitude of P
to cause motion of the crate. Note: All dimensions are in meters.
8. Use the method of sections to determine the forces in members DE, DI and IJ in the
truss shown below.
Indicate whether the member is in tension (T ) or compression (C)
when recording the forces in the table. You may use the given diagram as the FBD when
determining the reactions at the supports only.
Eng1010 - Sample Test 4
9. Use the method of sections to determine the forces in members DF , EF and EG in the truss
shown below.
Indicate whether the member is in tension (T ) or compression (C) when
recording the forces in the table below.
FDF = 28 (C) ( kN)
FEF = 5 (C) ( kN)
FEG = 32 (T ) ( kN)
10. Determine the forces in members AB, BC, BD, BE, CE and DE in the truss shown below.
One force is applied at C and one force is applied at E. Indicate whether the member is in
tension (T ) or compression (C) when recording the forces in the table below.
FAB = 320 (T ) ( kN)
FBD = 240 (C) ( kN)
FCE = 100 (C) ( kN)
FBC = 0
FBE = 400 (T ) ( kN)
FDE = 215 (C) ( kN)