www.wiziq.com/tests/mechanical%20engineering/1
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FE Review Manual - Michael R. Lindeburg, ,PE
Professional Publications, Inc; ISBN 1-888577-53-3 (pbk); www.ppi2pass.com
Belmont,CA 94002
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IBM Test
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www.identifythebest.com/WebAssessment/ibmCalibration.start?reqId=1026&ioId=dditest-IBM-1CFF33F5-7483-4B8E-BB94-29EE86F4B8A6
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Topic 1: Units and Fundamental Constants
Example 1.1
Calculate the weight, in lbf, of a 1.0 lbm object in a gravitational field of 27.5 ft/sec^2
Solution
From Eq. 1.4; F in lbf =
[(m in lbm)(a in ft/sec^2)]/gc in lbm-ft/lbf-sec^2
g(c) - conversion constant
F=ma/gc=
[(1 lbm)(27.5 ft/sec^2)]/[32.2(lbm-ft/lbf-sec^2)]
=0.854 lbf
Example 1.2
A rocket with a mass of 4000 lbm travels at 27,000 ft/sec. What is its kinetic energy in ft-lbf?
Solution
From Eq. 1.5; Kinetic Energy E = mv^2/2gc[in ft-lbf]
Ek = mv^2/2gc
= [(4000 lbm)(27,000 ft/sec)^2]/[(2)(32.2 lbm-ft/lbf-sec^2)]
= 4.53x10^10 ft-lbf
Example 1.3
A 10 kg block hangs from a cable. What is the tension in the cable?(Standart gravity is 9.81 m/s^2.)
Solution
F = mg = (10 kg)(9.81 m/s^2)
= 98.1 kg.m/s^2 = 98.1 N
Topic 7: Fluid Mechanics
Diagnostic Examination
1.10.0 L of an incopressible liquid exert a force of 20 N at the earth surface. What force would 2.3 L of this liquid exert on the surface of the moon? The gravitational acceleration of the surface of the moon is 1.67 m/s^2.
Solution 1:
The deencity of incompresible fluid is contant. It is not a function of pressure or temperature. The mass is calculated from Newton's second law.
m = F/a = 20 N/(9.81 m/s^2)= 2.04kg
The density is p = m/v = 2.04 kg/10 L
= 0.204 kg/L
The force of 2.3 L of this liquid on the moon is
F = m/a = pVa =
(0.204 kg/L)(2.3 L)(1.67 m /s^2) = 0.784 N
2. A Sliding-plate viscometer is used to measure the vilosity of a Newtonian fluid. A force of 25 N is required to keep the top plate moving at a constant velocity of 5 m/s. What is the viscosity of the fluid?
Solution 2: The force requireto maintain the velocity is given by
F = mvA/δ
Rearranging to solve or viscosity directly,
m = Fd/vA = (25 N)(0.001 m)/(5 m/s)(0.5 m)(0.25 m)
= 0.04 N.s/m^2
3. A 2 mm (inside diameter) glass tube is placed in a container of mercury. An angle of 40 degree is measured as illustrated. The density and surface tension of mercury are 13 550 kg/m^3 and 37.5 x 10^-2 N/m,resectively. How high will the mercury rise or be depressed in the tube as a result of capilarry action?
Solution 3:
The contact angle is the angle made by the liquid with the wetted tube wall. if this angle is greater thn 90 degree, the fluid level in the tube will be depressed. The equation of capillary action is
h = 4σ cos β/ρd(tube)g (d-diameter)(σ - surface tension)
=(4)(37.5 x 10 ^-2 N/m)cos 140°/(13 550 kg/m^3)(0.002 m)(9.81 m/s^2)=
-4,32x10^-3 m (-4.3 mm)(depression)
4. An open water manometer is used to measure the pressure in a tank. The tank is half filled with 50 000 kg of a liquid chemical, that is not miscible in water. The manometer tube is filled with liquid chemical. What is the pressure in the tank relative to the atmospheric pressure?
Solution 4:
Calculate the density of the chemical from the volume and mass. The total volume of the tank is
V = (4/3)pi r^3 + pi r^2(L - 2r) =
(4/3)pi(2m)^3 - pi(2m)^2(10 m - (2)(2 m)=108.9m^3
Since the contents have a mass of 50 000 kg and the tan is half-full, the density of the chemical is
ρ(chem) = 50 000 kg/(1/2)(108.9 m^3)
= 918.3 kg/m^3
The relative pressure is ρ(0) - ρ(2) = ρ(water)gh(2) - ρ(chem)gh(1) =
(1000 kg/m^3)(9.81 m/s^2)(0.40 m - 0.05 m) - (918.3kg/m^3)(9.81m/s^2)(0.225m)=1407 Pa (1.4kPa)
5.A gravity dam has the crosssection shown. What is the magnitude of the resultant water force (per meter of width) acting on the face of the dam?
Solution 5:
The maximum water pressure at the base of the dam is
p(max) = (ρ)gh = (1000 kg/m^3)(9.81 m/s^2)(50m)
= 490 500 Pa
The total horizontal force of the dam face is equal to the area of the pressure distribution.
F(x) = (1/2)p(max)A = (1/2)p(max)Lw
Per unit width, w,
F(x)/w = (1/2)(490 500 Pa)(20 m + 30 m)
= 12.3 x 10 ^6 N/m (12.3 MN/m)
The total verical force acting on the dam face is equal to the weight of the water column above the face.
F(y) = ρgV =ρgAw
F(y)/w = (1000 kg/m^3)(9.81 m /s^2) x [(30 m)(20 m)+(1/2)(20 m)(20 m)]=
7.85 x 10^6 N/m (7.85 MN/m)
The resultant force is
F = √[(F(x)^2)+ F(y)^2)]
= √[(12.3 MN/m)^2 + (7.85 MN/m)^2] = 14.6 MN/m
6. A 6m x 6m x6m vented cubical tank is half-filled with water,the remaining space is filled with oil (SG = 0.8). What is the total force on one side of the tank?
Solution 6:
ρ(oil) = (SG)ρ(water)
= (0.8)(1000 kg/m^3)= 800 kg/m^3
Since the oil has a lowe specific gravity than the water,it will float on the water. The pressure at the oil/water interface is due to the weight of the oil above the interface.
p1 = ρ(oil)g(h/2)= (800 kg/m^3)(9.81 m/s^2)(6 m/2)
= 23 544 Pa
The pressre at the bottom of the box is
p2 = p1 + ρ(water)g(h/2)
= 23 544 Pa + (1000 kg/m^3)(9.81 m /s^2)(6 m /2)
= 52 974 Pa
The total force acting on the face of the box is
F = (1/2)(23 544 Pa)(3 m)(6 m)
+ (23 544 Pa)(3 m)(6 m)+ (1/2)(52 974 Pa - 23 544 Pa)(3 m)(6 m)
= 9.01 x 10^5 N (900 kN)
7.
