Answer:
The weight of the station becomes [tex]3.756\times 10^{6}N[/tex]
Explanation:
Since the acceleration due to gravity decreases with increase in height we conclude that at a height of 350 kilometers the weight of the material will be lesser.
At the ground we have
[tex]W=mass\times g_{surface}\\\\\therefore mass=\frac{W}{g_{surface}}\\\\mass=\frac{4.22\times 10^{6}N}{9.81}\\\\\therefore mass=430173.292kg[/tex]
Now we know that the variation of acceleration due to gravity with height above surface of earth is given by
[tex]g(h)=g_{surface}(1-\frac{2h}{R})[/tex]
where R = 6371 km is Radius of earth
Applying values we get the value of 'g' at height of 350 kilometers equals
[tex]g(350)=9.81\times (1-\frac{2\times 350}{6371})=8.732ms^{-2}[/tex]
hence the weight in orbit becomes
[tex]W_{orbit}=mass\times g_{orbit}\\\\W_{orbit}=430173.292\times 8.732\\\\ \therefore W_{orbit}=3.756\times 10^{6}N\\[/tex]
convert
a) 760 miles/hour to meters/second
b) 921 kg/cubic meter to pound mass/cubic foot
c) 5.37 x 10^3 kJ/ min to hp.
Answer:
(a)[tex]1.308\times 10^{-4}m/sec[/tex]
(b)57.33831 pound/cubic feet
(c)120.1095 hp
Explanation:
We have
(a) 760 miles / hour
We know that [tex]1\ mile\ =0.00062m[/tex]
And 1 hour = 60×60=3600 sec
So [tex]760miles/hour=\frac{760\times 0.00062meter}{3600sec}=1.308\times 10^{-4}m/sec[/tex]
(b) 927 kg/cubic meter to mass/cubic foot
We know that 1 kg = 2.20 pound
So 921 kg = 921×2.20=2026.2 pound
We know that 1 cubic meter = 35.31 cubic feet
So 57.33831 pound / cubic feet
(c) We have to convert to hp
[tex]5.37\times 10^3kj/min=\frac{5.37\times 1000kj}{60sec}=89.5kj/sec[/tex]
We know that 1 kj /sec = 1.341 hp
So 89.5 kj/sec = 89.5××1.341=120.1095 hp
You are to define and illustrate the following in regards to a 4 bar linkage mechanism: a) Illustrate and briefly define an open-loop 4 bar linkage system. b) illustrate and briefly define a closed-loop 4 bar linkage system.
Answer:
Answered
Explanation:
Open- loop 4 bar linkage system:
Open- loop 4 bar linkages are not mechanically constrained, meaning in this type of linkages the degree of freedom is are more than 1.
DOF> 1 example = industrial robots, epicyclic gear trains etc.
Closed loop four bar linkage system:
These types are linkages are mechanically constrained and have degree of freedom as one. examples, four bar chains, slider crank mechanism.
DOF=1
Which two forms of energy are usually converted directly from solar radiation for houses in temperate climates. a. Thermal and Electrical b. Chemical and Nuclear c. Mechanical and Electrical
Answer:
The correct answer is option 'a': Thermal and Electrical energy
Explanation:
We know that the solar radiation that we receive from sun provides us heat. Thus the solar radiation is a natural source of thermal energy which can be utilized in solar cooker's to cook food or to warm water to generate steam in Solar thermal power plants. The steam generated is used to drive a turbine and hence produce electricity.
As we know that solar radiation is converted into directly usable forms of energy by the solar panels. The solar panels work on the principle of photo-electric effect in which light energy is directly converted into electrical energy to run our electrical devices at home such as light bulbs, fans, e.t.c. An excellent application of this principle is the international space station that orbits the earth and is fully powered by solar energy.
Describe the differences, if any, between fin efficiency and fin resistance.
Answer:
[tex]\eta =\dfrac{1}{mL}[/tex]
[tex]R=\sqrt {\dfrac{KA}{hP}}[/tex]
Explanation:
Fin efficiency:
Fin efficiency is the ratio of actual heat transfer through fin to the maximum heat transfer through fin.
For infinite long fin:
Actual heat transfer
[tex]q=\sqrt{hPKA}\Delta[/tex]
So the efficiency of fin
[tex]\eta =\dfrac{1}{mL}[/tex]
Where
[tex]m =\sqrt{\dfrac{hP}{KA}}[/tex]
h is heat transfer coefficient,P is the perimeter,K is the thermal conductivity and A is the cross sectional area.
[tex]q=\sqrt{hPKA}\Delta[/tex]\
From equation we can say that fin resistance
[tex]R=\sqrt {\dfrac{KA}{hP}}[/tex]
Thermal resistance offer resistance to flow of heat.
Briefly describe the operation and use of strain gauges
Answer:
Strain gauge:
Strain gauge is a sensor. It is use to measure the strain of the material with help of electric resistance. A strain gauge is attached to the work piece and when any change in dimensions of work piece occurs ,due to this electric resistance of strain gauge also changes. Then by using a proper electric circuit strain of a material can be measured. Depending upon the situations one or more than one strain gauges is attached to the work piece to measure the strain .
Use of strain gauges
1. To measure the stress of a structure.
2. Use for testing of ships ,vehicle ,dams etc.
A 25 lb sacrificial Mg anode is attached to the steel hull of a container ship. If the anode completely corrodes within 3 months, what is the average current produced by the anode?
Answer:
The average current will be of 6.36 A.
Explanation:
The anode capacity of magnesium is C = 550 A*h/lb
A month has 30 days.
A day has 24 hours.
Therefore 3 months have:
t = 3 * 30 * 24 = 2160 hours.
The average current is then:
I = C * m / t
I = 550 * 25 / 2160 = 6.36 A
The average current will be of 6.36 A.
The velocity of a point mass that moves along the s-axis is given by s' = 40 - 3t^2 m/s, where t is in seconds. Find displacement of the particle from t = 2 s to t 6 s.
Answer:
The displacement is -48m.
Explanation:
Velocity is the rate of change of displacement. So, the instantaneous displacement is calculated by the integration of velocity function with respect to time. Displacement in range of time is calculated by integrating the velocity function with respect to time with in time range.
Given:
Velocity along the s-axis is
[tex]s{}'=40-3t^{2}[/tex]
time range is t=2s to t=6s.
Calculation:
Step1
Displacement in the time range t=2s to t=6s is calculated as follows:
[tex]\frac{\mathrm{d}s}{\mathrm{d}t}=40-3t^{2}[/tex]
[tex]ds=(40-3t^{2})dt[/tex]
Step2
Integrate the above equation with respect to time with the lower limit as 2 and upper limit as 6 as follows:
[tex]\int ds=\int_{2}^{6}(40-3t^{2})dt[/tex]
[tex]s=(40\times6-40\times2)-3(\frac{1}{3})(6^{3}-2^{3})[/tex]
s=160-208
s=-48m
Thus, the displacement is -48m.
The displacement of a point mass moving along the s-axis from t = 2 s to t = 6 s, with a given velocity function s' = 40 - 3t^2 m/s, is calculated by integrating the velocity to get the position function and then evaluating it between the two time limits, resulting in a displacement of 152 meters.
Explanation:The question requires finding the displacement of a point mass moving along the s-axis between t = 2 s and t = 6 s. The velocity is given as s' = 40 - 3t^2 m/s. To find the displacement, we will integrate the velocity function with respect to time from 2 to 6 seconds.
To integrate s' = 40 - 3t^2, we get:
Integral of 40 dt is 40tIntegral of -3t^2 dt is -t^3The displacement (s) is thus
s = 40t - t^3 evaluated from t = 2 to t = 6. Inserting the limits, we subtract the value at t = 2 from the value at t = 6:
s(6) - s(2) = (40(6) - 6^3) - (40(2) - 2^3) = 240 - 216 - 80 + 8 = 152 m
So, the displacement of the particle from t = 2 s to t = 6 s is 152 meters.
The flow of a real fluid has (more —less - same ) complexity of an ideal fluid, owing to the phenomena caused by the existence of (viscosity-pressure drop- friction
Answer:
The flow of a real fluid has more complexity as compared to an ideal fluid owing to the phenomena caused by existence of viscosity
Explanation:
For a ideal fluid we know that there is no viscosity of the fluid hence the boundary condition need's not to be satisfied and the flow occur's without any head loss due to viscous nature of the fluid. The friction of the pipe has no effect on the flow of an ideal fluid. But for a real fluid the viscosity of the fluid has a non zero value, the viscosity causes boundary layer effects, causes head loss and also frictional losses due to pipe friction hugely make the analysis of the flow complex. The losses in the energy of the flow becomes complex to calculate as frictional losses depend on the roughness of the pipe and Reynolds number of the flow thus increasing the complexity of the analysis of flow.
A disk with radius of 0.4 m is rotating about a centrally located axis with an angular acceleration of 0.3 times the angular position theta. The disk starts with an angular velocity of 1 rad/s when theta = 0. Determine the magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
Answer:
a₁= 1.98 m/s² : magnitud of the normal acceleration
a₂=0.75 m/s² : magnitud of the tangential acceleration
Explanation:
Formulas for uniformly accelerated circular motion
a₁=ω²*r : normal acceleration Formula (1)
a₂=α*r: normal acceleration Formula (2)
ωf²=ω₀²+2*α*θ Formula (3)
ω : angular velocity
α : angular acceleration
r : radius
ωf= final angular velocity
ω₀ : initial angular velocity
θ : angular position theta
r : radius
Data
r =0.4 m
ω₀= 1 rad/s
α=0.3 *θ , θ= 2π
α=0.3 *2π= 0,6π rad/s²
Magnitudes of the normal and tangential components of acceleration of a point P on the rim of the disk when theta has completed one revolution.
We calculate ωf with formula 3:
ωf²= 1² + 2*0.6π*2π =1+2.4π ²= 24.687
ωf=[tex]\sqrt{24.687}[/tex] =4.97 rad/s
a₁=ω²*r = 4.97²*0.4 = 1.98 m/s²
a₂=α*r = 0,6π * 0.4 = 0.75 m/s²
A 400-m^3 storage tank is being constructed to hold LNG, liquefied natural gas, which may be assumed to be essentially pure methane. If the tank is to contain 90% liquid and 10% vapor, by volume, at 150 K, what mass of LNG (kg) will the tank hold? What is the quality in the tank?
Answer:
mass of LNG: 129501.3388 kg
quality: 0.005048662
Explanation:
Volume occupied by liquid:
400 m^3*0.9 = 360 m^3
Volume occupied by vapor
400 m^3*0.1 = 40 m^3
A figured with thermodynamic properties of saturated methane is attached. Notice that a liquid-gas mixture is present
For liquid phase specific volume (vf) at 150 K is 0.002794 m^3/kg and for vapor phase specific volume (vg) is 0.06118 m^3/kg
From specific volume definition:
vf = liquid volume/liquid mass
liquid mass = liquid volume/vf
liquid mass = 360 m^3/0.002794 m^3/kg
liquid mass = 128847.5304 kg
vg = vapor volume/vapor mass
vapor mass = liquid volume/vg
vapor mass = 40 m^3/0.06118 m^3/kg
vapor mass = 653.8084341 kg
total mass = 128847.5304 kg + 653.8084341 kg = 129501.3388 kg
Quality is defined as the ratio between vapor mass and total mass
quality = 653.8084341 kg/129501.3388 kg = 0.005048662
What system of units is superior, SI, BG, or EE? Why?
Answer:
SI unit of system is superior.
Explanation:
The metric system is used internationally for measurement of unit.This metric system of unit is easier to understand .
Metric system of unit also known as international system of unit.The international system of unit is also represent in short form like SI unit.This unit system is internationally accepted.
So SI unit of system is superior.
What is the governing ratio for thin walled cylinders?
Answer:
The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:
if [tex]\frac{radius}{thickness} >10[/tex] then you have a thin walled cylinderor using the diameter:
if [tex]\frac{diameter}{thickness} >20[/tex] then you have a thin walled cylinderWhat is the definition of a fluid?
Answer: Basically it's a substance with no shape and it's one of the different states of water.
Explanation:
A cylindrical specimen of some metal alloy having an elastic modulus of 123 GPa and an original cross-sectional diameter of 3.3 mm will experience only elastic deformation when a tensile load of 2340 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Answer:
maximum length of the specimen before deformation = 200 mm
Explanation:
Hi!
If we have a cylinder with length L₀ , and it is elasticaly deformed ΔL (so the final length is L₀ + ΔL), the strain is defined as:
[tex]\epsilon =\frac{\Delta L}{L_0}[/tex]
And the tensile stress is:
[tex]\sigma = \frac{F}{A}\\F = \text{tensile load}\\A = \text{ cross section area}[/tex]
Elastic modulus E is defined as:
[tex]E = \frac{\sigma}{\epsilon }[/tex]
In this case ΔL = 0.45 mm and we must find maximum L₀. We know that A=π*r², r=(3.3/2) mm. Then:
[tex]\sigma=\frac{2340N}{\pi (1.65 \;mm)^2}=273.68 MPa = [/tex]
[tex]E=123\;GPa=\frac{L_0 \;(273.68MPa)}{0.45\;mm } \\ L_0 = 200\; mm[/tex]
A rigid tank with a volume of 0.5 m3 contains air at 120 kPa and 300 K. Find the final temperature after 20 kJ of heat is added to the air using (a) constant specific heats and (b) ideal gas tables.
Answer:
T=340. 47 K
Explanation:
Given that
Volume of tank =0.5 [tex]m^3[/tex]
Pressure P=120 KPa
Temperature T=300 K
Added heat ,Q= 20 KJ
Given that air is treated as ideal gas and specific heat is constant.
Here tank is rigid so we can say that it is constant volume system.
We know that specific heat at constant volume for air
[tex]C_v=0.71\ \frac{KJ}{kg.K}[/tex]
We know that for ideal gas
P V = m R T
For air R=0.287 KJ/kg.K
P V = m R T
120 x 0.5 = m x 0.287 x 300
m=0.696 kg
[tex]Q=mC_v\Delta T[/tex]
Lets take final temperature of air is T
Now by putting the values
[tex]Q=mC_v\Delta T[/tex]
[tex]20=0.696\times 0.71\times (T-300)[/tex]
T=340. 47 K
So the final temperature of air will be 340.47 K.
The kinetic energy correction factor depends on the (shape — volume - mass) of the cross section Of the pipe and the (velocity — pressure — temperature distribution.
Answer:
The kinetic energy correction factor the depends on the shape of the cross section of the pipe and the velocity distribution.
Explanation:
The kinetic energy correction factor take into account that the velocity distribution over the pipe cross section is not uniform. In that case, neither the pressure nor the temperature are involving and as we can notice, the velocity distribution depends only on the shape of the cross section.
Which of the two materials (brittle vs. ductile) usually obtains the highest ultimate strength and why?
Answer:
Explanation:
Ductile materials typically have a higher ultimate strength because they stretch absorbing more energy before breaking. While fragile materials snap in half before larger deformations due to larger loads occur.
It should be noted that when ductile materials stretch their section becomes smaller, and in that reduced section the stresses concentrate.
A component is composed of 0.0633 moles of platinum and 0.044 moles of nickel. Determine (a) total mass of the component (b) weight % of platinum.
Answer:
(a) Total mass = 14.931 kg
(b) Weight % of platinum = 17.29 %
Explanation:
(a) We have given moles of platinum = 0.0633 moles
Molar mass of platinum = 195.084 unit
Now mass of platinum = number of moles × molar mass = 0.0633×195.084 = 12.3488 gram
We have given moles of nickle = 0.044
Molar mass of nickle = 58.6934 unit
So mass of nickle = 0.044×59.6394 =2.5825 gram
So total mass = 12.3488+2.5825 = [tex]14.931[/tex]
(b) Weight % of platinum = [tex]\frac{weight\ of\ platinum}{total\ weight}=\frac{2.5825}{14.93}=0.1729[/tex] = 17.29 %
Air is contained in a vertical piston–cylinder assembly such that the piston is in static equilibrium. The atmosphere exerts a pressure of 101 kPa on top of the 0.5-m-diameter piston. The gage pressure of the air inside the cylinder is 1.2 kPa. The local acceleration of gravity is g = 9.81 m/s2 . Subsequently, a weight is placed on top of the piston causing the piston to fall until reaching a new static equilibrium position. At this position, the gage pressure of the air inside the cylinder is 2.8 kPa. Determine (a) the mass of the piston, in kg, and (b) the mass of the added weight, in kg
Answer:
a) 24 kg
b) 32 kg
Explanation:
The gauge pressure is of the gas is equal to the weight of the piston divided by its area:
p = P / A
p = m * g / (π/4 * d^2)
Rearranging
p * (π/4 * d^2) = m * g
m = p * (π/4 * d^2) / g
m = 1200 * (π/4 * 0.5^2) / 9.81 = 24 kg
After the weight is added the gauge pressure is 2.8kPa
The mass of piston plus addded weight is
m2 = 2800 * (π/4 * 0.5^2) / 9.81 = 56 kg
56 - 24 = 32 kg
The mass of the added weight is 32 kg.
The 10mm diameter rod is made of Kevlar 49. Determine the change in
length and the change in diameter.
Lenght of the rod = 100mm
Therefore there are two 80KN forces are pulling the rod from both
sides.
Answer:
0.815 mm
Explanation:
The rod in made of Kevlar 49, so it has an Young's modulus of
E = 125 GPa
The stiffness of a rod is given by:
k = E * A / L
k = E * π/4 * d^2 / L
So:
k = 125*10^9 * π/4 * 0.01^2 / 0.1 = 98.17 MN/m
Of the pulling forces only one is considered because when you pull on something there is always another force on the other side of equal magnitude and opposite direction to maintain equilibrium.
Hooke's law:
Δl = P/k
Δl = 80*10^3 / 98.17*10^6 = 0.000815 m = 0.815 mm
Define factor of safety and its significance
Answer:
Factor of safety is defines as the ratio between the strength of a material and the maximum stress that is developed in the material as a result of any given loading.
Mathematically we can write
[tex]F.O.S=\frac{Strength}{Stress_{working}}[/tex]
Consider any member of a machine that is under state of stress due to a general external loading, now we know that the material has a definite value of strength meaning that there is a definite value of force/stress at which the material will fail by fracture or by yielding as an example say a rod of steel will fail if we apply a load of 1000 Newton's on it.
Since as a basic principle of design we do not want any chance of failure in the machine at any cost we limit the maximum value of the stress that is developed in the machine part by designing the part in such a way that during the application of maximum load in the machine part the maximum stresses that are developed in the machine are well below the strength of the material for safety considerations. The factor by which the working stresses are less than the strength is termed as factor of safety.
Another significance of factor of safety is that if we need to estimate the loads on the machine that may act on it and we are not absolutely sure about the magnitude of the loads then we tend to increase the loads by a factor so that we design the structure to withstand greater loads hence if in case in the lifetime of the structure the loads exceed the normal value our structure will still remain structurally safe.
A petrol engine produces 20 hp using 35 kW of heat transfer from burning fuel. What is its thermal efficiency, and how much power is rejected to the ambient surroundings?
Answer:
efficiency =42.62%
AMOUNT OF POWER REJECTED IS 20.080 kW
Explanation:
given data:
power 20 hp
heat energy = 35kW
power production = 20 hp = 20* 746 W = 14920 Watt [1 hp =746 watt]
[tex]efficiency = \frac{power}{heat\ required}[/tex]
[tex]efficiency = \frac{14920}{35*10^3}[/tex]
[tex]= 0.4262*10^100[/tex]
=42.62%
b) [tex]heat\ rejected = heat\ required - amount\ of\ power\ generated[/tex]
[tex]= 35*10^3 - 14920[/tex]
= 20.080 kW
AMOUNT OF POWER REJECTED IS 20.080 kW
What is the ideal cooling system for low horsepower motor? For example1hp motor
Answer:
Air cooling.
Explanation:
Low power motors are supposed to be low cost, and they dissipate little heat. Therefore a low cost solution is ideal.
Air cooling can be achieved with very little cost. Fins can be added to a cast motor casing and a fan can be places on the shaft to use a small amount of the motor power to move air to cool it.
One way to lower NOx emissions in diesel engines is to add water to the fuel. Since water is not mixable with diesel fuel, it has to been emulsified by a surfactant. Explain how it works and why.
After the emulsion is made, fuel droplet containing water droplets increase its temperature inside the engine, because of that, water droplets explode causing water rapid evaporation that break down fuel droplet, in consequence, smaller fuel droplets are generated. This makes the temperature of the flame during the combustion decrease, and that is why the reduction of NOx is accomplished by this method .
The pressure forces on a submersed object will be (A)- Tangential to the objects body (B)- Parallel (C)- Normal (D)- None of the above
Answer:
c) normal
Explanation:
The pressure forces on a submersed object will be normal to the surfaces of the object. Hydrostatic pressure cannot apply tangential forces, so only the normal components exist.
Also, it should be noted that in the hydrostatic case (submerged object) all pressures depend linearly of depth. For small object we can approximate this with equal pressures everywhere.
An open glass tube is inserted into a pan of fresh water at 20 °C. What tube diameter is needed to make the height of capillary rise equal to four times the tube diameter? State all assumptions.
Answer:
The tube diameter is 2.71 mm.
Explanation:
Given:
Open glass tube is inserted into a pan of fresh water at 20°C.
Height of capillary raise is four times tube diameter.
h = 4d
Assumption:
Take water as pure water as the water is fresh enough. So, the angle of contact is 0 degree.
Take surface tension of water at 20°C as [tex]72.53\times 10^{-3}[/tex] N/m.
Take density of water as 100 kg/m3.
Calculation:
Step1
Expression for height of capillary rise is gives as follows:
[tex]h=\frac{4\sigma\cos\theta}{dg\rho}[/tex]
Step2
Substitute the value of height h, surface tension, density of water, acceleration due to gravity and contact angle in the above equation as follows:
[tex]4d=\frac{4\times72.53\times10^{-3}\cos0^{\circ}}{d\times9.81\times1000}[/tex]
[tex]d^{2}=7.39\times10^{-6}[/tex]
[tex]d=2.719\times10^{-3}[/tex] m.
Or
[tex]d=(2.719\times10^{-3}m)(\frac{1000mm}{1m})[/tex]
d=2.719 mm
Thus, the tube diameter is 2.719 mm.
A closed, rigid tank is filled with a gas modeled as an ideal gas, initially at 27°C and a gage pressure of 300 kPa. If the gas is heated to 77°C, determine the final pressure, expressed as a gage pressure, in kPa. The local atmospheric pressure is 1 atm.
The final gage pressure of the gas after being heated from 27°C to 77°C in a closed, rigid tank can be found using Gay-Lussac's Law. After adjusting the initial and final temperatures to Kelvin and converting the gage pressure to absolute pressure, calculate the new absolute pressure and then convert back to gage pressure.
Explanation:To determine the final gage pressure of a gas modeled as an ideal gas in a closed, rigid tank after it is heated from an initial temperature of 27°C to a final temperature of 77°C, we can use the ideal gas law in a form that relates pressure and temperature while keeping the volume and number of moles constant (Gay-Lussac's Law). We're told that the initial gage pressure is 300 kPa, and we need to find the final gage pressure. The local atmospheric pressure is given as 1 atm, which is equivalent to approximately 101.325 kPa. The formula for the final pressure, assuming no changes in the amount of gas or volume, is P2 = P1 * (T2/T1), where pressures are absolute pressures.
The steps to solve the problem are:
Convert the initial and final temperatures from Celsius to Kelvin by adding 273.15 to each.Add the atmospheric pressure to the initial gage pressure to get the initial absolute pressure.Use the equation to calculate the final absolute pressure.Subtract the atmospheric pressure from the final absolute pressure to get the final gage pressure.The answer explains how to calculate the initial pressure of a gas using the combined gas law equation.
The initial pressure of the gas in the closed system can be found using the combined gas law equation:
P1V1/T1 = P2V2/T2
Substitute the given values to find the initial pressure, which is calculated to be 1.6 atm.
Therefore, the initial pressure of the gas was 1.6 atm.
Define the stress and strength? A material has yield strength 100 kpsi. A cantilever beam has length 10 in and a load of 100 Lbf is applied at the free end. The beam cross section is rectangular 2""x5’. Is the beam design acceptable or not for a factor of safety 2?
Answer:
Stress is a force that acts on a unit area of a material. The strength of a material is how much stress it can bear without permanently deforming or breaking.
Is the beam design acceptable for a SF of 2? YES
Explanation:
Your factor of safety is 2, this means your stress allowed is:
σall = YS/FS = 100kpsi/2 = 50kpsiWhere:
σall => Stress allowedYS => Yield StrengthFS => Factor of safetyNow we are going to calculate the shear stress and bending stresses of the proposed scenario. If the calculated stresses are less than the allowed stress, that means the design is adequate for a factor of safety of 2.
First off we calculate the reaction force on your beam. And for this you do sum of forces in the Y direction and equal to 0 because your system is in equilibrium:
ΣFy = 0-100 + Ry = 0 thus,Ry = 100 lbfKnowing this reaction force you can already calculate the shear stress on the cantilever beam:
τ = F/Aτ = 100lbf/(2in*5in) τ = 10 psiNow, you do a sum of moments at the fixed end of your cantilever beam, so you can cancel off any bending moment associated with the reaction forces on the fixed end, and again equal to 0 because your system is in equilibrium.
ΣM = 0-100lbf*10in + M = 0M = 1000 lbf-inKnowing the maximum bending moment you can now calculate your bending stress as follows:
σ = M*c/IxWhere:
σ => Bending StressM => Bending Momentc => Distance from the centroid of your beam geometry to the outermost fiber.Ix => Second moment area of inertiaOut of the 3 values needed, we already know M. But we still need to figure out c and Ix. Getting c is very straight forward, since you have a rectangle with base (b) 2 and height (h) 5, you know the centroid is right at the center of the rectangle, meaning that the distance from the centroid to the outermost fibre would be 5in/2=2.5in
To calculate the moment of Inertia, you need to use the formula for the second moment of Inertia of a rectangle and knowing that you will use Ix since you are bending over the x axis:
Ix = (b*h^3)/12 = (2in*5in^3)/12 = 20.83 in4Now you can use this numbers in your bending stress formula:
σ = M*c/Ixσ = 1000 lbf-in * 2.5in / 20.83 in4σ = 120 psiThe shear stress is 10psi and the bending stress is 120psi, this means you are way below the stress allowed which is 50,000 psi, thus the beam design is acceptable. You could actually use a different geometry to optimize your design.
25 gallons of an incompressible liquid exert a force of 70 lbf at the earth’s surface. What force in lbf would 6 gallons of this liquid exert on the surface of the moon? The gravitational acceleration on the surface of the moon is 5.51 ft/s2.
Answer:
froce by 6 gallon liquid on moon surface is 2.86 lbf
Explanation:
given data:
at earth surface
volume of an incompressible liquid = Ve = 25 gallons
force by liquid = 70 lbf
on moon
volume of liquid = Vm = 6 gallons
gravitational acceleration on moon is am = 5.51 ft/s2
Due to incompressibility , the density remain constant.
mass of liquid on surface of earth[tex]= \frac{ force}{ acceleration}[/tex]
[tex]mass = \frac{70lbf}{32.2 ft/s2}[/tex]
mass = 2.173 pound
[tex]density \rho = \frac{mass}{volume}[/tex]
[tex]= \frac{2.173}{25} = 0.0869 pound/ gallon[/tex]
froce by 6 gallon liquid on moon surface is
Fm = mass * acceleration
= density* volume * am
= 0.0869 *6* 5.51
= 2.86 lbf
What is the resultant force on one side of a 25cm diameter circular plate standing at the bottom of 3m of pool water?
The resultant force on one side of a 25cm diameter circular plate at the bottom of 3m of pool water is approximately 1447 Newtons, calculated using the principles of fluid pressure and hydrostatic force.
Explanation:The question asks for the resultant force on one side of a 25cm diameter circular plate at the bottom of 3m of pool water. To calculate this, we first need to understand that the pressure exerted by a fluid in a static situation is given by the equation P = ρgh, where ρ is the density of the fluid (water in this case, which is approximately 1000 kg/m³), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the depth of the fluid above the point of interest (3m in this case).
To find the force exerted by the water on the plate, we also need the area of the plate, which can be derived from its diameter (D = 0.25m). The area (A) of a circle is given by πD²/4, which in this case gives us an area of approximately 0.0491 square meters. The force (F) exerted by the water can then be calculated using the equation F = P*A.
Substituting the values into the equations gives us a pressure (P) of 29430 Pa and, subsequently, a resultant force of approximately 1447 Newtons. This represents the sum of all the hydrostatic forces acting perpendicularly to the surface area of the circular plate due to the water pressure at the depth of 3 meters.