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Papessa [141]
3 years ago
7

The beam is supported by a pin at A and a roller at B which has negligible weight and a radius of 15 mm. If the coefficient of s

tatic friction is mB = mC = 0.3, determine the largest angle u of the incline so that the roller does not slip for any force P applied to the beam.
Engineering
1 answer:
Anettt [7]3 years ago
5 0

Answer:

33.4

Explanation:

Step 1:

\sumMo=0 (moment about the origin)

Fb(15)-Fc(15)=0

Fb=Fc

Step 2:

\sumFx=0

-Fb-Fccos\theta+Ncsin\theta=0

Fc=0.3Nc=Fb

-0.3Nc-0.3Nccos\theta+Ncsin\theta=0

(-0.3-cos\theta+sin\theta)Nc=0----(1)

\sumFy=0

Nccos\theta+Fcsin\theta-Nb=0

Nccos\theta+0.3Ncsin\theta-Nc=0

Nc[cos\theta+0.3sin\theta-1]=0--------(2)

Solving eq (1) and eq (2)

\theta=33.4

Step 3:

As the roller is a two force member

2(90-\phi)+\theta=180

\phi=\theta/2

\phi=Tan(\muN/N)-1

\phi=16.7

\theta=2x16.7=33.4

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A 3.5-m3 rigid tank initially contains air whose density is 2 kg/m3 . The tank is connected to a high-pressure supply line throu
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Answer:

Explanation:

First, we find the mass of the air originally in the tank.

Density is given as mass divided by volume. It is given as:

Density = \frac{mass}{volume}

Therefore, mass is:

mass = denisty *volume

Density of air = 2 kg/m^3; Volume of the tank =  3.5 m^3

=> Mass = 3.5 * 2 = 7 kg

The mass of the air initially in the tank is 7 kg.

After air is allowed to enter, the mass changes.

New density = 6.5 kg/m^3

The new mass will be:

Mass = 6.5 * 3.5 = 22.75 kg

We can now find the mass of air that has entered the tank:

Mass of air that entered tank = New mass of air - Original mass of air

M = 22.75 - 7.0 = 15.75 kg

The mass of air that entered the tank is 15.75 kg.

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Suppose the working pressure for a boiler is 10 psig, then what is the corresponding absolute pressure?
yanalaym [24]

Answer:

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

Explanation:

From Fluid Mechanics, we remember that absolute pressure (p_{abs}), measured in pounds per square inch, is the sum of the atmospheric pressure and the working pressure (gauge pressure). That is:

p_{abs} = p_{atm}+p_{g} (1)

Where:

p_{atm} - Atmospheric pressure, measured in pounds per square inch.

p_{g} - Working pressured of the boiler (gauge pressure), measured in pounds per square inch.

If we suppose that p_{atm} = 14.696\,psi and p_{g} = 10\,psi, then the absolute pressure is:

p_{abs} = 14.696\,psi+10\,psi

p_{abs} = 24.696\,psi

The corresponding absolute pressure of the boiler is 24.696 pounds per square inch.

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A cylindrical brass rod has a length of 5.00cm extending from a holder and a diameter of 4.50mm. Its Young's modulus is 98.0GPa.
Galina-37 [17]

Answer:

elongation of the brass rod is 0.01956 mm

Explanation:

given data

length = 5 cm = 50 mm

diameter = 4.50 mm

Young's modulus = 98.0 GPa

load = 610 N

to find out

what will be the elongation of the brass rod in mm

solution

we know here change in length formula that is express as

δ = \frac{PL}{AE}    ................1

here δ is change in length and P is applied load  and A id cross section area and E is Young's modulus and L is length

so all value in equation 1

δ = \frac{PL}{AE}  

δ = \frac{610*50}{\frac{\pi}{4} * 4.50^2 * 98*10^3}  

δ = 0.01956 mm

so elongation of the brass rod is 0.01956 mm

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