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Ray Of Light [21]

3 years ago

11

The hoist supports the 125-kg engine. Determine the force the load creates in member DB and in member FB, which contains the hyd

raulic cylinder H.

1 answer:

natta225 [31]3 years ago

7 0

Answer:

hello the diagram attached to your question is missing attached below is the missing diagram

answer : Ffb = 1.9381 KN

Fdb = 2.6 KN

Explanation:

Given data

M = 125-kg

Determine The force on member FB

∅ = tan^-1 ( 3/1) = 71.56°

∑ Me = 0

= 125 * 9.81 * 3 - Ffb sin∅ * 2 = 0

hence : Ffb = 1.9381 KN

also

∑ Ex = 0

Ex = Ffb cos ( 71.56 ) = 1.9381 * cos 71.56 = 0.6132 KN

Determine the Force on member DB

∅ = tan^-1 ( 1/1 ) = 45°

∑ Mc = 0

Ex * 3 = Fdb * sin 45 * 1

Hence Fdb = 2.6 KN

attached below is the free body diagram

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Why Elastic Modulus is important and on what it depends?

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3 years ago

Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe

NemiM [27]

Answer:

(a) attached below

(b) $pf_{C}=0.85$ $lagging$

(c) $I_{C} =32.37 A$

(d) $X_{C} =49.37$ Ω

(e) $I_{cap} =9.72 A$ and $I_{line} =27.66 A$

Explanation:

Given data:

$P_{1}=15 kW$

$S_{2} =10 kVA$

$pf_{1} =0.6$ $lagging$

$pf_{2}=0.8$ $leading$

$V=480$ $Volts$

(a) Draw the power triangle for each load and for the combined load.

$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

$\alpha_{2}=cos^{-1} (0.8)=36.86$ °

$S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA$

$Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99$ ≅ $20kVAR$

$P_{2} =S_{2}*pf_{2} =10*0.8=8 kW$

$Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99$ ≅ $-6 kVAR$

The negative sign means that the load 2 is providing reactive power rather than consuming

Then the combined load will be

$P_{c} =P_{1} +P_{2} =15+8=23 kW$

$Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR$

(b) Determine the power factor of the combined load and state whether lagging or leading.

$S_{c} =P_{c} +jQ_{c} =23+14j$

or in the polar form

$S_{c} =26.92$ °

$pf_{C}=cos(31.32) =0.85$ $lagging$

The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

$I_{C} =S_{C}/\sqrt{3}*V$

$I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A$

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

$Q_{C} =3*V^2/X_{C}$

$X_{C} =3*V^2/Q_{C}$

$X_{C} =3*(480)^2/14*10^3$ Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

Line current flowing from the source is

$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

8 0

3 years ago

A steady, incompressible, two-dimensional velocity field is given by the following components in the x-y plane: u=1.85+2.05x+0.6

AnnZ [28]

Answer:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

Explanation:

The acceleration field is obtained by deriving the components in function of the time. That is to say:

$\frac{du}{dt}=2.05\cdot \frac{dx}{dt}+0.656\cdot \frac{dy}{dt}\\\frac{dv}{dt} = -2.18\cdot \frac{dx}{dt} -2.05\cdot \frac{dy}{dt}$

Where $\frac{dx}{dt} = u$ and $\frac{dy}{dt} = v$ .

The velocity components at given point are, respectively:

$u = 2.424$

$v=-5.266$

Lastly, the acceleration components are found:

$\frac{du}{dt} = 1.515$

$\frac{dv}{dt} = 5.511$

7 0

3 years ago