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aleksandr82 [10.1K]
3 years ago
8

The drawing shows a top view of a door that is free to rotate about an axis of rotation that is perpendicular to the screen. Fin

d the net torque (magnitude and direction) produced by the forces F1 and F2 about the axis.
Physics
1 answer:
dangina [55]3 years ago
4 0

Answer:

τ = 23.34 Nm

Direction = Anti-Clockwise (Positive)

Explanation:

The image of the question is attached in the answer.

First, we will calculate the torque due to force F₁.

τ₁ = F₁ d₁ Sin θ₁

where,

τ₁ = Torque due to F₁ = ?

F₁ = 20 N

d₁ = moment arm of F₁ = 0.5 m

θ₁ = Angle between F₁ and d₁ = 90°

Therefore,

τ₁ = (20 N)(0.5 m)Sin 90°

τ₁ = - 10 Nm

Negative sign due to clockwise direction.

Now, we will calculate the torque due to force F₂.

τ₂ = F₂ d₂ Sin θ₂

where,

τ₂ = Torque due to F₂ = ?

F₂ = 35 N

d₂ = moment arm of F₂ = 1.10 m

θ₂ = Angle between F₂ and d₂ = 120°

Therefore,

τ₁ = (35 N)(1.1 m)Sin 120°

τ₁ = 33.34 Nm

Positive sign due to anti-clockwise direction.

Now, we find net torque on door:

τ = τ₁ + τ₂

τ = - 10 Nm + 33.34 Nm

<u>τ = 23.34 Nm</u>

<u>Direction = Anti-Clockwise (Positive)</u>

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efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
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Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

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P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

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v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

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