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anygoal [31]
3 years ago
8

A square-thread power screw has a major diameter of 32 mm and a pitch of 4 mm with single threads, and it is used to raise a loa

d putting a force of 6.5 kN on the screw. The coefficient of friction for both the collar and screw is .08. If the torque from the motored used to raise the load is limited to 26 N*m, what is the largest value that the mean diameter of the collar can be?
Engineering
1 answer:
Valentin [98]3 years ago
5 0

Answer:

54mm.

Explanation:

So, we are given the following data or parameters or information that is going to assist in solving this type of question efficiently;

=> "A square-thread power screw has a major diameter of 32 mm"

=> "a pitch of 4 mm with single threads"

=> " and it is used to raise a load putting a force of 6.5 kN on the screw."

=> The coefficient of friction for both the collar and screw is .08."

=> "If the torque from the motored used to raise the load is limited to 26 N×M."

Step one: determine the lead angle. The lead angle can be calculated by using the formula below;

Lead angle = Tan^- (bg × T/ Jh × π ).

=> Jh = J - T/ 2. = 32 - 4/2. = 30mm.

Lead angle = Tan^- { 1 × 4/ π × 30} = 2.43°.

Step two: determine the Torque required to against thread friction.

Starting from; phi = tan^-1 ( 0.08) = 4.57°.

Torque required to against thread friction = W × Jh/2 × tan (lead angle + phi).

Torque required to against thread friction =( 6500 × 30/2) × tan ( 2.43° + 4.57°). = 11971.49Nmm.

Step three: determine the Torque required to against collar friction.

=> 2600 - 11971.49Nmm = 14028.51Nmm.

Step four = determine the mean collar friction.

Mean collar friction = 14028.51Nmm/0.08 × 6500 = 27mm

The mean collar diameter = 27 × 2 = 54mm.

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The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface
slamgirl [31]

Answer:

F = 641,771.52 \dfrac{lb-ft}{s^2}

Explanation:

Given that

R=8 ft

Width= 10 ft

We know that hydro statics force given as

  F=ρ g A X

ρ is the density of fluid

A projected area on vertical plane

X is distance of center mass of projected plane from free surface of water.

Here

X=8/2  ⇒X=4 ft

A=8 x 10=80  ft^2

So now putting the values

F=ρ g A X

F=62.4(32.14)(80)(4)

F = 641,771.52 \dfrac{lb-ft}{s^2}

   

4 0
3 years ago
Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th
Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine =1 - \frac{Tl}{Th} \\

where Th =temperature at which the heat enters the engine

Tl is the  temperature of the environment

since both engines have the same thermal capacities <em>n_{th} </em> therefore n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\

We have now that

\frac{-1400}{T}+\frac{T}{300}=0\\

multiplying through by T

-1400 + \frac{T^{2} }{300}=0\\

multiplying through by 300

-420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000}  \\T=648.07k

The temperature T= 648.07k

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3 years ago
How do u charge ur phone? :)
Pepsi [2]
With a phone charger.
7 0
2 years ago
Read 2 more answers
How do you extablish a chain of dimensions​
kap26 [50]

Answer:

Certamente você conhece três dimensões: comprimento, largura e profundidade. Além disso, quando se pensa um pouco fora da caixa também seria possível adicionar a dimensão do tempo.

Provavelmente, algumas pessoas viajam na maionese quando toca-se nesse assunto. Vem em suas mentes universos paralelos e até mesmo realidades alternativas. Mas também não se trata disso.

Explanation:

Basicamente as dimensões são as facetas do que nós percebemos a ser realidade. Existem muitos debates sobre dimensões na física. Um dos que mais chamam a atenção se chama Teoria das Cordas.

r

5 0
3 years ago
An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8
marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

a=18.66\ m/s^2

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0
3 years ago
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