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3 years ago

If the pressure of a confined liquid is doubled, the volume will do what?

Engineering

1 answer:

slava [35]3 years ago

3 0

Answer:

Boyle's Law simply describes the relationship between the pressure and volume of an enclosed gas when Temperature remains constant. Tripling the pressure will reduce its volume to 1/3, and so on. Alternatively, if you double the volume available to an enclosed gas, pressure is halved.

Explanation:

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3 years ago

Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces app

stiks02 [169]

Answer:

a) Fb= 275.77 lb Fc= 142.75 lb

b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)

c) Fax = 195 lb

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Fbx = 195 lb

Fby = 195 lb

Explanation:

Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.

(a) Determine FB and FC so that only couples are applied.

(b) Using your answers to Part (a), determine the resultant couple moment that is produced.

Solution:

<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>

<u />

a) For no translation condition

∑ $F_{x} = 0$ & ∑ $F_{y} = 0$

Hence,

$F_{A}cos(30) - F_{B}cos(45) - F_{C} = 0$

$F_{A}sin(30) - F_{B} sin(45) = 0$

and

$F_{A} = 390 lb$

Inserting the value of $F_{A}$ and solving the remaining equations simultaneously yields (magnitudes),

$F_{B} = 275.77 lb\\F_{C} = 142.75 lb$

b) Summing up moments

$M=45 ( -F_{Ay}-F_{Cy}) +5 (-F_{By})+22(-F_{Ax}-F_{Bx})\\ =45(-390cos(30)-142.75)+5(-275.77cos(45))+22 (-390sin(30)-275.77sin(45))$

$M = -779.97 lb.ft$ (i.e. 779.97 lb.ft clockwise)

$F_{Ax} = 390 sin(30) = 195 lb$

$F_{Ay} = 390 cos(30) = 337.75lb\\$

$F_{Bx} = 275.77 sin(45) = 195lb\\F_{By} = 275.77 cos(45) = 195 lb$

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