VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS
11th Edition
ISBN: 9781259633133
Author: BEER
Publisher: MCG
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Chapter 14, Problem 14.112RP
To determine

The force couple system that must be applied at A to hold the vane in place.

Expert Solution & Answer
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Answer to Problem 14.112RP

The x component of force (Ax) applied at A to hold the vane in place is 55.5lb()_.

The y component of force (Ay) applied at A to hold the vane in place is 20.2lb()_.

The couple (MA) applied at A to hold the vane in place is 41.4lbft(clockwise)_.

Explanation of Solution

Given information:

The nozzle discharges (Q) water at the rate of 200gal/min.

The stream of the water moves with a velocity (v) of magnitude is 100ft/s.

Assume unit weight of the water (γ) as 62.4lb/ft3.

Assume the acceleration due to gravity (g) as 32.2ft/s2.

Calculation:

Convert the gallons per minutes to cubic feet per second.

Q=200gal/min×1ft37.48gal×1min60sec=0.44563ft3/s

Calculate the rate of the water discharge (dmdt) per unit length along BC using the relation:

(dmdt)=γQg

Substitute 62.4lb/ft3 for γ, 32.2ft/s2 for g and 0.44563ft3/s for Q.

(dmdt)=62.4×0.4456332.2=0.8636lbs/ft

Express the velocity of water (vB) at point B in terms of vector:

vB=(100ft/s)j

Express the velocity of water (vC) at point C in terms of vector:

vC=(100ft/s)(sin40°i+cos40°j)

Consider right side and upward as positive.

Show the impulse momentum equation diagram as in Figure (1).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 14, Problem 14.112RP , additional homework tip  1

Calculate the x component of force (Ax) applied at A to hold the vane using Figure (1):

0+Ax(Δt)=(Δm)(vC)i

Substitute (100sin40°) for (vC)i.

Ax=(Δm)(Δt)(100sin40°)=(dm)(dt)(100sin40°)

Substitute 0.8636lbs/ft for (dmdt).

Ax=0.8636×(100sin40°)=55.5lb()

Calculate the y component of force (Ay) applied at A to hold the vane using Figure (1):

(Δm)vB+Ay(Δt)=(Δm)(vC)j

Substitute (100cos40°) for (vC)j and (100ft/s)j for vB.

(Δm)(100ft/s)+Ay(Δt)=(Δm)(100cos40°)Ay=(Δm)(100cos40°)(Δm)(100ft/s)(Δt)Ay=(Δm)(Δt)(100)(cos40°1)Ay=(dm)(dt)(100)(cos40°1)

Substitute 0.8636lbs/ft for (dmdt).

Ay=(0.8636)(100)(cos40°1)=20.2lb=20.2lb()

Show the given nozzle system as in Figure (2).

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS, Chapter 14, Problem 14.112RP , additional homework tip  2

Consider clockwise as negative.

Calculate the couple applied (MA) at the point A to hold the vane using Figure (1) and Figure (2).

(3in.×1ft12in.)(Δm)vB+MA(Δt)=[(9in.×1ft12in.)(Δm)((vC)j) (15in.×1ft12in.)(Δm)((vC)i)]

Substitute (100cos40°) for (vC)j,(100sin40°) for (vC)i and (100ft/s)j for vB.

(3in.×1ft12in.)(Δm)(100ft/s)+MA(Δt)=[(9in.×1ft12in.)(Δm)((100cos40°)) (15in.×1ft12in.)(Δm)((100sin40°))]25(Δm)+MA(Δt)=57.453(Δm)80.348(Δm)MA(Δt)=57.453(Δm)80.348(Δm)25(Δm)MA=47.895(Δm)(Δt)MA=47.895(dm)(dt)

Substitute 0.8636lbs/ft for (dmdt).

MA=47.895(0.8636)=41.362lbft41.4lbft=41.4lbft(clockwise)

Therefore, the x component of force (Ax) applied at A to hold the vane in place is 55.5lb()_.

Therefore, the y component of force (Ay) applied at A to hold the vane in place is 20.2lb()_.

Therefore, the couple (MA) applied at A to hold the vane in place is 41.4lbft(clockwise)_.

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Chapter 14 Solutions

VECTOR MECH...,STAT.+DYNA.(LL)-W/ACCESS

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