Explanation Given information: The length of the boom is 2.4 m . The 3.6 kN load applied at A . Calculation: Assumption: Apply the sign convention for calculating the equations of equilibrium as shown below. For the horizontal forces equilibrium condition, take the force acting towards right side as positive ( → + ) and the force acting towards left side as negative ( ← − ) . For the vertical forces equilibrium condition, take the upward force as positive ( ↑ + ) and the downward force as negative ( ↓ − ) . For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive. Sketch the Free Body Diagram as shown in Figure 1. Calculate the position vector ( r ) as shown below. The position of A is r A = ( 2.4 m ) k . The position of B is r B = ( 1.2 m ) k . The position of D is r D = − ( 0.8 m ) i + ( 0.6 m ) j . The position of E is r E = ( 0.8 m ) i + ( 1.2 m ) j . Calculate the position of cable AD as shown below. A D → = r D − r A Substitute − ( 0.8 m ) i + ( 0.6 m ) j for r D and ( 2.4 m ) k for r A . A D → = [ − ( 0.8 m ) i + ( 0.6 m ) j ] − [ ( 2.4 m ) k ] = − ( 0.8 m ) i + ( 0.6 m ) j − ( 2.4 m ) k Calculate the length of cable AD as shown below. A D = ( − 0.8 ) 2 + ( 0.6 ) 2 + ( − 2.4 ) 2 = 6.76 = 2.6 m Calculate the position of cable AE as shown below. A E = r E − r A Substitute ( 0.8 m ) i + ( 1.2 m ) j for r E and ( 2.4 m ) k for r A . A E = [ ( 0.8 m ) i + ( 1.2 m ) j ] − [ ( 2.4 m ) k ] = ( 0.8 m ) i + ( 1.2 m ) j − ( 2.4 m ) k Calculate the length of cable AE as shown below. A E = ( 0.8 ) 2 + ( 1.2 ) 2 + ( − 2.4 ) 2 = 7.84 = 2.8 m Calculate the tension in cable AD ( T A D ) as shown below. T A D = T A D A D → A D Substitute − ( 0.8 m ) i + ( 0.6 m ) j − ( 2.4 m ) k for A D → and 2.6 m for A D . T A D = T A D − 0.8 i + 0.6 j − 2.4 k 2.6 = T A D 2.6 ( − 0.8 i + 0.6 j − 2.4 k ) Calculate the tension in cable AE ( T A E ) as shown below. T A E = T A E A E → A E Substitute ( 0.8 m ) i + ( 1.2 m ) j − ( 2.4 m ) k for A E → and 2.8 m for A E . T A E = T A E 0.8 i + 1.2 j − 2.4 k 2.8 = T A E 2.8 ( 0.8 i + 1.2 j − 2.4 k ) Apply the Equations of Equilibrium as shown below. Apply the Equilibrium equation of moment about point C and equate to zero for equilibrium. ∑ M C = 0 r A × T A D + r A × T A E + r A × ( − 3 j ) = 0 Substitute ( 2.4 m ) k for r A , T A D 2.6 ( − 0.8 i + 0.6 j − 2.4 k ) for T A D , and T A E 2.8 ( 0.8 i + 1

Vector Mechanics for Engineers: Statics, 11th Edition

11th Edition

ISBN: 9780077687304

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek

Publisher: McGraw-Hill Education

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Chapter 4.3, Problem 4.109P

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Find the tension in each cable and the reaction at C.

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Chapter 4.3, Problem 4.108P

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You are asked to design a unit to condense ammonia. The required condensation rate is 0.09kg/s. Saturated ammonia at 30 o C is passed over a vertical plate (10 cm high and 25 cm wide).The properties of ammonia at the saturation temperature of 30°C are hfg = 1144 ́10^3 J/kg andrv = 9.055 kg/m 3 . Use the properties of liquid ammonia at the film temperature of 20°C (Ts =10 o C):Pr = 1.463 rho_l= 610.2 kf/m^3 liquid viscosity= 1.519*10^-4 kg/ ms kinematic viscosity= 2.489*10^-7 m^2/s Cpl= 4745 J/kg C kl=0.4927 W/m CCalculate the surface temperature required to achieve the desired condensation rate of 0.09 kg/s( should be 688 degrees C) a) Show that if you use a bigger vertical plate (2.5 m-wide and 0.8 m-height), the requiredsurface temperature would be now 20 o C. You may use all the properties given as an initialguess. No need to iterate to correct for Tf. b) What if you still want to use small plates because of the space constrains? One way to getaround this problem is to use small…

Chapter 4 Solutions

Vector Mechanics for Engineers: Statics, 11th Edition

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Find the tension in each cable and the reaction at C .

Find the tension in each cable and the reaction at C.

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Chapter 4 Solutions