Home Chapter 6 Multiple Choice

# Multiple Choice

This activity contains 17 questions.

## Compute the force in each member of the Warren truss and indicate whether the members are in tension or compression. All the members are 3 m long.

 AB = CD = 8.00 kN C, AE = DE = 6.92 kN T, BE = CE = 8.00 kN T, BC = 6.92 kN C AB = CD = 4.62 kN C, AE = DE = 2.31 kN T, BE = CE = 4.62 kN T, BC = 2.31 kN C AB = CD = 8.00 kN T, AE = DE = 6.92 kN C, BE = CE = 8.00 kN C, BC = 6.92 kN T AB = CD = 4.62 kN T, AE = DE = 2.31 kN C, BE = CE = 4.62 kN C, BC = 2.31 kN T

## Determine the force in each member of the truss and indicate whether the members are in tension or compression.

 CB = 447 N C, CD = 200 N T, DB = 800 N C, DE = 200 N T, BE = 447 N T, BA = 894 N C, AE = 800 N T CB = 447 N T, CD = 200 N C, DB = 800 N T, DE = 200 N C, BE = 447 N C, BA = 894 N T, AE = 800 N C CB = 894 N T, CD = 800 N C, DB = 800 N T, DE = 800 N C, BE = 894 N C, BA = 1789 N T, AE = 800 N C CB = 894 N C, CD = 800 N T, DB = 800 N C, DE = 800 N T, BE = 894 N T, BA = 1789 N C, AE = 800 N T

A sign is subjected to a wind loading that exerts horizontal forces of 300 lb on joints B and C of one of the side supporting trusses. Determine the force in members BC, CD, DB, and DE of the truss and state whether the members are in tension or compression.

 CB = 720 lb C, CD = 780 lb T, DB = 0, DE = 780 lb T CB = 720 lb T, CD = 780 lb C, DB = 0, DE = 780 lb C CB = 125 lb T, CD = 325 lb C, DB = 0, DE = 325 lb C CB = 125 lb C, CD = 325 lb T, DB = 0, DE = 325 lb T

## The Pratt bridge truss is subjected to the loading shown. Determine the force in members CD, CL and ML, and indicate whether these members are in tension or compression.

 CL = 100 kN T, ML = 150 kN T, CD = 150 kN C CL = 50 kN C, ML = 112.5 kN C, CD = 112.5 kN T CL = 100 kN C, ML = 150 kN C, CD = 150 kN T CL = 50 kN T, ML = 112.5 kN T, CD = 112.5 kN C

## The Warren truss is used to support a staircase. Determine the force in members CE, ED, and DF, and state whether the members are in tension or compression. Assume all joints are pinned.

 ED = 3.60 kN T, DF = 1.70 kN C, CE = 6.22 kN C ED = 2.00 kN C, DF = 2.26 kN C, CE = 2.26 kN T ED = 0.800 kN C, DF = 1.131 kN T, CE = 2.83 kN C ED = 0.400 kN C, DF = 2.26 kN T, CE = 4.53 kN C

A tower used in an electrical substation supports a power line which exerts a horizontal tension of T = 500 lb on each side truss of the tower as shown. Determine the force in members BC, CM, and LM of a side truss and indicate whether the members are in tension or compression.

 CM = 833 lb T, CB = 3330 lb C, LM = 2670 lb T CM = 581 lb T, CB = 2050 lb C, LM = 2000 lb T CM = 325 lb T, CB = 3040 lb C, LM = 2670 lb T CM = 58 lb T, CB = 1646 lb C, LM = 1600 lb T

A crane is constructed from two side trusses. If a load of 4 kN is suspended from one of these trusses as shown, determine the force in members FG, GK, and <KJ. State whether the members are in tension or compression. Assume the joints are pin-connected.

 KG = 2.66 kN C, KJ = 4.00 kN C, FG = 5.96 kN T KG = 5.66 kN T, KJ = 4.00 kN C, FG = 17.89 kN T KG = 0, KJ = 4.00 kN C, FG = 8.94 kN T KG = 1.886 kN C, KJ = 4.00 kN C, FG = 5.96 kN T

## Determine the force in members FF, FB, and BC of the Fink truss and indicate whether the members are in tension or compression.

 BF = 693 lb T, FG = 1800 lb C, BC = 1212 lb T BF = 8660 lb C, FG = 3600 lb C, BC = 1212 lb T BF = 3810 lb T, FG = 3600 lb C, BC = 1212 lb T BF = 1732 lb T, FG = 2400 lb C, BC = 1212 lb T

## Determine the force in members GF, CF, CD of the symmetric roof truss and indicate whether the members are in tension or compression.

 CD = 1.783 kN C, CF = 0, GF = 1.427 kN T CD = 3.48 kN C, CF = 00904 kN T, GF = 1.783 kN T CD = 2.79 kN C, CF = 0.723 kN T, GF = 1.427 kN T CD = 2.23 kN C, CF = 0, GF = 1.783 kN T

The principles of a differential chain block are indicated schematically in the figure. Determine the magnitude of force P needed to support the 800-N force. Also compute the distance x where the cable must be attached to bar AB so the bar remains horizontal. All pulleys have a radius of 60 mm.

 P = 80 N, x = 240 mm P = 80 N, x = 180 mm P = 40 N, x = 180 mm P = 40 N, x = 240 mm

## Determine the horizontal and vertical components of force at pins A and C of the two-member frame.

 Ax = -212 N, Ay = 388 N, Cx = 212 N, Cy = 212 N Ax = -300 N, Ay = 300 N, Cx = 300 N, Cy = 300 N Ax = -849 N, Ay = 149 N, Cx = 849 N, Cy = 849 N Ax = -1200 N, Ay = 1200 N, Cx = 1200 N, Cy = -600 N

## Determine the horizontal and vertical components of force at pin A.

 Ax = -46.7 lb, Ay = 45.0 lb Ax = -53.3 lb, Ay = 40.0 lb Ax = 13.3 lb, Ay = 10.0 lb Ax = -106.7 lb, Ay = 80.0 lb

## Determine the magnitude of the forces in pins B and D of the four-member frame.

 B = 503 lb, D = 225 lb B = 225 lb, D = 503 lb B = 56.3 lb, D = 125.8 lb B = 125.8 lb, D = 112.5 lb

The floor beams AB and BC are stiffened using the two tie rods CD and AD. Determine the force along each rod. Assume the three contacting members at B are smooth and the joints at A, C, and D are pins.

 T = 480 lb T = 520 lb T = 1248 lb T = 1152 lb

## The hoist supports the 125-kg engine. Determine the force the load creates in member DB and in member FB, which contains the hydraulic cylinder H.

 FB = 5.82 kN C, BD = 7.80 kN T FB = 1.939 kN C, BD = 2.60 kN T FB = 1.839 kN C, BD = 2.47 kN T FB = 1.839 kN C, BD = 2.60 kN T

The scissors lift consists of two sets of symmetrically placed cross members (one in front that is shown and one behind that is not shown) and two hydraulic cylinders (front [labeled DE] and back [not shown]). The uniform platform has a mass of 60 kg with a center of gravity at G1. The 85 kg load (center of gravity at G2) is centered front to back. Determine the force in each of the hydraulic cylinders necessary to maintain equilibrium. These are rollers at B and D.

 DE = 1.067 kN C DE = 1.139 kN C DE = 0.606 kN C DE = 1.207 kN C

## The jack shown supports a 350-kg automobile engine. Determine the compression in the hydraulic cylinder C and the magnitude of force that pin B exerts on the horizontal member BDE.

 FC = 8.75 kN, FB = 5.43 kN FC = 5.15 kN, FB = 5.15 kN FC = 8.58 kN, FB = 5.15 kN FC = 5.25 kN, FB = 7.36 kN