Home Chapter 4 Multiple Choice

# Multiple Choice

This activity contains 44 questions.

## Determine the magnitude and direction of the moment of the force of the movement if the force at A about point P.

 Mp = 400 Nm Mp = 400 Nm Mp = 200 Nm Mp = 200 Nm

## Determine the magnitude and direction of the moment of the force of the movement if the force at A about point P.

 Mp = 1440 Nm Mp = 600 Nm Mp = 1440 Nm Mp = 600 Nm

## Determine the magnitude and direction of the moment of the force of the movement if the force at A about point P.

 Mp = 17.86 kNm Mp = 1.072 kNm Mp = 9.86 kNm Mp = 14.93 kNm

## Determine the magnitude and direction of the moment of the force at A about point P.

 Mp = 143.5 ft-lb Mp = 1191 ft-lb Mp = 1104 ft-lb Mp = 1200 ft-lb

## Determine the magnitude and direction of the moment of the force at A about point P.

 Mp = 0.707 kN-m Mp = 5.66 kN-m Mp = 5.70 kN-m Mp = 4.50 kN-m

## Determine the moment of force F1 about point A on the beam.

 M1 = 1600 ft-lb M1 = 100 ft-lb M1 = 100 ft-lb M1 = 1600 ft-lb

## Determine the moment of force F2 about point A on the beam.

 M2 = 800 ft-lb M2 = 50 ft-lb M2 = 600 ft-lb M2 = 37.5 ft-lb

## Determine the moment of force F3 about point A on the beam.

 M3 = 6930 ft-lb M3 = 6000 ft-lb M3 = 3000 ft-lb M3 = 4000 ft-lb

Determine the direction (0° 180°) of the 30-lb force F so that the moment of F about point A has the maximum magnitude.

 = 36.9° = 53.1° = 127.0° = 90.0°

Determine the direction (0° 180°) of the 30-lb force F so that the moment of F about point A has the minimum magnitude.

 = 36.9° = 53.1° = 127.0° = 90.0°

## Determine the moment of the force at A about point P. Use a vector analysis and express the result in Cartesian vector form.

 MP = (160i+240j+40k) N-m MP = (380i+160j+400k) N-m MP = (280i+200j+400k) N-m MP = (40i+80k) N-m

## Determine the moment of the force at A about point P. Use a vector analysis and express the result in Cartesian vector form.

 MP = (-6i+6j-4k) N-m MP = (24i+8j+9k) N-m MP = (-6i-6j-4k) N-m MP = (24i-8j +9k) N-m

## Determine the moment of the force at A about point P. Use a vector analysis and express the result in Cartesian vector form.

 MP = (600i-200k) N-m MP = (800i-200j+1200k) N-m MP = (1200i+100k) N-m MP = (400i-200j+1200k) N-m

## The man at B exerts a force of 140N on the rope attached to the end of beam AC as shown. Determine the moment of this force about the base of the beam at A.

 MA = (720i-360k) N-m MA = (-720i+360k) N-m MA = (-720i-360k) N-m MA = (720i+360k) N-m

## The pole supports a 22-lb traffic light. Using Cartesian vectors, determine the moment of the weight of the traffic light about the base of the pole at A.

 MA = 216k lb-ft MA = (-132i-229j-216k) lb-ft MA = (-229i+121j) lb-ft MA = (-132i+229j) lb-ft

## Determine the magnitude of the projection of the moment cause by the force about the aa axis.

 Maa = 80.0 N-m Maa = 56.6 N-m Maa = 28.3 N-m Maa = 100.0 N-m

## Determine the magnitude of the projection of the moment cause by the force about the aa axis.

 Maa = 16.97 kN-m Maa = 12.00 kN-m Maa = 6 kN-m Maa = 8.48 kN-m

## A force of 50 N is applied to the handle of the door as shown. Determine the projection of the moment of this force about the hinged axis z. Neglect the size of the doorknob. Suggestion: Use a scaler analysis.

 Mz = 15.9 N-m Mz = -15.9 N-m Mz = -27.6 N-m Mz = 27.6 N-m

## Determine the magnitude and direction of the couple moment.

 M = 3900 lb-ft CW M = 3900 lb-ft CCW M = 3120 lb-ft CW M = 3120 lb-ft CCW

## Determine the magnitude and direction of the couple shown.

 M = 1200 N-m CW M = 400 N-m CCW M = 400 N-m CW M = 1200 N-m CCW

## Determine the magnitude and direction of the couple shown.

 M = 22.6 kN-m CCW M = 22.6 kN-m CW M = 21.9 kN-m CCW M = 21.9 kN-m CW

## A twist of 4 N-m is applied to the handle of the screwdriver. Resolve this couple moment into a pair of couple forces F exerted on the handle.

 F = 133 N F = 120 N F = 60 N F = 266 N

## A twist of 4 N-m is applied to the handle of the screwdriver. Resolve this couple moment into a pair of couple forces F exerted on the handle.

 P = 20 N P = 10 N P = 800 N P = 1600 N

The main beam along the wing of an airplane is swept back at an angle of 25°. From load calculations it is determined that the beam is subjected to couple moments aMx = 25,000 lb • ft and My = 17,000 lb • ft. Determine the equivalent couple moments created about the x' and y' axis.

 Mx' = 26.0 kip-ft, My' = 29.8 kip-ft Mx' = 29.8 kip-ft, My' = -4.84 kip-ft Mx' = 26.0 kip-ft, My' = 15.47 kip-ft Mx' = 29.8 kip-ft, My' = 4.84 kip-ft

## Determine the couple moment. Use a vector analysis and express the result as a Cartesian vector.

 M = (800i-4800j-800k) lb-ft M = (4000i-2000j+4000k) lb-ft M = (-4000i+7000j-4000k) lb-ft M = (-800i-4800j-800k) lb-ft

## Express the moment of the couple acting on the pipe on Cartesian vector form. What is the magnitude of the couple moment?

 Mc = (37.5i-25j) N-m, Mc = 45.1 N-m Mc = (25i-37.5j) N-m, Mc = 45.1 N-m Mc =45.1kN-m, Mc = 45.1 N-m Mc =-45.1kN-m, Mc = 45.1 N-m

## Replace the force at A by an equivalent force and couple moment at P.

 F = 15 kN @ = 36.9° , Mp = 67.1 kN-m F = 15 kN @ = 36.9° , Mp = 30 kN-m F = 15 kN @ = 143.1° , Mp = 67.1 kN-m F = 15 kN @ = 143.1° , Mp = 30 kN-m

## This structural connection is subjected to the 8,000-lb force. Replace this force by an equivalent force and couple acting at the center of the bolt group, O.

 F = (6400i+4800j) lb, M = (-3.20i+7.20j) kip-ft F = (-6400i-4800j) lb, M = -400k kip-ft F = (-6400i-4800j) lb, M = (-3.20i-7.20j) kip-ft F = (6400i+4800j) lb, M = 400k kip-ft

## Replace the force and couple system by an equivalent single force and couple acting at point P.

 F = (-8.66i-15.00j) N, M = 50.0 N-m F = (8.66i-15.00j) N, M = 42.0 N-m F = (8.66i-15.00j) N, M = 50.0 N-m F = (-8.66i-15.00j) N, M = 42.0 N-m

## Replace the force and couple system by an equivalent single force and couple acting at point P.

 F = (-0.0858i-1.184j) kN, M = 36.6 kN-m F = (-0.0858i-1.184j) kN, M = 21.6 kN-m F = (-0.0858i-1.184j) kN, M = 13.61 kN-m F = (-0.0858i-1.184j) kN, M = 35.7 kN-m

## Replace the loading system acting on the post by an equivalent force and couple system at point O.

 F = (-8i-66j) lb, M = 220 lb-ft F = (-8i-66j) lb, M = 160 lb-ft F = (4i78j) lb, M = 220 lb-ft F = (4i78j) lb, M = 100 lb-ft

## Replace the force at A by an equivalent force and couple moment at P. Express the results on Cartesian vector form.

 F = (-50i+20j+30k) N, M = (50i-130j+170k) N-m F = (-50i+20j+30k) N, M = (-50i+60j+60k) N-m F = (50i-20j-30k) N, M = (-50i+130j-170k) N-m F = (50i-20j-30k) N, M = (50i-60j-60k) N-m

## Replace the force at A by an equivalent force and couple moment at P. Express the results on Cartesian vector form.

 F = (60j-80k) lb, M = (20i+160j+120k) lb-ft F = (60j-80k) lb, M = (300j+560k) lb-ft F = (60j-80k) lb, M = 20i lb-ft F = (60j-80k) lb, M = (-160i-400j+120k) lb-ft

## The resultant force of a wind loading acts perpendicular to the face of the sign as shown. Replace this force by an equivalent force and couple moment acting at point O.

 F = -120i lb, M = (-3000j+1800k) lb-ft F = -120i lb, M = 3500i lb-ft F = -120i lb, M = -3500j lb-ft F = -120i lb, M = 3500k lb-ft

## Replace the force F, having a magnitude of F = 40 lb and acting at B, by an equivalent force and couple moment at A.

 F = (32i-24k) lb, M = (-120i+96j-160k) lb-ft F = (32i-24k) lb, M = 233k lb-ft F = (32i-24k) lb, M = -72k lb-ft F = (32i-24k) lb, M = (-120i+160j+40k) lb-ft

## Replace the two forces acting on the tree branches by an equivalent force and couple moment acting at point O.

 R = (60i+198j-36k) N, M = (747i+18j-924k) N-m R = (60i+18j-324k) N, M = (747i+18j-924k) N-m R = (60i+198j-36k) N, M = (360i+1260j-414k) N-m R = (60i+18j-324k) N, M = (360i+1260j-414k) N-m

## The three forces acting on the water tank represent the effect of the wind. Replace this system by a single resultant force and specify its vertical location from point O.

 R = 600 lb, d = 65.0 ft below O R = 600 lb, d = 130.4 ft below O R = 600 lb, d = 65.0 ft above O R = 600 lb, d = 130.4 ft above O

The boys A, B and C stand near the edges of a raft as shown. Determine the location (x, y) of boy D so that all four boys create a single resultant force acting through the raft's center O. Provided the raft itself is symmetric, this would keep the raft afloat in a horizontal plane. the mass of each boy is indicated in the diagram.

 x = 4.5 m, y = 1.5 m x = 3.0 m, y = 3.0 m x = 1.5 m, y = 4.5 m x = 3.0 m, y = 4.0 m

A force and couple act on the pipe assembly. Replace this system by an equivalent single resultant force. Specify the location of the resultant force along the y acis, measured from A. The pipe lies in the x-y plane.

 R = 25k lb, y = -0.9 ft R = 25k lb, y = 0.9 ft R = 25k lb, y = 3.9 ft R = 25k lb, y = -3.9 ft

Three parallel forces act on the rim of the tube. If it is required that the resultant force FR of the system have a line of action that coincides with the central z axis, determine the magnitude of FC and its location on the rim. What is the magnitude of the resultant force FR?

 Fc = 361 lb, = 56.3°, R = 861 lb Fc = 500 lb, = 54.0°, R = 1000 lb Fc = 500 lb, = 36.0°, R = 1000 lb Fc = 361 lb, = 36.9°, R = 861 lb

## Replace the loading by an equivalent force and couple moment acting at point O.

 R = 90 kN, M = 473 kN-m CW R = 90 kN, M = 338 kN-m CW R = 45 kN, M = 203 kN-m CW R = 135 kN, M = 270 kN-m CW

The wind has blown sand over a platform such that the intensity of load can be approximated by the function . Simplify this distributed loading to a single concentrated force and specify the magnitude and location of the force measured from A.

 R = 1250 N, x = 6.67 m R = 2500 N, x = 6.67 m R = 1250 N, x = 8 m R = 2500 N, x = 8 m

## The distributed loadings of soil pressure on the sides and bottom of a spread footing are shown. Simplify this system to a single resultant force and couple moment acting at A. What is the resultant force F and the couple moment M?

 F = (72i + 125j)N, M = 83.5 N-m CCW F = (144i + 125j)N, M = 135.7 N-m CCW F = (144i + 125j)N, M = 83.5 N-m CCW F = (72i + 125j)N, M = 102.1 N-m CCW

The bricks on top of the beam and the supports at the bottom create the distributed loading shown in the second figure. Determine the required intensity w and dimension d of the right support so that the resultant force and couple moment about point A of the system are both zero.

 w = 175.0 N/m, d = 1.5 m w = 138.2 N/m, d = 1.9 m w = 125.0 N/m, d = 2.1 m w = 154.4 N/m, d = 1.7 m