Home Chapter 2 Multiple Choice

# Multiple Choice

This activity contains 29 questions.

## Determine the magnitudes of the resultant force and its direction measured from the positive x axis.

 R = 12.49 kN, = 43.9° CW R = 13.6 kN, = 21.5° CW R = 14.00 kN, = 60.0° CW R = 10.80 kN, = 68.2° CW

## Determine the magnitude of the x and y components of the 2—kN force.

 Fx = -1.414 kN, Fy = -1.414 kN Fx = -1.000 kN, Fy = -1.732 kN Fx = -1.732 kN, Fy = -1.000 kN Fx = -4.000 kN, Fy = -2.312 kN

## Determine the magnitude of the x and y components of the 700-lb force.

 Fx = -350 lb, Fy = 606 lb Fx = -404 lb, Fy = 571 lb Fx = -606 lb, Fy = 350 lb Fx = -571 lb, Fy = 404 lb

Determine the magnitude and direction of F so that this force has components of 40 lb acting from A toward B and 60lb acting from A toward C on the frame.

 F = 44.7 lb, = 22.9° F = 80.3 lb, = 46.2° F = 62.9 lb, = 37.1° F = 72.1 lb, = 56.3°

Determine the design angle ( <90°) bretween the two struts so that the 500-lb horizontal force has a component of 600lb directed from A toward C. That is the component of force acting along member AB?

 FAB = 321 lb, = 40.0° FAB = 383 lb, = 36.9° FAB = 171.1 lb, = 20.0° FAB = 215 lb, = 52.7°

If = 20° and = 35°, determine the magnitudes of F1 and F2 so that the resultant force has a magnitude of 20 lb and is directed along the positive x axis.

 F1 = 20.0 lb, F2 = 22.9 lb F1 = 25.6 lb, F2 = 26.6 lb F1 = 28.5 lb, F2 = 11.91 lb F1 = 14.00 lb, F2 = 8.35 lb

The gusset plate G of a bridge joint is subjected to the two member forces at A and B. If the force at B is horizontal and the force at A is directed at = 30°, determine the magnitude and direction of the resultant force.

 R = 458 N, = 97.5° CCW R = 252 N, = 82.5° CCW R = 252 N, = 97.5° CCW R = 458 N, = 82.5° CCW

## Determine the design angle for connecting member A to the plate if the resultant force is to be directed vercially upward. Also, what is the magnitude of the resultant?

 R = 400 N, = 53.5° R = 250 N, = 30.0° R = 300 N, = 36.9° R = 640 N, = 38.6°

## Determine the magnitude of the resultant force by adding the rectangular components of the three forces.

 R = 29.7 N R = 54.2N R = 90.8 N R = 24.0 N

## Determine the magnitude and direction of the resultant force.

 R = 251 N, = 85.5° CCW R = 251 N, = 94.5° CCW R = 421 N, = 67.7° CCW R = 421 N, = 112.3° CCW

## Determine the magnitude and direction of the resultant force.

 R = 80.3 lb, = 106.2° CCW R = 80.3 lb, = 73.8° CCW R = 72.1 lb, = 63.6° CCW R = 72.1 lb, = 116.4° CCW

If F1 = F2 = 30lb, determine the angles and so that the resultant force is directed along the positive x axis and has a magnitude of FR = 20 lb.

 = = 70.5° = = 41.4° = = 19.47° = = 18.43°

## Express each force in Cartesian vector form.

 F1 = (2.50 i + 3.54 j + 2.50 k) kN, F2 = -2 j kN F1 = (4.33 i + 3.54 j + 4.33 k) kN, F2 = -2 j kN F1 = (2.17 i + 3.75 j + 4.33 k) kN, F2 = -2 j kN F1 = (4.33 i + 3.75 j + 2.17 k) kN, F2 = -2 j kN

## Express the force F1 in Cartesian vector form.

 F1 = (200 i - 200 j + 283 k) lb F1 = (200 i + 200 j + 283 k) lb F1 = (-200 i + 200 j + 565 k) lb F1 = (-200 i + 200 j + 283 k) lb

## Express the Force F2 in Cartesian vector form.

 F2 = (155 i + 155 j + 300 k) lb F2 = (212 i + 212 j - 519 k) lb F2 = (155 i + 155 j - 300 k) lb F2 = (367 i + 367 j - 300 k) lb

## The ball joint is subjected to the three forces shown. Find the magnitude of the resultant force.

 R = 5.30 kN R = 5.74 kN R = 5.03 kN R = 6.20 kN

## Determine the magnitude and direction angles of F2, so that the resultant of the two forces acts upward along the z axis of the pole and has a magnitude of 275 N.

 F2 = 246 N, = 54.9°, = 114.0°, = 44.7° F2 = 200 N, = 45.0°, = -60.0°, = 29.0° F2 = 200 N, = 45.0°, = 120.0°, = 29.0° F2 = 246 N, = 54.9°, = 66.0°, = 44.7°

## Force F acts on peg A such that one of its components, lying in the x-y plane, has a magnitude of 50 lb. Express F as a Cartesian vector.

 F = (43.3 i + 25.0 j + 25.0 k) lb F = (43.3 i + 25.0 j + 28.9 k) lb F = (43.3 i - 25.0 j + 25.0 k) lb F = (43.3 i - 25.0 j + 28.9 k) lb

Specify the magnitude and direction angles 1, 1 and 1 of F1 so that the resultant of the three forces acting on the post is FR = {-350k}lb. Note that F3 lies in the xy plane.

 F1 = 429 lb, = 62.2°, = 70.0°, = 35.3° F1 = 429 lb, = 62.2°, = 110.0°, = 144.7° F1 = 447 lb, = 30.0°, = 120.0°, = 141.5° F1 = 447 lb, = 30.0°, = 60.0°, = 38.5°

## Determine the magnitude and direction of the position vector r which points from point A to point B.

 r = 12 ft, = 70.5°, = 48.2°, = 131.8° r = 12 ft, = 109.5°, = 131.8°, = 48.2° r = 12 ft, = 70.5°, = 48.2°, = 48.2° r = 12 ft, = 109.5°, = 131.8°, = 131.8°

## The cord is attached between two walls. If it is 8 m long, determine the distance x to the point of attachment at B.

 x = 7.75 m x = 7.68 m x = 6.93 m x = 7.94 m

## Express force F as a Cartesian vector; then determine its direction angles.

 F = (-2i +j+2k) kN, = 48.2°, = 70.5°, = 48.2° F = (-2i +j+2k) kN, = 131.8°, = 70.5°, = 48.2° F = (-4i +j+4k) kN, = 48.2°, = 70.5°, = 48.2° F = (-4i +2j+4k) kN, = 131.8°, = 70.5°, = 48.2°

## The antenna tower is supported by three cables. The forces in these cables are as follows: FB = 520 N, FC = 680 N, and FD = 560 N. Write the resultant of these three forces as a vector.

 R = (4i +16j-72k) N R = (-120i +40 j-960k) N R = (560i +720j+1440k) N R = (80i +320j-1440k) N

## The cable AO exerts a force on the top of the pole of F = {—120i — 90j — 80k} lb. If the cable has a length of 34 ft, determine the height z of the pole and the location (x,y) of its base.

 x = 16 ft, y = 16 ft, z = 25 ft x = 12 ft, y = 9 ft, z = 8 ft x = 20 ft, y = 10 ft, z = 14 ft x = 24 ft, y = 18 ft, z = 16 ft

## Determine the angle between the pole AC and the wire AB.

 = 131.8° = 70.5° = 109.5° = 48.2°

## Determine the projection of the position vector r along the aa axis.

 raa = 35.3 m raa = 6.28 m raa = 5.42 m raa = 5.61 m

## Cable BC exerts a force of F = 28 N on the top of the flagpole. Determine the projection of this force along the positive z axis of the pole.

 F = 24 N F = -24 N F = 12 N F = 12 N

## Two forces act on a block. Determine the angle between them.

 = 135.0° = 65.1° = 45.0° = 114.9°

## What is the projection of the force F2 along the positive axis?

 F2y = 170.0 N F2y = —80.0 N F2y = 90.0 N F2y = 120.0 N