Statics Quiz: Equivalent Resultant 3d
6 questions · exam conditions
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Equivalent Resultant 3dQuestion 1 of 6

Two parallel forces F1=150F_1 = 150 N and F2=100F_2 = 100 N act in the same direction along the z-axis. F1F_1 is applied at point (3, 2, 0) m and F2F_2 is applied at point (1, 5, 0) m. At what coordinates should the equivalent single resultant force be applied to maintain the same moment about the origin?

(2.2,3.2,0)(2.2, 3.2, 0) m with magnitude 250250 N
(2.0,3.5,0)(2.0, 3.5, 0) m with magnitude 250250 N
(2.2,3.2,0)(2.2, 3.2, 0) m with magnitude 200200 N
(1.8,4.0,0)(1.8, 4.0, 0) m with magnitude 250250 N
(2.5,2.8,0)(2.5, 2.8, 0) m with magnitude 275275 N
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Statics Quiz

Statics Quiz: Equivalent Resultant 3d

Practice Equivalent Resultant 3d in Statics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Equivalent Resultant 3d, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics.

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Question 1

Two parallel forces F1=150F_1 = 150 N and F2=100F_2 = 100 N act in the same direction along the z-axis. F1F_1 is applied at point (3, 2, 0) m and F2F_2 is applied at point (1, 5, 0) m. At what coordinates should the equivalent single resultant force be applied to maintain the same moment about the origin?

  1. (2.2,3.2,0)(2.2, 3.2, 0) m with magnitude 250250 N (correct answer)
  2. (2.0,3.5,0)(2.0, 3.5, 0) m with magnitude 250250 N
  3. (2.2,3.2,0)(2.2, 3.2, 0) m with magnitude 200200 N
  4. (1.8,4.0,0)(1.8, 4.0, 0) m with magnitude 250250 N
  5. (2.5,2.8,0)(2.5, 2.8, 0) m with magnitude 275275 N
Explanation: When you encounter parallel forces acting at different points, you need to find an equivalent single force that produces the same resultant force and moment about any reference point. This is a fundamental principle of force systems in statics. To solve this, start by finding the resultant force magnitude: FR=F1+F2=150+100=250F_R = F_1 + F_2 = 150 + 100 = 250 N in the z-direction. Next, calculate the moment each force creates about the origin. Since both forces act along the z-axis, only their x and y coordinates matter for moments about the origin. The moment of the resultant force about the origin must equal the sum of individual moments: M⃗R=M⃗1+M⃗2\vec{M}_R = \vec{M}_1 + \vec{M}_2. For the equivalent system: r⃗eq×F⃗R=r⃗1×F⃗1+r⃗2×F⃗2\vec{r}_{eq} \times \vec{F}_R = \vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 This gives us: xeq=F1x1+F2x2FR=150(3)+100(1)250=550250=2.2x_{eq} = \frac{F_1 x_1 + F_2 x_2}{F_R} = \frac{150(3) + 100(1)}{250} = \frac{550}{250} = 2.2 m Similarly: yeq=F1y1+F2y2FR=150(2)+100(5)250=800250=3.2y_{eq} = \frac{F_1 y_1 + F_2 y_2}{F_R} = \frac{150(2) + 100(5)}{250} = \frac{800}{250} = 3.2 m Answer A gives the correct coordinates (2.2, 3.2, 0) and magnitude 250 N. Answer B has the wrong y-coordinate (3.5 instead of 3.2). Answer C has the correct coordinates but wrong magnitude (200 N instead of 250 N). Answer D has completely wrong coordinates (1.8, 4.0). Remember: when combining parallel forces, the resultant magnitude is always the algebraic sum, and the location is found using the weighted average of positions based on force magnitudes.

Question 2

A space frame has forces applied at three joints: F⃗1=60i^−40j^+30k^\vec{F}_1 = 60\hat{i} - 40\hat{j} + 30\hat{k} N at joint A(1, 2, 1) m, F⃗2=−30i^+80j^−20k^\vec{F}_2 = -30\hat{i} + 80\hat{j} - 20\hat{k} N at joint B(3, 0, 2) m, and F⃗3=−20i^−30j^+50k^\vec{F}_3 = -20\hat{i} - 30\hat{j} + 50\hat{k} N at joint C(0, 3, 0) m. If these are replaced by a single equivalent force acting along the line from point (2, 1, 1) m to point (4, 3, 3) m, what constraints must be satisfied?

  1. The resultant force R⃗=10i^+10j^+60k^\vec{R} = 10\hat{i} + 10\hat{j} + 60\hat{k} N must be parallel to direction (2,2,2)(2, 2, 2), which it is not (correct answer)
  2. The resultant force R⃗=10i^+10j^+60k^\vec{R} = 10\hat{i} + 10\hat{j} + 60\hat{k} N must be parallel to direction (1,1,1)(1, 1, 1), which it is not
  3. The resultant force R⃗=10i^+10j^+60k^\vec{R} = 10\hat{i} + 10\hat{j} + 60\hat{k} N is parallel to direction (1,1,6)(1, 1, 6), satisfying the constraint
  4. The system can be replaced by any force along the given line with appropriate moment compensation
  5. The resultant force magnitude 62.062.0 N can act along the line if properly positioned for moment equilibrium
Explanation: When you encounter problems about replacing multiple forces with a single equivalent force, you're dealing with the principle of force equivalence in 3D space. The key constraint is that the equivalent force must not only have the correct magnitude and direction as the resultant, but it must also be parallel to the specified line of action. First, let's find the resultant force by adding all three forces: R⃗=F⃗1+F⃗2+F⃗3=(60−30−20)i^+(−40+80−30)j^+(30−20+50)k^=10i^+10j^+60k^\vec{R} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = (60-30-20)\hat{i} + (-40+80-30)\hat{j} + (30-20+50)\hat{k} = 10\hat{i} + 10\hat{j} + 60\hat{k} N Next, determine the direction of the given line from (2,1,1) to (4,3,3): Direction vector = (4-2, 3-1, 3-1) = (2, 2, 2) For the system to be replaceable by a single force along this line, the resultant force R⃗\vec{R} must be parallel to the direction vector (2, 2, 2). Two vectors are parallel if one is a scalar multiple of the other. Checking: Is 10i^+10j^+60k^10\hat{i} + 10\hat{j} + 60\hat{k} parallel to (2,2,2)(2, 2, 2)? The ratios would need to be equal: 102=102=602\frac{10}{2} = \frac{10}{2} = \frac{60}{2}, which gives us 5=5=305 = 5 = 30. Since 5≠305 ≠ 30, the vectors are not parallel. Option A correctly identifies this constraint violation. Option B uses the wrong direction vector (1,1,1). Option C incorrectly claims the forces are parallel. Option D ignores the fundamental constraint that equivalent forces must preserve both force and moment effects. Remember: Force equivalence requires both the correct resultant magnitude/direction AND compatibility with the specified line of action.

Question 3

A distributed loading system is equivalent to three concentrated forces: F⃗1=120k^\vec{F}_1 = 120\hat{k} N at (2, 3, 0) m, F⃗2=80k^\vec{F}_2 = 80\hat{k} N at (4, 1, 0) m, and F⃗3=60k^\vec{F}_3 = 60\hat{k} N at (1, 5, 0) m. Where should a single equivalent force of magnitude 260 N be placed to maintain the same moment about the y-axis?

  1. At (2.38,y,0)(2.38, y, 0) m for any value of y, maintaining moment about y-axis (correct answer)
  2. At (2.69,y,0)(2.69, y, 0) m for any value of y, using weighted centroid method
  3. At (x,2.69,0)(x, 2.69, 0) m for any value of x, considering moment arm perpendicularity
  4. At (2.38,2.69,0)(2.38, 2.69, 0) m as the unique point for complete moment equilibrium
  5. At (1.85,y,0)(1.85, y, 0) m for any value of y, using force magnitude weighting
Explanation: When you encounter equivalent force systems in statics, you need to preserve both the total force and the moment about specified axes. Since all forces act in the z-direction and we only care about the moment about the y-axis, this becomes a problem about finding the correct x-coordinate. First, verify the equivalent force magnitude: 120+80+60=260120 + 80 + 60 = 260 N in the k^\hat{k} direction. Now find where to place this equivalent force to preserve the moment about the y-axis. The moment about the y-axis depends only on the x-coordinates and z-components of forces. For the original system: My=F1⋅x1+F2⋅x2+F3⋅x3=120(2)+80(4)+60(1)=240+320+60=620M_y = F_1 \cdot x_1 + F_2 \cdot x_2 + F_3 \cdot x_3 = 120(2) + 80(4) + 60(1) = 240 + 320 + 60 = 620 N⋅m For the equivalent force: My=260⋅xeq=620M_y = 260 \cdot x_{eq} = 620 Therefore: xeq=620260=2.38x_{eq} = \frac{620}{260} = 2.38 m Since moment about the y-axis doesn't depend on the y-coordinate, the equivalent force can be placed anywhere along the line x=2.38x = 2.38 m. A is correct - the equivalent force maintains the same moment about the y-axis when placed at (2.38,y,0)(2.38, y, 0) for any y-value. B uses an incorrect x-coordinate (2.69 instead of 2.38), likely from calculation error. C incorrectly places the constraint on the y-coordinate rather than x-coordinate, misunderstanding which coordinate affects moment about the y-axis. D unnecessarily constrains both coordinates when only the x-coordinate matters for y-axis moments. Study tip: For moments about a specific axis, identify which coordinates actually contribute to that moment - this determines which coordinates must be fixed for equivalency.

Question 4

A force system in 3D space produces a resultant force R⃗=100i^+50j^−75k^\vec{R} = 100\hat{i} + 50\hat{j} - 75\hat{k} N and a resultant moment about point A(2, 3, 1) m of MA⃗=30i^−60j^+90k^\vec{M_A} = 30\hat{i} - 60\hat{j} + 90\hat{k} N⋅m. Determine the perpendicular distance from point A to the line of action of the equivalent single resultant force.

  1. 0.720.72 m
  2. 0.890.89 m (correct answer)
  3. 1.051.05 m
  4. 0.960.96 m
Explanation: The perpendicular distance from a point to the line of action of a force is given by d=∣MA⃗∣∣R⃗∣d = \frac{|\vec{M_A}|}{|\vec{R}|}, where MA⃗\vec{M_A} is the moment about that point. First, find the magnitudes: ∣R⃗∣=1002+502+(−75)2=10000+2500+5625=18125=134.6|\vec{R}| = \sqrt{100^2 + 50^2 + (-75)^2} = \sqrt{10000 + 2500 + 5625} = \sqrt{18125} = 134.6 N. ∣MA⃗∣=302+(−60)2+902=900+3600+8100=12600=112.2|\vec{M_A}| = \sqrt{30^2 + (-60)^2 + 90^2} = \sqrt{900 + 3600 + 8100} = \sqrt{12600} = 112.2 N⋅m. Therefore, d=112.2134.6=0.89d = \frac{112.2}{134.6} = 0.89 m. Choice A uses only the x-component of moment, C incorrectly inverts the ratio, D uses an approximation error in the magnitude calculations.

Question 5

Three concurrent forces meet at point P(2, 3, 1) m: F1⃗=80i^\vec{F_1} = 80\hat{i} N, F2⃗=60j^\vec{F_2} = 60\hat{j} N, and F3⃗=−50k^\vec{F_3} = -50\hat{k} N. These forces are to be replaced by an equivalent system consisting of a force F⃗\vec{F} at the origin and a couple M⃗\vec{M}. What is the magnitude of the required couple?

  1. 180.3180.3 N⋅m
  2. 156.5156.5 N⋅m
  3. 203.7203.7 N⋅m
  4. 174.9174.9 N⋅m (correct answer)
Explanation: The resultant force is R⃗=80i^+60j^−50k^\vec{R} = 80\hat{i} + 60\hat{j} - 50\hat{k} N (this remains the same). To move the force from point P to the origin, we need a couple equal to the moment of the original force system about the origin. Since all forces are concurrent at P, the moment about origin is: M⃗=rP⃗×R⃗=(2i^+3j^+k^)×(80i^+60j^−50k^)\vec{M} = \vec{r_P} \times \vec{R} = (2\hat{i} + 3\hat{j} + \hat{k}) \times (80\hat{i} + 60\hat{j} - 50\hat{k}). Computing: M⃗=∣i^j^k^2318060−50∣=i^(3×(−50)−1×60)−j^(2×(−50)−1×80)+k^(2×60−3×80)=i^(−150−60)−j^(−100−80)+k^(120−240)=−210i^+180j^−120k^\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 80 & 60 & -50 \end{vmatrix} = \hat{i}(3 \times (-50) - 1 \times 60) - \hat{j}(2 \times (-50) - 1 \times 80) + \hat{k}(2 \times 60 - 3 \times 80) = \hat{i}(-150 - 60) - \hat{j}(-100 - 80) + \hat{k}(120 - 240) = -210\hat{i} + 180\hat{j} - 120\hat{k} N⋅m. The magnitude is ∣M⃗∣=(−210)2+(180)2+(−120)2=44100+32400+14400=90900=174.9|\vec{M}| = \sqrt{(-210)^2 + (180)^2 + (-120)^2} = \sqrt{44100 + 32400 + 14400} = \sqrt{90900} = 174.9 N⋅m. Choice A incorrectly sums force magnitudes, B uses only two components, C makes sign errors in cross product.

Question 6

A 3D force system produces a resultant R⃗=90i^+120j^+40k^\vec{R} = 90\hat{i} + 120\hat{j} + 40\hat{k} N and a resultant moment about point A(1, 2, 0) m of MA⃗=200i^−150j^+180k^\vec{M_A} = 200\hat{i} - 150\hat{j} + 180\hat{k} N⋅m. If this system is equivalent to a single force acting along a line parallel to vector v^=0.6i^+0.8j^\hat{v} = 0.6\hat{i} + 0.8\hat{j}, what constraint must be satisfied?

  1. The given system is already equivalent to such a force since R⃗⋅k^≠0\vec{R} \cdot \hat{k} ≠ 0
  2. The system cannot be equivalent to such a force because R⃗\vec{R} has a non-zero z-component (correct answer)
  3. The system is equivalent if the moment about any point on the line has no component perpendicular to v^\hat{v}
  4. The system requires an additional couple in the z-direction to satisfy the constraint
Explanation: For a 3D force system to be equivalent to a single force acting along a specific direction v^\hat{v}, the resultant force must be parallel to that direction. The given direction vector v^=0.6i^+0.8j^\hat{v} = 0.6\hat{i} + 0.8\hat{j} lies entirely in the xy-plane (no z-component). However, the resultant force R⃗=90i^+120j^+40k^\vec{R} = 90\hat{i} + 120\hat{j} + 40\hat{k} N has a non-zero z-component (40 N). Since the resultant force is not parallel to the specified direction, the system cannot be equivalent to a single force acting along that line. The presence of the z-component in R⃗\vec{R} makes this reduction impossible. Choice A incorrectly suggests the system is already equivalent, C describes a condition for moment equilibrium but ignores force direction requirements, D suggests adding couples which would change the resultant force.