Statics Quiz: Moment About A Point 3d

2 questions · exam conditions

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Moment About A Point 3dQuestion 1 of 2

A force $\vec{F} = 20\hat{i} + 30\hat{j} - 15\hat{k}$ N is applied at point P(4, -2, 3) m. What is the magnitude of the moment of this force about point O(1, 2, -1) m?

75.0

N⋅m

125.6

N⋅m

98.7

N⋅m

156.2

N⋅m

89.4

N⋅m

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Statics Quiz

Statics Quiz: Moment About A Point 3d

Practice Moment About A Point 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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Question 1

A force $\vec{F} = 20\hat{i} + 30\hat{j} - 15\hat{k}$ N is applied at point P(4, -2, 3) m. What is the magnitude of the moment of this force about point O(1, 2, -1) m?

$75.0$ N⋅m
$125.6$ N⋅m (correct answer)
$98.7$ N⋅m
$156.2$ N⋅m
$89.4$ N⋅m

Explanation: When you encounter a moment calculation problem in 3D space, you're dealing with the cross product of a position vector and a force vector. The moment

\vec{M} = \vec{r} \times \vec{F}

, where

\vec{r}

is the position vector from the reference point to the point of force application. First, find the position vector from O to P:

\vec{r} = (4-1)\hat{i} + (-2-2)\hat{j} + (3-(-1))\hat{k} = 3\hat{i} - 4\hat{j} + 4\hat{k}

m. Now calculate the cross product

\vec{M} = \vec{r} \times \vec{F}

= \hat{i}[(-4)(-15) - (4)(30)] - \hat{j}[(3)(-15) - (4)(20)] + \hat{k}[(3)(30) - (-4)(20)]

= \hat{i}[60 - 120] - \hat{j}[-45 - 80] + \hat{k}[90 + 80]

= -60\hat{i} + 125\hat{j} + 170\hat{k}

N⋅m The magnitude is:

|\vec{M}| = \sqrt{(-60)^2 + (125)^2 + (170)^2} = \sqrt{3600 + 15625 + 28900} = \sqrt{48125} = 125.6

N⋅m. Answer (B) is correct. Answer (A) likely comes from incorrectly calculating just one component of the cross product. Answer (C) might result from sign errors in the cross product calculation. Answer (D) could come from arithmetic mistakes in the final magnitude calculation or incorrectly setting up the position vector. Always double-check your cross product components and remember that the position vector goes from the reference point TO the force application point.

Question 2

A wrench applies a force $\vec{F} = 20\hat{j}$ N at point P(0, 0.3, 0) m on a bolt. If the moment about the bolt axis (z-axis) must not exceed 4 N⋅m in magnitude, what is the maximum distance the point of application can be moved in the x-direction while maintaining the same force direction?

0.2 m (correct answer)
0.5 m
0.8 m
1.0 m

Explanation: The moment about the z-axis is given by

M_z = \vec{r} \times \vec{F} \cdot \hat{k}

. If the force is moved to point (x, 0.3, 0), then

\vec{r} = x\hat{i} + 0.3\hat{j}

and

\vec{F} = 20\hat{j}

. The cross product is

\vec{r} \times \vec{F} = (x\hat{i} + 0.3\hat{j}) \times (20\hat{j}) = x(20)(\hat{i} \times \hat{j}) = 20x\hat{k}

. The z-component is

M_z = 20x

. For

|M_z| \leq 4

, we need

|20x| \leq 4

, so

|x| \leq 0.2

. The maximum distance in the x-direction is 0.2 m. Choice B incorrectly uses the force magnitude instead of the moment arm. Choice C assumes the constraint applies to the total moment magnitude. Choice D neglects the force magnitude in the calculation.