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Chapter 1: Problem 4

According to the distributive law for vector cross products $$ \mathbf{r} \times \mathbf{F}_{1}+\mathbf{r} \times \mathbf{F}_{2}=\mathbf{r} \times\left(\mathbf{F}_{1}+\mathbf{F}_{2}\right) $$ This states that the sum of the moments of two concurrent forces about a point is equal to the moment of the vector sum of the forces about the same point. Verify this by simple geometry in the special case where \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\) lie in the \(x y\) plane and intersect at the point \(P\), and \(\mathbf{r}\) is the displacement vector \(O P\).

Short Answer

Expert verified
With the diagram and the steps followed, it is visually and arithmetically demonstrated the validity of the distributive law for vector cross products in the context of moments of forces about a point.

Step by step solution

01

Set up the scenario using a diagram

Draw a Cartesian coordinate system in three dimensions. On the xy plane, define two vectors \(\mathbf{F}_1\) and \(\mathbf{F}_2\) that intersect at point P. Also, draw the displacement vector \(\mathbf{r}\) from origin O to point P.
02

Calculate the individual cross products

Calculate \(\mathbf{r} \times \(\mathbf{F}_1\) and \(\mathbf{r} \times \(\mathbf{F}_2\). In a cross product, the resultant vector is orthogonal (perpendicular) to the plane containing the original vectors. Therefore, these cross products yield vectors that are perpendicular to the xy plane.
03

Compute the vector sum of the two forces and its cross product with \(\mathbf{r}\)

Next, add together the two force vectors \(\mathbf{F}_1\) and \(\mathbf{F}_2\) to get the vector sum \(\mathbf{F}_1 + \mathbf{F}_2\). Then, calculate the cross product \(\mathbf{r} \times (\mathbf{F}_1 + \mathbf{F}_2)\) using the same principles described in Step 2.
04

Compare the results

The final step is to compare the results of Step 2 and Step 3. According to the given distributive law, the sum of the moments (which are the cross products) of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) about point P should be equal to the moment of the vector sum of \(\mathbf{F}_1\) and \(\mathbf{F}_2\) about the same point. If the vectors obtained in Step 2 and Step 3 are equivalent, this verifies the law.

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Most popular questions from this chapter

A rigid rod with negligible weight and small transverse dimensions carries a load \(W\) whose position is adjustable. The rod rests on a small roller at \(A\) and bears against the vertical wall at \(B\). Determine the distance \(x\) for any given value of \(\theta\) such that the rod will be in equilibrium. Assume that friction is negligible.

The angles between the vector \(\mathbf{F}=F_{x} \mathbf{i}+F_{y} \mathbf{j}+F_{\mathbf{z}} \mathbf{k}\) and the coordinate axes are \(\theta_{x}, \theta_{y}\), and \(\theta_{z}\) The cosines of these angles are known as direction cosines. Evaluate the direction cosines in terms of the components of \(\mathbf{F}\). Show that $$ \cos ^{2} \theta_{x}+\cos ^{2} \theta_{y}+\cos ^{2} \theta_{z}=1 $$

A block of weight \(W\) rests on an inclined plane which makes an angle \(\theta=\tan ^{-1} 3 / 4\) as shown. A force \(P\), parallel to the \(x\) axis, is applied to the block and gradually increased from zero; when \(P\) reaches the value \(0.4 W\) the block begins to slide. What is the coefficient of friction between the block and the inclined plane?

According to the distributive law for vector cross products $$ \mathbf{r} \times \mathbf{F}_{1}+\mathbf{r} \times \mathbf{F}_{2}=\mathbf{r} \times\left(\mathbf{F}_{1}+\mathbf{F}_{2}\right) $$ This states that the sum of the moments of two concurrent forces about a point is equal to the moment of the vector sum of the forces about the same point. Verify this by simple geometry in the special case where \(\mathbf{F}_{1}\) and \(\mathbf{F}_{2}\) lie in the \(x y\) plane and intersect at the point \(P\), and \(\mathbf{r}\) is the displacement vector \(O P\).

The top of a tin can is removed, and the empty can is inverted over a pair of billiard balls on a table as shown in the sketch. For certain combinations of sizes and weights the configuration shown is stable. For other combinations the can tips over when released. It is proposed to set up a demonstration for a temperance lecture by using in sequence a frozen-orange-juice can and a beer can with the same pair of billiard balls. Investigate whether tipping will occur for the following sizes and weights. \(\begin{array}{lcl} & \text { Orange juice } & \text { Beer } \\ \text { Diameter of ball } & 45 \mathrm{~mm} & 45 \mathrm{~mm} \\ \text { Weight of ball } & 2.0 \mathrm{~N} & 2.0 \mathrm{~N} \\ \text { Diameter of can } & 50 \mathrm{~mm} & 70 \mathrm{~mm} \\ \text { Weight of empty can with lid removed } & 0.57 \mathrm{~N} & 1.0 \mathrm{~N}\end{array}\)
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