- #1

zenterix

- 530

- 74

- Homework Statement
- A force ##\vec{F}_1=A\hat{j}## is applied to ##\vec{r}_1=a\hat{i}## and a force ##\vec{F}_2## is applied to ##\vec{r}_2=b\hat{j}##.

Find the arm ##l## of the resultant force relative to the origin.

- Relevant Equations
- The solution below coincides with the solution manual but at this point it is pretty clear that I arrived at this answer by chance (and incorrect methods).

We have

The torque about the origin is

$$\vec{\tau}=(aA-bB)\hat{k}\tag{1}$$

The resultant force is

$$\vec{F}_R=B\hat{i}+A\hat{j}\tag{2}$$

At this point, all I did was compute

$$|\vec{r}\times\vec{F}_R|=|\vec{r}||\vec{F}_R|\sin{\theta}=l|\vec{F}_R|=|\vec{\tau}|\tag{3}$$

which led to

$$l=\frac{|\vec{\tau}|}{|\vec{F}_R|}=\frac{\sqrt{(aA-bB)^2}}{\sqrt{A^2+B^2}}\tag{4}$$

which coincides with the solution manual

My question is about (3).

I wrote ##|\vec{r}\times\vec{F}_R|=|\vec{\tau}|## without thinking much about why I did that and now it seems pretty incorrect.

Here is my current thinking

1) We have a system of two particles of certain unknown masses ##m_1## and ##m_2##.

2) The resultant force on the system works on the center of mass, which is at an unknown position ##\vec{r}## which appears in (3).

$$\vec{r}=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}$$

$$\vec{r}\times\vec{F}_R=\frac{(m_1\vec{r}_1+m_2\vec{r}_2)\times (\vec{F}_1+\vec{F}_2)}{m_1+m_2}$$

$$=\frac{m_1\vec{\tau}_1+m_2\vec{\tau}_2+m_1\vec{r}_1\times\vec{F}_2+m_2\vec{r}_2\times \vec{F}_1}{m_1+m_2}$$

The two last terms in the numerator are zero in this problem.

Thus,

$$\vec{r}\times\vec{F}_R=\frac{m_1\vec{\tau}_1+m_2\vec{\tau}_2}{m_1+m_2}$$

$$|\vec{r}\times\vec{F}_R|=\frac{m_1aA-m_2bB}{m_1+m_2}=l|\vec{F}_R|$$

$$l=\frac{m_1aA-m_2bB}{(m_1+m_2)\sqrt{A^2+B^2}}$$

How did the solution manual arrive at the rhs of (4) as the solution? Ie, why are there no masses in their solution?

The torque about the origin is

$$\vec{\tau}=(aA-bB)\hat{k}\tag{1}$$

The resultant force is

$$\vec{F}_R=B\hat{i}+A\hat{j}\tag{2}$$

At this point, all I did was compute

$$|\vec{r}\times\vec{F}_R|=|\vec{r}||\vec{F}_R|\sin{\theta}=l|\vec{F}_R|=|\vec{\tau}|\tag{3}$$

which led to

$$l=\frac{|\vec{\tau}|}{|\vec{F}_R|}=\frac{\sqrt{(aA-bB)^2}}{\sqrt{A^2+B^2}}\tag{4}$$

which coincides with the solution manual

My question is about (3).

I wrote ##|\vec{r}\times\vec{F}_R|=|\vec{\tau}|## without thinking much about why I did that and now it seems pretty incorrect.

Here is my current thinking

1) We have a system of two particles of certain unknown masses ##m_1## and ##m_2##.

2) The resultant force on the system works on the center of mass, which is at an unknown position ##\vec{r}## which appears in (3).

$$\vec{r}=\frac{m_1\vec{r}_1+m_2\vec{r}_2}{m_1+m_2}$$

$$\vec{r}\times\vec{F}_R=\frac{(m_1\vec{r}_1+m_2\vec{r}_2)\times (\vec{F}_1+\vec{F}_2)}{m_1+m_2}$$

$$=\frac{m_1\vec{\tau}_1+m_2\vec{\tau}_2+m_1\vec{r}_1\times\vec{F}_2+m_2\vec{r}_2\times \vec{F}_1}{m_1+m_2}$$

The two last terms in the numerator are zero in this problem.

Thus,

$$\vec{r}\times\vec{F}_R=\frac{m_1\vec{\tau}_1+m_2\vec{\tau}_2}{m_1+m_2}$$

$$|\vec{r}\times\vec{F}_R|=\frac{m_1aA-m_2bB}{m_1+m_2}=l|\vec{F}_R|$$

$$l=\frac{m_1aA-m_2bB}{(m_1+m_2)\sqrt{A^2+B^2}}$$

How did the solution manual arrive at the rhs of (4) as the solution? Ie, why are there no masses in their solution?

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