Answer to Question #345067 in Analytic Geometry for guy

Question

A 9N force F1 and a 10N force F2 act in the same directions 2i+j-2k and 4i-3j respectively.

a) Find the resultant of the two forces

b)Find the total work done that the forces make the object move 2m along the vector s= i+j+k

c)find F1•F2

d)find F1 * F2

e)find the unit vector alone which the object will move to achieve maximum work done

f)what is the maximum work done that can be achieve if the tje forces displace the object by 1m

Answer 1

a)

$$\sqrt{(2)^2+(1)^2+(-2)^2}=3$$ $$\sqrt{(4)^2+(-3)^2+(0)^2}=5$$

$$\vec F_1=6\vec{i}+3\vec{j}-6\vec{k}$$ $$\vec F_2=8\vec{i}-6\vec{j}$$

The resultant of the two forces is

$$\vec{R}=\vec F_1+\vec F_2=14\vec{i}-3\vec{j}-6\vec{k}$$

b)

$$\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3}$$ $$\vec{r}=\dfrac{2}{\sqrt{3}}\vec{i}+\dfrac{2}{\sqrt{3}}\vec{j}+\dfrac{2}{\sqrt{3}}\vec{k}$$ $$W=\vec{R}\cdot\vec{r}=14(\dfrac{2}{\sqrt{3}})-3(\dfrac{2}{\sqrt{3}})-6(\dfrac{2}{\sqrt{3}})=\dfrac{10}{\sqrt{3}}$$

c)

$$\vec F_1\cdot \vec F_2=6(8)+3(-6)+0=30$$

d)

$$\vec F_1\times \vec F_2=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 6 & 3 & -6 \\ 8 & -6 & 0 \\ \end{vmatrix}$$ $$=\vec{i}\begin{vmatrix} 3 & -6 \\ -6 & 0 \end{vmatrix}-\vec{j}\begin{vmatrix} 6 & -6 \\ 8 & 0 \end{vmatrix}+\vec{k}\begin{vmatrix} 6 & 3 \\ 8 & -6 \end{vmatrix}$$ $$=-36\vec{i}+48\vec{j}-60\vec{k}$$

e) The maximum work will be done when the scalar product of vector of the resultant force and the vector of the displacement are collinear and have the same direction. Let $\vec{m}$ be a vector of displacement. Given that $|\vec{m}|=1.$

Then

$$|\vec{R}|=\sqrt{(14)^2+(-3)^2+(-6)^2}=\sqrt{241}$$

Unit vector is

$$\vec{m}=\dfrac{14}{\sqrt{241}}\vec{i}-\dfrac{3}{\sqrt{241}}\vec{j}-\dfrac{6}{\sqrt{3}}\vec{k}$$

f) The maximum work will be done when the scalar product of vector of the resultant force and the vector of the displacement are collinear and have the same direction. Let $\vec{m}$ be a vector of displacement. Given that $|\vec{m}|=1.$

Then

$$|\vec{R}|=\sqrt{(14)^2+(-3)^2+(-6)^2}=\sqrt{241}$$ $$\vec{m}=\dfrac{14}{\sqrt{241}}\vec{i}-\dfrac{3}{\sqrt{241}}\vec{j}-\dfrac{6}{\sqrt{3}}\vec{k}$$ $$W_{max}=\vec{R}\cdot\vec{m}$$ $$=14(\dfrac{14}{\sqrt{241}})-3(\dfrac{-3}{\sqrt{241}})-6(\dfrac{-6}{\sqrt{3}})=\sqrt{241}$$