Answer to Question #345393 in Analytic Geometry for rey

Question

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

a) Find the resultant of the two forces

b)find the total work done if the forces make the object move 2m along the vector S=i+j+k

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

d)what is unit vector along which the object will move to achieve maximum work done?

f) find F1.F2

g) find F1*F2

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) 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

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

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

$$W_{max}=\vec{R}\cdot\vec{m}$$

$$=14(\dfrac{14}{\sqrt{241}})-3(\dfrac{-3}{\sqrt{241}})-6(\dfrac{-6}{\sqrt{3}})=\sqrt{241}$$

d)

Unit vector is

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

f)

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

g)

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

$$=\vec{i}\begin{vmatrix} 3 & -6 \\ -6 & 0 \end{vmatrix}-\vec{j}\begin{vmatrix} 6 & -6 \\ 8 & 0 \end{vmatrix}+\vec{k}\begin{vmatrix} 6 & 3 \\ 8 & -6 \end{vmatrix}$$

$$=-36\vec{i}+48\vec{j}-60\vec{k}$$