Question

Force F = (-6.0 N)ihat + (2.0 N) jhatacts on a particle with position vector r =(4.0 m) ihat+ (4.0 m) jhat.

(a) What is the torque on the particle aboutthe origin?
1 N·m ihat+ 2 N·m jhat+ 3 N·m khat

(b) What is the angle between the directions of rand F? (If there is no torque, enter 0.)
4°

Answer 1

(a)
`\overrightarrow{\tau }=32\widehat{K}  Nm`

(b) 116.6°

Step-by-step explanation:

`\overrightarrow{F}=-6\widehat{i}+2\widehat{j}`

`\overrightarrow{r}=4\widehat{i}+4\widehat{j}`

(a) Torque is defined as

`\overrightarrow{\tau }=\overrightarrow{r}* \overrightarrow{F}`

`\overrightarrow{\tau }=\left (4\widehat{i}+4\widehat{j}  \right )* \left (-6\widehat{i}+2\widehat{j}  \right )`

`\overrightarrow{\tau }=32\widehat{K}  Nm`

(b) Let θ be the angle between the r and F.

magnitude of r =
`\sqrt{4^(2)+4^(2)}=√(32)`

magnitude of F =
`\sqrt{6^(2)+2^(2)}=√(40)`

`Cos\theta =\frac{\overrightarrow{r}.\overrightarrow{F}}{rF}`

`Cos\theta =(-24+8)/(√(32*40))`

θ = 116.6°

Answer 2

`\tau=(-32k)\ N-m`

`\theta=116.55^(\circ)`

Step-by-step explanation:

Given that.

Force acting on the particle,
`F=(-6i + 2.0j)\ N`

Position of the particle,
`r=(4i+4j)\ m`

To find,

(a) Torque on the particle about the origin.

(b) The angle between the directions of r and F

Solution,

(a) Torque acting on the particle is a scalar quantity. It is given by the cross product of force and position. It is given by :

`\tau=F* r`

`\tau=(-6i + 2.0j)* (4i+4j)`

`\tau=\begin{pmatrix}0&0&-32\end{pmatrix}`

`\tau=(-32k)\ N-m`

So, the torque on the particle about the origin is (32 N-m).

(b) Magnitude of r,
`|r|=√(4^2+4^2)=5.65\ m`

Magnitude of F,
`|F|=√((-6)^2+2^2)=6.324\ m`

Using dot product formula,

`F{\circ}\ r=|F|.|r|\ cos\theta`

`cos\theta=\frac{F{\circ} r}r`

`cos\theta=(-24+8)/(6.324* 5.65)`

`\theta=116.55^(\circ)`

Therefore, this is the required solution.