Question

A force of 100 N is acting through points A (1, 2) and B (2, 1). The moment of force about point O (0, 0) is:

Answer 1

Answer:

According to the definition of torque,

τ

=

r

×

F

Given that,the force is

F

=20

i

^

N

and the arm vector is

r

=(0−3) +(2−0)

j

^

+(0−0)

)m

Therefore,

−3

20

=(0−0)

−(0−0)

+(0−40)

τ

∣=40N−m

Answer 2

Step-by-step Explanation:

Given: Force F = 100 N

Coordinates on which the force is acting = A (1, 2) and B (2, 1)

To Find: the moment of the force about point O (0, 0)

Solution:

• Determining the moment of the force about origin O (0,0)

The moment of force (torque) τ is given by

$\vec{\tau} = \vec{r} \times \vec{F}$ . . . . . . . . . . (1)

where $\vec{r}$ is the position vector.

With the coordinates we have,

$\vec{r} = (x_B - x_A) \hat{i} + (y_B - y_A) \hat{j} = \hat{i}-\hat{j}$  . . . . . . . . (2)

Now, considering the attached figure, we have AB = $\sqrt{2}$ and BC = 1 such that, $cos \theta = \frac{BC}{AB} = \frac{1}{\sqrt{2}} \Rightarrow \theta = 45^{o}$ . . . . . . . . . (3)

Further, resolving components of the force F, we get;

$\vec{F} = F cos \theta \hat{i} + F sin \theta \hat{j}$ . . . . . . . (4)

using (2), (3), and (4) in (1), we get

$\vec{\tau} = \vec{r} \times \vec{F} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\1 & -1 & 0\\100 \ cos45^o & 100 \ sin45^o & 0\end{vmatrix} =141.42 \hat{k}$

$\Rightarrow \tau = 141.42 \ N.m$

Hence, the moment of the force about point O (0, 0) is 141.42 N.m