Question

Numerical
1- Calculate the moment of force of 8N
applied on a body at a distance of 30 cm from a pivoted point.
who will tell the right answer i will mark as brainlist

Answer 1

• Force=8N
• Perpendicular distance=30cm=0.3m

We know

$\boxed{\sf Moment\:of\:Force=Force\times Perpendicular distance}$

$\\ \sf\longmapsto Moment\:of\:Force=8N\times 0.3m$

$\\ \sf\longmapsto Moment\:of\:Force=\dfrac{8\times 3}{10}$

$\\ \sf\longmapsto Moment\:of\:Force=\dfrac{24}{10}$

$\\ \sf\longmapsto Moment\:of\:Force=2.4Nm$

Answer 2

Answer:

Force=8N

Perpendicular distance=30cm=0.3m

We know

\boxed{\sf Moment\:of\:Force=Force\times Perpendicular distance}

MomentofForce=Force×Perpendiculardistance

\begin{gathered}\\ \sf\longmapsto Moment\:of\:Force=8N\times 0.3m\end{gathered}

⟼MomentofForce=8N×0.3m

\begin{gathered}\\ \sf\longmapsto Moment\:of\:Force=\dfrac{8\times 3}{10}\end{gathered}

⟼MomentofForce=

10

8×3

\begin{gathered}\\ \sf\longmapsto Moment\:of\:Force=\dfrac{24}{10}\end{gathered}

⟼MomentofForce=

10

24

\begin{gathered}\\ \sf\longmapsto Moment\:of\:Force=2.4Nm\end{gathered}

⟼MomentofForce=2.4Nm