Question

The moment of force of 25N about a point is 2.5Nm. Find the perpendicular distance of force from that point.​

Answer 1

${\underline {\mathfrak { Question } }}$

The moment of force of 25N about a point is 2.5Nm. Find the perpendicular distance of force from that point.

${\underline {\mathfrak { Given } }}$

• Moment of force = 2.5Nm
• Force applied = 25N

${\underline { \mathfrak { To \: Find } }}$

• Perpendicular (⊥) distance from the point of rotation.

${\underline { \mathfrak { Solution }}}$

Let the perpendicular (⊥) distance from the point of rotation be x.

We know that,

• ${ \boxed {\underline {\underline {\tt {\sf {\red { Moment \: of \: force = Force \times ⊥ distance }}}}}}}$

$★$ So, after substituting values-

$☞ \sf\green { \: 2.5 = 25 \times x}$

$☞ \: \therefore \sf { \: x = ⊥ \: distance = \frac{2.5}{25}}$

$☞ \sf {⊥ \: distance = \frac{\cancel {25}^{1} }{\cancel25 \times 100}}$

$☞ \sf\green { ⊥ \: distance = \frac{1}{10} m}$

Or in cm-

$☞ \sf\green { ⊥ \: distance = \frac{1}{10} \times 100 = 10cm}$

Therefore, the perpendicular distance of force from that point is $\sf { \frac{1}{10}m }$ or 10 cm.

____________________

Answer 2

Answer:

10cm

Explanation:

perpendicular distance = moment of force / force applied

= 2.5/25

=1/10m

=(1/10×100)cm =10cm