Question

A body is pivoted at a point. A force of 20N is applied at a distance of 30cm from the pivot. The moment of force about the pivot is
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Answer 1

» To Find :

The Moment of Force .

» Given :

• Force Applied = 20 N

• Perpendicular Distance = 30 cm = 0.3 m

» We Know :

Formula Moment For Force :

$\sf{\underline{\boxed{Torque(\tau) = Force \times Perpendicular\:Distance}}}$

» Concept :

According to the question , we have to Find the Moment of Force or torque , so we can simply use the Formula for finding the moment of force.

» Solution :

Given :

• Force Applied = 20 N
• Perpendicular Distance = 30 cm = 0.3 m

Using the formula and substituting the values in it, we get :

$\sf{\underline{\boxed{Torque(\tau) = Force \times Perpendicular\:Distance}}}$

$\sf{\Rightarrow \tau = 20 N \times 0.3 m}$

$\sf{\Rightarrow \tau = 20 N \times \dfrac{3}{10} m}$

$\sf{\Rightarrow \tau = \cancel{20} N \times \dfrac{3}{\cancel{10}} m}$

$\sf{\Rightarrow \tau = 2 N \times 3 m}$

$\sf{\Rightarrow \tau = 6 N\:m}$

Hence ,the Moment of Force or Torque is 6 N m.

» Additional information :

• Dimensional Formula of Work = $\sf{[M^{1}L^{2}T^{-2}]}$

• Dimensional Formula of Weight = $\sf{[M^{1}L^{1}T^{-2}]}$

• Dimensional Formula of Impulse = $\sf{[M^{1}L^{1}T^{-1}]}$

• Dimensional Formula of Force = $\sf{[M^{1}L^{1}T^{-2}]}$

Answer 2

Given :-

• Force Applied = 20 N
• Perpendicular Distance = 30 cm = 0.3 m

To Find :-

We Know that,

$\tt{\underline{\boxed{Torque(\tau) = Force \times Perpendicular\:Distance}}}$

Solution :-

Using this formula,

$\tt{\underline{\boxed{Torque(\tau) = Force \times Perpendicular\:Distance}}}$

Put the values,

$\tt{\implies \tau = 20 N \times 0.3 m}$

$\tt{\implies \tau = 20 N \times \dfrac{3}{10} m}$

$\tt{\implies \tau = \cancel{20} N \times \dfrac{3}{\cancel{10}} m}$

$\tt{\implies \tau = 2 N \times 3 m}$

$\tt{\implies \tau = 6 N\:m}$

Hence, the Moment of Force or Torque is 6 N m.