Torque about an axis

A force is applied perpendicular to the edge of a cube of side length 'a' as shown in the diagram. The torque about edge C D CD and A B AB respectively is

0, Fa 0, 0 Fa, Fa Fa, 0

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2 solutions

Md Zuhair
Mar 26, 2017

The perpendicular distance from the point of rotation when cross producted with Force given us Torque

So τ S u r f a c e = F × r \tau_{Surface} = \vec{F} \times \vec{r} .

Now In these case Angle between r \vec{r} and F \vec{F} is 9 0 o 90^{o}

SO τ C D = F × 0 = 0 \tau_{CD} = \vec{F} \times 0 = \boxed{0}

And τ A B = F × a = F a \tau_{AB} = \vec{F} \times a = \boxed{Fa} where a a is the edge of the cube

The formula for torque you used is incorrect. It should be τ = r × F \vec \tau = \vec r \times \vec F .

Rohit Gupta - 4 years, 2 months ago

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Yes you are correct

Md Zuhair - 4 years, 2 months ago
Swayanprava Giri
Oct 26, 2020

ABOUT CD ZERO AND ABOUT AB Fa

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