While calculating torque, we assume that the force to the moment is perpendicular to it.
I was wondering, what would happen if its not perpendicular.
Please help!

If the force is not perpendicular, there are two ways of calculating the torque. I refer to this picture.

Splitting the force (\(F\)) into two components, one perpendicular to the axis (\(F_h\)) and one directly towards the axis (\(F_v\)). Then the torque \(\tau\) becomes \(\tau=F_h l=Fl\sin(\theta)\).

Projecting the arm so it's perpendicular to the force. Here, the torque becomes \(\tau=Fl'=Fl\sin(\theta)\).
As you see, the two methods agree, and are actually the same thing.

In more advanced physics, you usually calculate torque using vectors instead: \(\vec \tau=\vec l\times \vec F\), where \(\|\vec \tau\|=\|\vec l\| \| \vec F\| \sin(\theta)\).

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## Comments

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TopNewestUse VECTORS.

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Torque is r X F. Take the cross-product.

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if it is parallel,and u happen to be strong enough as to break the axis of rotation,linear motion takes place

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If the force is not perpendicular, there are two ways of calculating the torque. I refer to this picture.

Splitting the force (\(F\)) into two components, one perpendicular to the axis (\(F_h\)) and one directly towards the axis (\(F_v\)). Then the torque \(\tau\) becomes \(\tau=F_h l=Fl\sin(\theta)\).

Projecting the arm so it's perpendicular to the force. Here, the torque becomes \(\tau=Fl'=Fl\sin(\theta)\). As you see, the two methods agree, and are actually the same thing.

In more advanced physics, you usually calculate torque using vectors instead: \(\vec \tau=\vec l\times \vec F\), where \(\|\vec \tau\|=\|\vec l\| \| \vec F\| \sin(\theta)\).

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