# Offset Torque Concept

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1. Mar 28, 2017

### DarkBlitz

Hello guys,

I am doing some basic design work and can't remember how offset torques work.

I am doing the sum of the torques should be equal to 0 in order to find the tipping point around either of the wheels.

My problem is with the 1000N applied at to top taking the pivot point around the left wheel.

Should I be using the component of the 1000N which is at right angles to the distance to the pivot point? (the green line) and take the green line as the distance? should I use 'h' as the distance to the pivot point with 1000N as the force? or a combination of both?

The same for m*g in the diagram. Should I assume that the centre of mass is where the weight is applied and take the perpendicular component?

Thanks very much for your help guys! Been a long time since I looked into this.

2. Mar 28, 2017

### jack action

By definition, the torque is the result of a cross product of 2 vectors, i.e. $\mathbf{T} = \mathbf{r} \times \mathbf{F}$. The magnitude of this torque (what you are looking for) is $T = rF\sin\theta$ where $\theta$ is the angle between the 2 vectors. This basically means that only the components of the vectors that are perpendicular to each other are considered.

You can see it as $r_{\perp} F$ or $r F_{\perp}$ as you wish, since both will give you the same answer: $rF\sin\theta$.

The following is a representation of $r F_{\perp}$:

The following is a representation of $r_{\perp} F$:

So in your case, you take 1000 N times h ( or $r_{\perp} F$). You could use $r F_{\perp}$ by using the length of the green line and by finding the perpendicular component of the 1000 N force (the orange line), it would give the same answer. It is just more complicated in this particular case.

Similarly, for the weight, you take mg times x2. Where x2 is the perpendicular component of the distance between the wheel and the center of mass.

3. Apr 7, 2017

### DarkBlitz

Thanks a lot! cleared it up!