Offset Torque: Solving the 1000N Problem

[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.

 

[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.

 

[DarkBlitz]

Thanks a lot! cleared it up!