Finding the Pivot Point of a Chassis: Need Help!

[MrAJMemphis]

I am having some real trouble with this, could anybody help, attached is a picture of a chassis with some weights on etc, i need to find out where the pivot point would be. i can't move any of the objects around, does anybody know how to do it?

 

[HallsofIvy]

The torque (turning force) about any point at distance x from a mass is its weight times that distance: Wx. The total torque of a group of mass is the total of the torques for each one. In particular, if x is measured from the far left, the torque due to the first 6N weight is 6(x- .5), due to the next 10N weight, 10(x- 3), due to the next weight, 2(x- 10), and the last, 8(x- 5), so the total torque is
6(x- .5)+ 10(x- 3)+ 2(x- 10)+ 8(x- 5). in order that the this balance, the total torque must be 0. Set that equal to 0 and solve for x.

Note that, because the pivot applies a force vertically, only the horizontal distance is important. (Which is pretty much implied by the way to problem is given: they do not tell you the height of that "up" section on the right.)

 

[MrAJMemphis]

So if the vertical was dimensioned would this change the any part of the equation?

 

[HallsofIvy]

No, it wouldn't. The only forces here are the weights which act only vertically. As far as the pivot point is concerned only horizontal distances matter. My point was that if the vertical distance were necessary, since it is not given, you would not be able to do the problem. The fact that the vertical distance is not given, in a "book" problem, is a hint that it is not necessary.

 

[MrAJMemphis]

The answer i have got for that then is the pivot is 3.577M in from the left. cool, well that is a massive help, i can not tell you how much i appreciate it, thanks very much HallsofIvy :)