I'm trying to show how you can cornerweight a car with bathroom scales and 2x4s but it's late and I'm not thinking straight.

Given the force at B, how do you calculate the the actual force at the tire?

 

Ok, I think I've got this figured out...

This is prob simpler since it's just similar triangles of different sizes. If anyone wants to add input and correct my logic, feel free. If all goes well, this goes to DIY forum. h:

BTW, "Will-style" MS Paint drawing ownerages. :exnbp:

 

The similar triangles approach works as far as the math goes...but I can't see how you can get the car on there evenly. The angle at the ground, no matter how small (unless we're talking a really long 2x4, but then I doubt you'd get a good reading), would still create uneven weight distribution over that wheel as the whole car sits slightly at an angle. Maybe I'm just not seeing something...and now I'll spend the day at classes thinking about solutions...great creativity by the way.

 

"WiLL style"? :fawk:

i dont understand cornerweighing. wont the bathroom scales have to support at least 6-700lbs? those are huge scales.

 

I don't know if you meant that the scales are made to support 6-700 lbs or that using a technique of cornerweighing, that each corner will have that much weight.
agent87's technique is using much less weight on the scale and then using geometry to solve for a supposed equivalent weight of car's corner.

Let's say the bathroom scale reads up to 300lbs. No way can it give a reading on the car's corner, so he's "lightening" the load on the scale by using the 2x4 beam. I just drew some diagrams up, but I keep coming up with that the weight on the scale will have to be more than the weight of the corner.

Similiar Triangles as in agent87's diagram:
a/b = A/B
given lengths of a=5 and b=15, if there's 500 lbs on b, there has to be 1500 lbs on B (scale end). Therefore, having the opposite of the intended result.

But Torque is telling me the opposite...since Torque = force * distance, a weight of 500 lbs at point b along the horizontal, at a distance of 5, would give a total of 2500lb-ft of torque at the origin. At B, the same torque of 2500 would require roughly 166 lbs at 15 ft.

Maybe I'm totally looking at this wrong, and my brain is totally not functioning right...(kinda burnt out from test week)...I'll keep thinking about this. It's starting to bother me now...at least I'll have something to do in the next lecture.

 

the only problem with your 2x4 method is because of the width of the tire it leaves a lot of room for error.

 

:chuckles:

I hear ya. Thanks for helping me try to figure this out.

Does this help?

Another problem that's come up includes getting the car to sit evenly on all 4 corners.