# The Force of Impact

• Newtons-law

#### Newtons-law

A Iron flat bar 1 inch wide resting on the edge of a 1/4 inch thick plate, there is 40 lbs of force applied by the iron bar on the metal plate 1 inch by 1/2 inch. The plate that is 1/4 inch thick can roll forward by gravity because the plate is resting on ball bearings that allow the plate to roll forward 1 1/2 inch.
Question: If you let a 10 lbs ball roll 5ft at a 45 degree angle down hill and on to the plate that is resting on ball bearings held in place by 40 lbs of force 1 inch wide by 1/2 inch deep by the iron bar. (The plate is the Impact point) the ball will stop on the plate. Will the force of Impact move the plate forward 1 1/2 inch?

Hi Newtons-law!

Hint: how much work has to be done against the iron bar to move it 1 1/2 inches?

The plate is on a 45 degree angle and wants to slam forward but can not because of the 40 lbs of force keeping it in place. The 1 inch flat bar rests on the plate 1/2 inch deep which both are smooth on the surface. speed of 5 ft x 10 lbs = > amount of work to move the plate forward 1 1/2 inch. If lubricant added 5 lbs is enough.

The plate is on a 45 degree angle and wants to slam forward but can not because of the 40 lbs of force keeping it in place. The 1 inch flat bar rests on the plate 1/2 inch deep which both are smooth on the surface. speed of 5 ft x 10 lbs = > amount of work to move the plate forward 1 1/2 inch. If lubricant added 5 lbs is enough.

Hi Newtons-law!

This is physics … you need equations … just "yes" or "no" isn't enough.

(can you check the question … is the plate on the slope, or is it on a flat horizontal surface?)
(and lubricant is irrelevant … this is frictionless anyway … ball-bearings mean no friction!)

I think you don't understand what the iron bar is doing. It's pressing up against the edge of the plate … that's the vertical side-edge, not the top … and the plate has to push it back 1 1/2 inches, against a resisting force of 40lbs.

The ball rolls down the slope faster and faster … its speed is not proportional to the 5 ft distance.

Now you have to put all that into equations.

as they say … go figure!

The bar is sitting on top of the plate with 40 lbs applied, the plate slides forward the bar never moves forward or backwards, but can fall through the gap or opening the force of impact on the plate can create. The force of impact doesn't have to move 40 lbs just simply slide under it. And I don't know how to write the equation. That why I ask smarter people. Like you.