# Inclined Plane Physics (Force to Push Object and Distance Explained)

VAM! Physics & Engineering
Published at : 24 Oct 2020
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In this video we will be going over another simple machine the inclined plane. We will be covering how to calculate the force needed to push a crate up an incline (Assuming no friction) and how distance changes relative to raising an object straight up
If you are to lift a box straight up that weights 49 newtons. It would require a force of greater than 49 newtons to raise the box.
Now if you were to put that box on a frictionless inclined plane with an angle of 30 degrees it would only require a force greater than half the force required to life the box straight up or 24.5 newtons.
We can calculate the force needed to push up the crate by taking the sine value of the angle times the force due to gravity in this case 49 newtons. We are left with a force of 24.5 newtons. The way I always remember to calculate forces on an inclined is to draw a triangle with the hypotenuse or longest side representing the acceleration due to gravity times the mass going straight down. The angle of the incline then goes in the top corner. We can then solve the right triangle for the force needed to move the object by completing SOH CAH TOA.
Another way to find the force needed to push an object up an incline is to take the height of ramp in this case 10 meters and divide by the length traveled on the ramp (or hypotenuse) in this case 20 meters. This gives us the answer that half the force of lifting the box straight up is needed when pushing the box up the incline
As stated in previous videos what happens when you reduce the amount of force needed to move an object? the displacement increases. In this example lifting the box straight up we only have to move the box 10 meters. In the inclined plane example we have to push the box 20 meters to gain the same height of 10 meters.
When pushing the box up the incline we have traded off reducing the force by half at the cost of increasing the distance by 2
Notice that if we take the formula for work (force times distance) we get the same amount of work. 20 meters times 24.5 newtons equals 490 joules and 10 times 49 newtons equals 490 joules