a person exerts 35 lbs of force to turn a door handle ( radius = 4 cm). How much force is exerted on the inner "axle" with a radius of 5 mm?
To find out how much force is exerted on the inner "axle" with a radius of 5 mm, we can use the principle of mechanical advantage. The mechanical advantage of a simple machine, such as a lever, is defined as the ratio of the output force to the input force.
In this scenario, the door handle acts as a lever, with the outer "axle" being the point where the input force is applied and the inner "axle" being the point where the output force is exerted.
To determine the force on the inner "axle" (output force), we need to find the mechanical advantage of the lever system. The mechanical advantage of a lever can be calculated using the formula:
Mechanical Advantage = Radius of Outer Axle / Radius of Inner Axle
In this case, the radius of the outer axle is 4 cm, which is equivalent to 40 mm, and the radius of the inner axle is 5 mm. Substituting these values into the formula, we get:
Mechanical Advantage = 40 mm / 5 mm = 8
This means that the mechanical advantage of the lever system is 8, which implies that the output force exerted on the inner "axle" will be 8 times greater than the input force.
Given that the person exerts a force of 35 lbs on the door handle, we can calculate the force exerted on the inner "axle" as follows:
Force on Inner Axle = Mechanical Advantage x Input Force
Force on Inner Axle = 8 x 35 lbs
Hence, the force exerted on the inner "axle" is 280 lbs.