No, the statement is incorrect. When using an inclined plane, the effort (force applied to move an object) is typically less than the resistance (force opposing the motion). This is due to the mechanical advantage provided by the inclined plane.
The mechanical advantage of an inclined plane can be calculated using the formula:
\[Mechanical\: Advantage (MA) = \frac{Length\: of\: inclined\: plane (L)}{Height\: of\: inclined\: plane (h)}\]
Since the inclined plane length is typically longer than the height, the mechanical advantage is greater than 1. That means the effort force (De) required to move an object up the inclined plane is less than the resistance force (Dr) acting against the motion.
Whenever you use an inclined plane. De is greater than dr?
3 answers
De is less 5han dr.
Apologies for the confusion in the previous response. In the case of an inclined plane, the effort force (De) required to move an object up the incline is greater than the resistance force (Dr) acting against the motion. This is because the inclined plane reduces the vertical height that the object needs to be lifted, but it lengthens the distance it needs to be moved horizontally. Therefore, more effort force is needed to overcome the resistance force and move the object up the inclined plane.
2 answers