A doorknob is located as far as possible from the door’s hinge line for a good reason. If you want to open a heavy door, you must certainly apply a force; that alone, however, is not enough. Where you apply that force and in what direction you push are also important. If you apply your force nearer to the hinge line than the knob, or at any angle other than 90° to the plane of the door, you must use a greater force to move the door than if you apply the force at the knob and perpendicular to the door’s plane.
To determine how F results in a rotation of the body around the rotation axis, we resolve F into two components . One component, called the radial component Fr’ points along Y. This component does not cause rotation, because it acts along a line that extends through O. (If you pull on a door parallel to the plane of the door, you do not rotate the door.).
The ability of F to rotate the body depends not only on the magnitude of its tangential component Ft, but also on just how far from 0 the force is applied. To include both these factors, we define a quantity called torque T as the product of the two factors and write it as
In the next chapter we shall discuss torque in a general way as being a vector. quantity. Here, however, because we consider only rotation around a single axis, we do not need vector notation-Instead, a torque has either a positive or negative value depending on the direction of rotation it would give a body initially at rest. If the body would rotate counterclockwise.
CHECKPOINT 6: The figure shows an overhead view of a meter stick that can pivot about the dot at the position marked 20 (for 20 em). All five horizontal forces on the stick have the same magnitude.