8.2. Torque http://www.ck12.org
FIGURE 8.4
(a.) The force is perpendicular to the lever arm soτ=rF= 14. 47 m∗N. Since the doorknob is 5.7 cm from the left
edge of the door, the distance from the axis of rotation isr= 0. 81 m− 0. 057 m= 0. 753 m. The force required is then:
F=^140 ..^47753 mm∗N= 19. 216 → 19. 2 N
(b.) Since the doorknob is now in the center of the door, it is^0.^812 m= 0. 405 mfrom the axis of rotation. Therefore,
the force required is^140 ..^47405 mm∗N= 35. 73 → 35. 7 N
How much force would be necessary to open the door if the doorknob were placed along the axis of rotation?
http://demonstrations.wolfram.com/TorqueExertedOpeningADoor/
Illustrative Example 8.1.2
InFigure8.5, a forceFof 95.0 N is applied to a hinged rod of lengthr= 2. 2 m. The angle betweenFandris
130-degrees. Find the magnitude of the torque that the forceFexerts upon the rod.
Answer: