Two equal forces are applied to a door. The first force is applied at the midpoint of the door, the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?A) both exert equal non-zero torquesB) the first at the midpointC) both exerts zero torquesD) the second at the doorknobE) additional information is needed

Respuesta :

Answer:

D) the second at the doorknob

Explanation:

The torque exerted by a force is given by:

[tex]\tau = Fdsin \theta[/tex]

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation

[tex]\theta[/tex] is the angle between the direction of the force and d

In this problem, we have:

- Two forces of equal magnitude F

- Both forces are perpendicular to the door, so [tex]\theta=90^{\circ}, sin \theta=1[/tex]

- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force

--> therefore, the 2nd force exerts a greater torque