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) the first at the midpoint.
b) the second at the doorknob.
c) both exert equal non-zero torques.
d) both exert zero torques.
e) Additional information is needed.

b) the second at the doorknob.

Explanation:

The torque applied by a force is given by the equation

where

is the torque

is the component of the force perpendicular to the direction between the axis of rotation and the point of application of the force

d is the arm (the distance between the axis of rotation and the point of application of the force)

In this problem, we have two equal forces F both applied perpendicular to a door, so

The first force is applied at the midpoint of the door; if we call L the width of the door, then the arm of this force is , so the torque applied by this first force is

The 2nd force instead is applied at the doorknob, so the arm in this case is L:

d = L

So, the torque exerted by the second force is

Therefore, we see that the 2nd force exerts a greater torque.

the answer is 3.1 his answer is wrong mines is right

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west