Answer:
The second force at the doorknob
Explanation:
The magnitude of the torque is given by the following formula:
[tex]\tau=rFsin\theta[/tex]
F is the applied force, r is the distance of the line of action of force from the axis of rotation and [tex]\theta[/tex] is the angle between them, in this case is [tex]90^\circ[/tex] for both points.
The door's midpoint it's half the distance from the doorknob. Thus, the torque at the doorknob is twice the torque at the door's midpoint.
The torque applied by the second force at the knob of the door will be twice.
Torque is the momentum of force. It is also known as the tendency of the force to rotate a body. It is the product of the force and the distance between the force and the point about which the torque will be calculated.
As we want to know which force exerts the greater torque, therefore, let the length of the gate will be d. Also, the force that if applied to the door is F. Therefore,
Torque applied by the first force at the midpoint of the door
[tex]T_1 = F \times \dfrac{d}{2}[/tex]
Torque applied by the second force at the knob of the door
[tex]T_2 = F \times d[/tex]
If we calculate the ratio of the two torques,
[tex]\dfrac{T_1}{T_2} = \dfrac{F \times \frac{d}{2}}{F \times d}\\\\\dfrac{T_1}{T_2} = \dfrac{F \times d}{F \times d\times 2}\\\\\dfrac{T_1}{T_2} = \dfrac{1}{2}\\\\T_2 = 2T_1[/tex]
Hence, the torque applied by the second force at the knob of the door will be twice.
Learn more about Torque:
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