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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.

Respuesta :

Answer:

B

Explanation:

Torque exerted on an object depends upon three things.

1. Moment arm

2. Magnitude of applied force

3. Direction of force

T=r x F

T= r * F sinФ

Let T1 is force due to force applied at mid point of door.

let d is the width of door, so

T1= [tex]\frac{d}{2}* F * sin(90[/tex]°)                  (∴ moment arm =d)

T1=[tex]\frac{d* F}{2}[/tex]  __________(2)       (∴ sin 90°=1)

Let T2 is the torque due to force applied at the doorknob.

T2=  [tex]d* F * sin(90)[/tex]                                (∴ moment arm =d)

T2=[tex]d*F[/tex] __________(2)                      (∴ sin 90°=1)

From equation (1) and equation (2)

2T1=T2

or

T1= T2/2

As T2 is greater so option B is the correct answer.

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