A 20 kg uniform door has a width of 1.2 m and a height of 2.5 m. The door is mounted on a post by a pair of hinges, marked 1 and 2, at the top and bottom of the door. An external force of 60 N, at an angle of 30° above the horizontal, is applied to the doorknob, as shown. The doorknob is 1.0 m above the bottom of the door.
-In the figure,the x-component of the force,exerted on the door at the top by hinge 1,is closest to:
A)zero
B)+55 N
C)+80 N
D)-55 N
E)-80 N

Question

A 20 kg uniform door has a width of 1.2 m and a height of 2.5 m. The door is mounted on a post by a pair of hinges, marked 1 and 2, at the top and bottom of the door. An external force of 60 N, at an angle of 30° above the horizontal, is applied to the doorknob, as shown. The doorknob is 1.0 m above the bottom of the door. 
-In the figure,the x-component of the force,exerted on the door at the top by hinge 1,is closest to:
A)zero 
B)+55 N 
C)+80 N 
D)-55 N 
E)-80 N

Quizplus · Accepted Answer

Since the door is in equilibrium, the sum of the moments about any point must be equal to zero. Let us choose the point where hinge 2 meets the ground as our point of reference. Then, we have:
(clockwise moments) - (counterclockwise moments) = 0
Clockwise moments:
- The weight of the door, which acts at the center of the door, is equal to 20 kg x 9.81 m/s^2 = 196.2 N. Its lever arm is half the height of the door, or 1.25 m, since it acts at the center.
- The force applied by hinge 1 acts horizontally at a distance of 1.2/2 = 0.6 m from the point of reference.
- The normal force exerted by hinge 2 acts vertically at a distance of 1.25 m from the point of reference.
Counterclockwise moments:
- The external force of 60 N applied at the doorknob acts at a distance of 1 m from the point of reference and at an angle of 30 degrees above the horizontal. Its lever arm can be found using trigonometry:
cos(30) = adjacent/hypotenuse
adjacent = hypotenuse x cos(30) = 1 x sqrt(3)/2
Therefore, the lever arm of the force is 0.5 m.
- The force applied by hinge 1 acts horizontally at a distance of 0.6 m from the point of reference, but in the opposite direction as the external force.

Putting it all together, we get:
- 196.2 N x 1.25 m - 60 N x 0.5 m - Fhinge1 x 0.6 m + Fhinge2 x 1.25 m = 0
Solving for Fhinge1, we get:
Fhinge1 = -55.4 N
Therefore, the x-component of the force, exerted on the door at the top by hinge 1, is closest to -55 N (choice D).

A 20 Kg Uniform Door Has a Width of 1