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 sum of the y-components of the forces,exerted on the door by hinges 1 and 2,is closest to:
A)145 N
B)165 N
C)195 N
D)225 N
E)250 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 sum of the y-components of the forces,exerted on the door by hinges 1 and 2,is closest to:
A)145 N 
B)165 N 
C)195 N 
D)225 N 
E)250 N

Quizplus · Accepted Answer

First, we need to find the x and y components of the force applied on the doorknob.

F_x = 60 cos 30° = 51.96 N 
F_y = 60 sin 30° = 30 N

Now, we can use the fact that the door is in static equilibrium to find the sum of the y-components of the forces exerted on the door by hinges 1 and 2.

∑F_x = 0 
∑F_y = 0

Let's start by considering the forces acting on the top hinge (hinge 1). We have the following forces: 
- The force exerted on hinge 1 by the door, with a magnitude of F_d. 
- The force exerted on hinge 1 by the bottom hinge (hinge 2), with a magnitude of F_b.
- The vertical component of the force applied on the doorknob, with a magnitude of F_y.

Since the door is in static equilibrium, the sum of the forces acting on hinge 1 must be equal to zero.

F_d - F_b - F_y = 0

Now let's consider the forces acting on the bottom hinge (hinge 2). We have the following forces: 
- The force exerted on hinge 2 by the door, with a magnitude of F_d.
- The force exerted on hinge 2 by the top hinge (hinge 1), with a magnitude of F_b.
- The weight of the door, with a magnitude of F_w = mg = 20 kg * 9.81 m/s^2 = 196.2 N.
- The horizontal component of the force applied on the doorknob, with a magnitude of F_x.

Again, since the door is in static equilibrium, the sum of forces acting on hinge 2 must be equal to zero.

F_d + F_b - F_w - F_x = 0

We now have two equations with two unknowns (F_d and F_b). We can solve for these unknowns by adding the two equations:

2F_d - F_w = F_y + F_x

Plugging in the values we know, we get:

2F_d - 196.2 N = 30 N + 51.96 N 
2F_d = 278.16 N + 196.2 N = 474.36 N 
F_d = 237.18 N

Using the equation F_d - F_b - F_y = 0, we can solve for F_b:

F_b = F_d - F_y 
F_b = 237.18 N - 30 N = 207.18 N

Finally, to find the sum of the y-components of the forces exerted on the door by hinges 1 and 2, we add F_y, F_b, and the weight of the door (since it is acting downward):

∑F_y = F_y + F_b + F_w = 30 N + 207.18 N + 196.2 N = 433.38 N

The closest answer choice is B) 165 N.

A 20 Kg Uniform Door Has a Width of 1