To solve the problem, we need to determine the forces acting on the door at both the top and bottom hinges. We'll analyze the forces using the conditions for static equilibrium: the sum of forces in both the $x$- and $y$-directions must be zero, and the sum of the torques about any point must also be zero.

Given Data:

• Height of the door, $h = 2.2 \, \text{m}$
• Width of the door, $w = 0.828 \, \text{m}$
• Mass of the door, $m = 33 \, \text{kg}$
• Distance of hinges from the top and bottom of the door, $d = 0.50 \, \text{m}$
• Gravitational acceleration, $g = 9.8 \, \text{m/s}^2$
• The weight of the door, $W = mg = 33 \times 9.8 = 323.4 \, \text{N}$
• Force at the top hinge: $\vec{F}_\text{T} = F_{\text{Tx}}\hat{i} + F_{\text{Ty}}\hat{j}$
• Force at the bottom hinge: $\vec{F}_\text{B} = F_{\text{Bx}}\hat{i} + F_{\text{By}}\hat{j}$

Step 1: Force Balance in the $y$-direction

Since the door is in static equilibrium, the sum of the vertical forces must be zero. The door's weight $W$ is shared equally between the two hinges: $F_{\text{Ty}} + F_{\text{By}} = W$ Given that the door's weight is evenly distributed: $F_{\text{Ty}} = F_{\text{By}} = \frac{W}{2} = \frac{323.4 \, \text{N}}{2} = 161.7 \, \text{N}$

Step 2: Force Balance in the $x$-direction

The sum of the horizontal forces must be zero as there are no external horizontal forces acting on the door: $F_{\text{Tx}} + F_{\text{Bx}} = 0$ This implies: $F_{\text{Tx}} = -F_{\text{Bx}}$

Step 3: Torque Balance about the Bottom Hinge

Let's take the bottom hinge as the pivot point to sum torques. The torque due to the force at the top hinge $\vec{F}_\text{T}$, the door's weight $W$, and the force at the bottom hinge $\vec{F}_\text{B}$ should sum to zero.

The torque due to the door's weight $W$ is calculated at the center of the door: $\text{Torque due to } W = W \times \frac{h}{2} = 323.4 \times \frac{2.2}{2} = 355.74 \, \text{N}\cdot\text{m}$

The horizontal component of the force at the top hinge $F_{\text{Tx}}$ creates a counterclockwise torque: $\text{Torque due to } F_{\text{Tx}} = F_{\text{Tx}} \times h = F_{\text{Tx}} \times 2.2$

Setting the sum of torques about the bottom hinge equal to zero: $F_{\text{Tx}} \times 2.2 = 355.74$ $F_{\text{Tx}} = \frac{355.74}{2.2} = 161.7 \, \text{N}$

Since $F_{\text{Tx}} = -F_{\text{Bx}}$: $F_{\text{Bx}} = -161.7 \, \text{N}$

Step 4: Final Forces in Unit-Vector Notation

• Top Hinge Force: $\vec{F}_\text{T} = (161.7\hat{i} + 161.7\hat{j}) \, \text{N}$
• Bottom Hinge Force: $\vec{F}_\text{B} = (-161.7\hat{i} + 161.7\hat{j}) \, \text{N}$

Final Answer:

(a) The force on the door at the top hinge is: $\vec{F}_\text{T} = (161.7\hat{i} + 161.7\hat{j}) \, \text{N}$

(b) The force on the door at the bottom hinge is: $\vec{F}_\text{B} = (-161.7\hat{i} + 161.7\hat{j}) \, \text{N}$

Would you like further details on this solution or have any questions? Here are some related questions:

1. How would the forces change if the door's mass were doubled?
2. What would be the forces on the hinges if the door's width were increased?
3. How would the solution differ if the door had three hinges instead of two?
4. What role does the distance of the hinges from the top and bottom of the door play in the calculation?
5. How do you calculate the torque if the door were at an angle instead of vertical?

Tip: Always consider torque balance along with force balance when dealing with rotational equilibrium problems.