A uniform door (0.81 m wide and 2.1 m high) weighs 150 N and is hung on two hinges that fasten the long left side of the door to a vertical wall. The hinges are 1.9 m apart. Assume that the lower hinge bears all the weight of the door.
Determine the magnitude and direction of the force applied by the door to the upper hinge.
Find the magnitude and direction of the horizontal component of the force applied to the door by the lower hinge.
determine the magnitude and direction (below the horizontal) of the force applied by the door to the lower hinge.
To solve the problem, we need to apply the principles of equilibrium and analyze the forces acting on the door.
1. Magnitude and direction of the force applied by the door to the upper hinge:
Since the door is in equilibrium, the net torque acting on it must be zero. The torque produced by the door's weight is countered by the torque produced by the force at the upper hinge.
Let's start by calculating the torque due to the door's weight. The weight acts vertically downwards at the center of the door, which is 1.05 m from the hinges (halfway between the two hinges).
Torque due to weight (τ_weight) = weight * distance from hinge
τ_weight = 150 N * 1.05 m = 157.5 N·m (magnitude)
Since the net torque is zero, the torque produced by the force at the upper hinge must balance this.
Torque due to the force at the upper hinge (τ_upper) = force at upper hinge * distance from hinge
Since τ_upper = τ_weight, we can rearrange the formula to find the force at the upper hinge:
Force at upper hinge = τ_weight / distance from hinge
Force at upper hinge = 157.5 N·m / 1.9 m = 82.89 N (magnitude)
The direction of the force applied by the door to the upper hinge is downwards (opposite to the force of gravity).
2. Magnitude and direction of the horizontal component of the force applied to the door by the lower hinge:
The lower hinge bears all the weight of the door, so the magnitude of the force applied by the lower hinge is equal to the weight of the door, which is 150 N.
To find the horizontal component of the force applied by the lower hinge, we need to consider the geometry of the door. Since the door is hung on the left side, the force applied by the lower hinge is not purely vertical. It has a horizontal component as well.
Using trigonometry, we can find the horizontal component:
Horizontal component = Force applied by lower hinge * cos(angle)
The angle can be found using the triangle formed by the door, wall, and a horizontal line:
tan(angle) = height of the door / distance between hinges
angle = atan(height of the door / distance between hinges)
angle = atan(2.1 m / 1.9 m) = 49.78°
Horizontal component = 150 N * cos(49.78°) = 97.86 N
The direction of the horizontal component of the force applied to the door by the lower hinge is towards the wall.
3. Magnitude and direction (below the horizontal) of the force applied by the door to the lower hinge:
Since the upper hinge is at a higher position than the lower hinge, the door exerts a force on the lower hinge directed upwards.
The magnitude of this force can be determined by applying vertical equilibrium. The vertical component of the force applied by the door to the lower hinge must balance the weight of the door.
Vertical component of force at lower hinge = weight of the door = 150 N
The direction of the force applied by the door to the lower hinge is upwards.