uniform door (0.81 m wide and 2.1 m high) weighs 122 N and is hung on two hinges that fasten the long left side of the door to a vertical wall. The hinges are 1.5 m apart. Assume that the lower hinge bears all the weight of the door.
(a) Find the magnitude and direction of the horizontal component of the force applied to the door by the upper hinge.
magnitude N
direction ---Select--- to the left to the right upwards downwards
(b) Find the magnitude and direction of the horizontal component of the force applied to the door by the lower hinge.
magnitude N
direction ---Select--- to the left to the right upwards downwards
(c) Determine the magnitude and direction of the force applied by the door to the upper hinge.
magnitude N
Direction ---Select--- to the left to the right upwards downwards
(d) determine the magnitude and direction (below the horizontal) of the force applied by the door to the lower hinge.
magnitude N
direction °
This is far to difficult to do in ASCII.
STart with a drawing, at each hinge you have a vertical force and a horizontal force. The sum of the vertical forces equals weight of the door. The sum of the horizontal forces is zero, and the sum of moments about any point is zero. Those relations will yield the answer.
To solve this problem, we can use the principle of moments and the fact that the door is in equilibrium.
Let's start by finding the weight of the door.
Weight of the door = mass x gravity
Given that the weight of the door is 122 N, we can find the mass by using the equation weight = mass x gravity. Assuming the acceleration due to gravity is approximately 9.8 m/s², we have
122 N = mass x 9.8 m/s²
Rearranging the equation, we find
mass = 122 N / 9.8 m/s² = 12.45 kg
Now, we can find the point where the weight of the door is acting. The weight acts at the midpoint of the door, so its distance from the left hinge is half the width of the door.
Distance of weight from left hinge = 0.81 m / 2 = 0.405 m
(a) Finding the magnitude and direction of the horizontal component of the force applied to the door by the upper hinge:
Since the door is in equilibrium, the sum of the moments about any point must be zero. We can choose the left hinge as the pivot point.
Taking moments about the left hinge, we have:
Sum of clockwise moments = Sum of anticlockwise moments
The only forces acting anticlockwise are the horizontal components of the forces from both hinges. The horizontal component of the force from the lower hinge is equal to the horizontal component of the force from the upper hinge.
Let's denote the magnitude of the horizontal component of the force from the upper hinge as F_upper.
Therefore, F_upper x 1.5 m = (Weight of the door) x (Distance of weight from left hinge)
F_upper = (122 N) x (0.405 m) / (1.5 m)
F_upper ≈ 32.76 N
The magnitude of the horizontal component of the force applied to the door by the upper hinge is approximately 32.76 N.
Since the force applied by the upper hinge is in the opposite direction to the horizontal component of the force applied by the lower hinge, the direction of the force applied by the upper hinge is to the left.
Direction: to the left
(b) Finding the magnitude and direction of the horizontal component of the force applied to the door by the lower hinge:
We already found the magnitude of the horizontal component of the force applied by the upper hinge, which is approximately 32.76 N.
Since the door is in equilibrium, the sum of the horizontal components of the forces from both hinges must be zero.
Therefore, the magnitude of the horizontal component of the force applied by the lower hinge is also 32.76 N.
Since the force applied by the lower hinge is in the opposite direction to the force applied by the upper hinge, the direction of the force applied by the lower hinge is to the right.
Direction: to the right
(c) Determining the magnitude and direction of the force applied by the door to the upper hinge:
Since the door is in equilibrium, the sum of the vertical forces acting on it must be zero.
The vertical component of the force applied by the upper hinge is equal to the weight of the door.
Therefore, the magnitude of the force applied by the door to the upper hinge is equal to the weight of the door, which is 122 N.
Since the weight of the door acts downwards, the force applied by the door to the upper hinge also acts downwards.
Direction: downwards
(d) Determining the magnitude and direction (below the horizontal) of the force applied by the door to the lower hinge:
The magnitude of the vertical component of the force applied by the upper hinge is equal to the weight of the door, which is 122 N.
Since the door is in equilibrium, the sum of the vertical forces acting on it must be zero.
Therefore, the magnitude of the vertical component of the force applied by the lower hinge is also 122 N.
The vertical component of the force applied by the door to the lower hinge is equal in magnitude and opposite in direction to the vertical component of the force applied by the upper hinge. Therefore, it acts upwards.
Direction: upwards
In summary:
(a) The magnitude of the horizontal component of the force applied to the door by the upper hinge is approximately 32.76 N. The direction is to the left.
(b) The magnitude of the horizontal component of the force applied to the door by the lower hinge is approximately 32.76 N. The direction is to the right.
(c) The magnitude of the force applied by the door to the upper hinge is 122 N. The direction is downwards.
(d) The magnitude of the force applied by the door to the lower hinge is 122 N. The direction is upwards.
To solve this problem, we need to analyze the forces acting on the door and use the principles of equilibrium.
(a) The magnitude of the horizontal component of the force applied to the door by the upper hinge can be determined by recognizing that the sum of the horizontal forces acting on the door must be zero for equilibrium. Since the lower hinge bears all the weight of the door, the horizontal component of the force applied by the upper hinge must balance out the horizontal component of the weight of the door.
The weight of the door can be determined by multiplying the mass of the door by the acceleration due to gravity (9.8 m/s^2). The mass can be calculated using the formula weight = mass * gravity. Rearranging the formula, we get mass = weight / gravity.
Given that the weight of the door is 122 N, we can calculate the mass as follows:
mass = 122 N / 9.8 m/s^2 = 12.4489 kg
Now, we can find the horizontal component of the weight of the door. Using trigonometry, we determine that the angle between the weight vector and the horizontal is given by: angle = atan(height / width) = atan(2.1 m / 0.81 m).
Using a calculator, we find that angle ≈ 69.115°.
To find the horizontal component of the weight, we can use the formula: horizontal component = weight * cos(angle).
horizontal component = 122 N * cos(69.115°) ≈ 41.527 N
Therefore, the magnitude of the horizontal component of the force applied to the door by the upper hinge is approximately 41.527 N.
The direction of this force is to the right since the lower hinge is closer to the left side of the door.
(b) The magnitude of the horizontal component of the force applied to the door by the lower hinge can be determined using the same approach as in part (a). The only difference is that now we need to find the horizontal component of the weight that acts at the lower hinge.
Using the formulas and calculations described in part (a), we find that the magnitude of the horizontal component of the force applied to the door by the lower hinge is approximately 41.527 N.
The direction of this force is to the left since the lower hinge is closer to the right side of the door.
(c) The force applied by the door to the upper hinge must balance out the force applied by the upper hinge to the door. Since the horizontal component of the weight of the door is pushing the door towards the right, the upper hinge needs to exert a force to the left to counteract this.
Therefore, the magnitude of the force applied by the door to the upper hinge is also approximately 41.527 N.
The direction of this force is to the left, opposite to the direction of the horizontal component of the force applied by the upper hinge.
(d) The magnitude of the force applied by the door to the lower hinge can be determined by recognizing that this force must balance out the vertical component of the weight of the door (since the lower hinge bears all the weight).
Here, we need to find the vertical component of the weight, which is given by: vertical component = weight * sin(angle).
vertical component = 122 N * sin(69.115°) ≈ 111.196 N
Therefore, the magnitude of the force applied by the door to the lower hinge is approximately 111.196 N.
The direction of this force is downwards, as it counteracts the vertical component of the weight of the door.