A 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 °
I will be happy to critique your thinking. Hint: find the cg of the door first. Then you can sum moments using that cg and distances.
(a) The magnitude of the horizontal component of the force applied to the door by the upper hinge is 0 N because the upper hinge does not exert any horizontal force on the door.
(b) The magnitude of the horizontal component of the force applied to the door by the lower hinge is also 0 N. This is because the lower hinge only bears the weight of the door in the vertical direction.
(c) The force applied by the door to the upper hinge is equal in magnitude to the weight of the door, which is 122 N. The direction of this force is downwards.
(d) The magnitude of the force applied by the door to the lower hinge is also equal to the weight of the door, which is 122 N. The direction of this force is downwards.
To solve this problem, we can use the principles of torque and equilibrium.
(a) To find the magnitude and direction of the horizontal component of the force applied to the door by the upper hinge, we need to calculate the torque exerted by the door about the lower hinge and then determine the horizontal component of that torque.
- First, calculate the torque exerted by the door about the lower hinge:
The weight of the door is 122 N and its center of gravity is halfway up the door, so the torque exerted by the weight about the lower hinge is (122 N) * (1.05 m) = 127.05 N·m.
- Second, determine the horizontal component of this torque by considering the 1.5 m distance between the hinges:
Since the lower hinge bears all the weight of the door, the horizontal component of torque exerted by the upper hinge must balance this torque.
Let F_upper be the magnitude of the horizontal component of the force applied by the upper hinge.
Then, the torque from the upper hinge is (F_upper) * (1.5 m).
Setting up the equation for torque balance:
(F_upper) * (1.5 m) = 127.05 N·m
Solving for F_upper:
F_upper = 127.05 N·m / 1.5 m
F_upper ≈ 84.7 N
Therefore, the magnitude of the horizontal component of the force applied to the door by the upper hinge is approximately 84.7 N.
The direction of this force can be determined by noting that it pulls on the door toward the right to counterbalance the torque exerted by the weight of the door. Therefore, the direction of the force applied by the upper hinge is to the right.
(b) To find the magnitude and direction of the horizontal component of the force applied to the door by the lower hinge, we need to consider that the lower hinge bears all the weight of the door and must balance the torque from the weight.
Since the weight of the door is 122 N, the magnitude of the horizontal component of the force applied by the lower hinge is equal to the weight of the door, which is 122 N.
The direction of this force can be determined from the fact that it supports the weight of the door and counterbalances the torque exerted by the weight. Therefore, the direction of the force applied by the lower hinge is upwards.
(c) To determine the magnitude of the force applied by the door to the upper hinge, we can use Newton's third law: for every action, there is an equal and opposite reaction.
Since the force applied by the upper hinge on the door is 84.7 N to the right, the force applied by the door to the upper hinge will be equal in magnitude but in the opposite direction. Therefore, the magnitude of the force applied by the door to the upper hinge is 84.7 N.
Since the force applied by the upper hinge is to the right, the direction of the force applied by the door to the upper hinge is to the left.
(d) To determine the magnitude and direction (below the horizontal) of the force applied by the door to the lower hinge, we can use Newton's third law again.
Since the force applied by the lower hinge on the door is 122 N upwards, the force applied by the door to the lower hinge will be equal in magnitude but in the opposite direction. Therefore, the magnitude of the force applied by the door to the lower hinge is 122 N.
Since the force applied by the lower hinge is upwards, the direction of the force applied by the door to the lower hinge is downwards. The angle below the horizontal is 90 degrees. Therefore, the direction of the force applied by the door to the lower hinge is downwards at an angle of 90 degrees.
To find the answers to these questions, we'll need to analyze the forces acting on the door.
First, let's consider the weight of the door. The weight is given as 122 N and acts vertically downward. This force is equal to the total force applied by both hinges.
(a) To find the magnitude and direction of the horizontal component of the force applied to the door by the upper hinge, we need to determine the vertical component of the force at the lower hinge. Since the hinges are 1.5 m apart and the door is 2.1 m high, the distance from the upper hinge to the line of action of the weight force is 0.6 m (2.1 m - 1.5 m).
Using this distance and the weight of the door, we can find the vertical component of the force at the lower hinge using the equation:
Vertical force = Weight * (distance to line of action of the weight force / height of the door)
Vertical force = 122 N * (0.6 m / 2.1 m) = 34.857 N
The magnitude of the horizontal component of the force at the upper hinge is the same as the magnitude of the horizontal component at the lower hinge because the weight is evenly distributed across the door.
Therefore, the magnitude of the horizontal component of the force applied to the door by the upper hinge is 34.857 N.
Since the weight force is acting vertically downward, the direction of the horizontal component of the force applied to the door by the upper hinge will be to the right (opposite to the direction of the force applied by the lower hinge).
(a) magnitude: 34.857 N
direction: to the right
Next, let's calculate the magnitude and direction of the horizontal component of the force applied to the door by the lower hinge.
(b) The horizontal component of the force applied by the lower hinge is equal to the total weight of the door since the lower hinge is bearing all the weight.
The magnitude of the horizontal component of the force applied to the door by the lower hinge is 122 N.
Since the weight force is acting vertically downward, the direction of the horizontal component of the force applied to the door by the lower hinge will be to the left (opposite to the direction of the force applied by the upper hinge).
(b) magnitude: 122 N
direction: to the left
Moving on to the force applied by the door to the hinges, we need to consider the equal and opposite reaction forces.
(c) The magnitude of the force applied by the door to the upper hinge is equal to the magnitude of the force applied by the upper hinge to the door. This force is the horizontal component we calculated earlier.
Therefore, the magnitude of the force applied by the door to the upper hinge is 34.857 N.
Since this force is acting to the right, the direction will also be to the right.
(c) magnitude: 34.857 N
direction: to the right
Lastly, let's determine the magnitude and direction (below the horizontal) of the force applied by the door to the lower hinge.
(d) The magnitude of the force applied by the door to the lower hinge is equal to the magnitude of the total weight of the door.
Therefore, the magnitude of the force applied by the door to the lower hinge is 122 N.
Since this force is acting vertically downward, the direction will be downwards.
(d) magnitude: 122 N
direction: downwards
Note: The direction is provided as "downwards" because it's given that the lower hinge is bearing all the weight of the door, resulting in a downward force.