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 was able to find the direction but not sure what is mean by the magnitude
To find the magnitude, we need to calculate the numerical value of the force.
We can start by analyzing the forces acting on the door:
1. Weight of the door: The weight of the door is given as 122 N, which acts downward.
2. Force applied by the upper hinge: This force exerts a horizontal component and a vertical component on the door.
3. Force applied by the lower hinge: This force exerts a horizontal component and a vertical component on the door.
Now let's proceed with finding the magnitude and direction of the forces:
(a) To determine the magnitude and direction of the horizontal component of the force applied by the upper hinge:
The door is in equilibrium, so the sum of the moments around any point must be zero.
Since the door rotates around the lower hinge, we can choose that point to analyze the moments. The weight of the door creates a downward moment, and the horizontal force creates a clockwise moment.
Thus, the moment equation becomes:
(122 N * 2.1 m) + (Force_upper_hinge_horizontal * 1.5 m) = 0
Solving for Force_upper_hinge_horizontal, we get:
Force_upper_hinge_horizontal = -(122 N * 2.1 m) / 1.5 m
The negative sign indicates that the force is in the opposite direction of the clockwise moment. This force acts to the left. The magnitude is the absolute value of the calculated force.
(b) To determine the magnitude and direction of the horizontal component of the force applied by the lower hinge:
Since the door is in equilibrium, the horizontal forces on the door must balance each other out.
Therefore, the force applied by the lower hinge must have an equal magnitude but in the opposite direction to the force applied by the upper hinge. The magnitude would be the same as in part (a), but the direction would be to the right.
(c) To determine the magnitude and direction of the force applied by the door to the upper hinge:
The force exerted by the door on the upper hinge can be determined using Newton's third law: every action has an equal but opposite reaction.
Thus, the magnitude and direction of the force applied by the door to the upper hinge are the same as the magnitude and direction of the force applied by the upper hinge to the door, which was determined in part (a).
(d) To determine the magnitude and direction of the force applied by the door to the lower hinge:
Similarly to part (c), the magnitude and direction of the force applied by the door to the lower hinge are the same as the magnitude and direction of the force applied by the lower hinge to the door, which was determined in part (b).
Remember to substitute the values given in the problem (such as the weight of the door and the distance between the hinges) into the calculations to obtain specific numerical results.