Two strong magnets on opposite sides of a small table are shown. The long-range attractive force between the magnets keeps the lower magnet in place. Suppose the weight of the table is 26.7 N, the weight of each magnet is 6.55 N, and the magnetic force on the lower magnet is 2.16 times its weight.
A)What is the magnitude of the normal force of the table on the upper magnet?
B)What is the magnitude of the normal force of the table on the lower magnet?
C)What is the normal force of the ground on the table?
... I tried balancing the vertical forces in each scenario to solve for the normal force but I'm not getting the right answer.
For the upper magnet, I had the normal force of the table (up), the weight of the magnet mg (down) and the magnetic force of the lower magnet (down)
For the lower magnet I had the same forces but the normal force and the magnetic forces switched direction
For the table I had, the weights of the table and the two magnets being exerted down
... Am I missing something?
you are right on the ground force (table+2magnets).
The normal force on the table is the magnet weight +magnetic force.
The normal force of the table on the lower magnet is magnetic force- magnet weight.
Okay, I get it now. Thanks Bob
It seems like you are on the right track in analyzing the forces in each scenario. However, it's important to consider the direction of each force correctly while balancing the vertical forces.
A) To determine the magnitude of the normal force of the table on the upper magnet:
The forces acting on the upper magnet are:
1. Weight of the magnet (6.55 N), directed vertically downward.
2. Magnetic force from the lower magnet (2.16 times the weight of the magnet), directed vertically upward.
3. Normal force from the table on the upper magnet.
Since the upper magnet is in equilibrium (not accelerating), the sum of the vertical forces acting on it must be zero. Therefore, the magnitude of the normal force on the upper magnet is the sum of the weight of the magnet and the magnetic force from the lower magnet:
Normal force on the upper magnet = Weight of the magnet + Magnetic force from the lower magnet
= 6.55 N + (2.16 * 6.55 N)
B) To determine the magnitude of the normal force of the table on the lower magnet:
The forces acting on the lower magnet are:
1. Weight of the magnet (6.55 N), directed vertically downward.
2. Magnetic force from the upper magnet (2.16 times the weight of the magnet), directed vertically downward.
3. Normal force from the table on the lower magnet.
Again, since the lower magnet is in equilibrium, the sum of the vertical forces acting on it must be zero. This means that the normal force on the lower magnet should balance the weight of the magnet and the magnetic force from the upper magnet:
Normal force on the lower magnet = Weight of the magnet + Magnetic force from the upper magnet
= 6.55 N + (2.16 * 6.55 N)
C) To determine the normal force of the ground on the table:
The forces acting on the table are:
1. Weight of the table (26.7 N), directed vertically downward.
2. Normal force from the upper magnet, directed vertically upward.
3. Normal force from the lower magnet, directed vertically upward.
Since the table is in equilibrium, the sum of the vertical forces acting on it must be zero. Therefore, the normal force of the ground on the table is the sum of the weight of the table and the normal forces from the magnets:
Normal force of the ground on the table = Weight of the table + Normal force from the upper magnet + Normal force from the lower magnet
I hope this clarifies your confusion. Let me know if you have any further questions!
Based on your explanation, it seems like you are on the right track by considering the vertical forces acting on each object. Let's go through each scenario step by step to see if we can identify any mistakes.
A) Magnitude of the normal force of the table on the upper magnet:
- The weight of the table is 26.7 N, which acts downwards.
- The weight of each magnet is 6.55 N, which also acts downwards.
- The magnetic force on the lower magnet is 2.16 times its weight, so the magnetic force acting on the lower magnet is 2.16 * 6.55 N = 14.154 N, directed upwards.
To find the magnitude of the normal force of the table on the upper magnet, we need to sum up all the vertical forces acting on the upper magnet and set it equal to zero since it is not accelerating in the vertical direction.
- The weight of the table (26.7 N) and the weight of each magnet (6.55 N + 6.55 N = 13.1 N) act downwards.
- The magnetic force on the lower magnet (14.154 N) acts upwards.
Therefore, the normal force of the table on the upper magnet is:
Normal force = weight of the table + weight of each magnet - magnetic force on the lower magnet
Normal force = 26.7 N + 13.1 N - 14.154 N
Normal force = 25.646 N
B) Magnitude of the normal force of the table on the lower magnet:
Based on your explanation, the forces you mentioned for the lower magnet seem correct. To find the magnitude of the normal force of the table on the lower magnet, we need to sum up all the vertical forces acting on the lower magnet and set it equal to zero since it is not accelerating in the vertical direction.
Therefore, the normal force of the table on the lower magnet is:
Normal force = weight of the table + weight of each magnet + magnetic force on the lower magnet
Normal force = 26.7 N + 13.1 N + 14.154 N
Normal force = 54.954 N
C) Normal force of the ground on the table:
To find the normal force of the ground on the table, we need to consider the weight of the table. Since the table is not accelerating vertically, the normal force exerted by the ground on the table must cancel out the weight of the table.
Therefore, the normal force of the ground on the table is equal to the weight of the table. So, the normal force of the ground on the table is 26.7 N.
Please double-check your calculations and make sure you consider the direction and sign of the forces correctly.