Please help with this magnet physics problem

Question

On the internet, I came across an interesting physics problem. Can you help with this:

The weight of a small, flat refrigerator magnet is "M". We place the magnet on the refrigerator's vertical side, then start pulling it in some direction in a vertical plane perpendicular to the metal plate. The smallest force with which we can move the magnet vertically downwards is F1 and upwards is F2

Questions:

When not pulled anywhere, how much pulling force does the metal plate exert on the magnet?

What is the coefficient of friction between the magnet and the side of the refrigerator?

I tried like this:

I started with pulling downwards first. Because I pull the magnet in a plane perpendicular to the surface, at an α angle, and this is where the frictional force is created. this will act upwards on the surface of the refrigerator. And vertically downwards, the gravitational force acts on the magnet (mg=M). Since we can move the magnet F1+M>S1 (S1...friction). At the base of the α angle ---> cos(α)=M/F1. However, I don't get results for α this way.

We know M=0.10 N, F1=0.06 N and F2=0.25 N.

Thanks in advance!

Answer 1

When the magnet is not being pulled anywhere, it remains stationary, which means the net force acting on it is zero. This implies that the force exerted by the metal plate on the magnet balances out the gravitational force acting on it. Let's denote the gravitational force as F_g ​ . Then the force exerted by the metal plate on the magnet, denoted as F_p​ , must be equal in magnitude but opposite in direction to F_g.

so, f_p = f_g

The force of gravity acting on the magnet can be calculated using the formula: F_g = m x g

Where:

1. m is the mass of the magnet
2. g is the acceleration due to gravity (approximately 9.8 m/s 2 9.8m/s 2 )

So, the force exerted by the metal plate on the magnet is equal to the weight of the magnet, which is M in this case.

Now, to determine the coefficient of friction between the magnet and the side of the refrigerator ( μ), we can use the relationship between the minimum force required to move the magnet (the force of static friction) and the normal force exerted by the metal plate on the magnet.

The force of static friction ( friction F friction ​ ) can be expressed as:

F_{friction =} μ×F_normal​

Since the magnet is not moving vertically, the normal force exerted by the metal plate on the magnet ( normal F normal ​ ) is equal to the weight of the magnet ( M).

F_friction = × F _friction ​ =μ×M

Thus, the coefficient of friction (μ) between the magnet and the side of the refrigerator is

μ=F1/

F1 ​ is the smallest force required to move the magnet upwards and F2 ​ is the smallest force required to move the magnet downwards, the force of friction is likely equal to either F1 ​ or F2 ​ , depending on the direction of the motion.

Thus, the coefficient of friction ( μ) between the magnet and the side of the refrigerator is:

μ=F1/M = F2/M

​

Using the given values: M=0.10N, F1 ​ =0.06N, F2 ​ =0.25N

Substituting the given values: μ= 0.6N/ 0.10N ​ = 0.25N/ 0.10N

μ=0.6

Therefore, the coefficient of friction between the magnet and the side of the refrigerator isμ=0.6.