A refrigerator magnet weighing \(0.14 \mathrm{N}\) is used to hold up a photograph weighing \(0.030 \mathrm{N}\). The magnet attracts the refrigerator door with a magnetic force of 2.10 N. (a) Identify the interactions between the magnet and other objects. (b) Draw an FBD for the magnet, showing all the forces that act on it. (c) Which of these forces are long-range and which are contact forces? (d) Find the magnitudes of all the forces acting on the magnet.

There are three objects interacting with the magnet: the refrigerator door, the photograph, and the Earth. The magnet attracts the refrigerator door with a magnetic force, it holds the photograph with a frictional force, and it experiences gravitational force (i.e., weight) due to the Earth.

In the free-body diagram for the magnet, we need to show all the forces acting on it: 1. \(F_{m}\): The magnetic force exerted by the refrigerator door on the magnet (pointing horizontally). 2. \(F_{g_{m}}\): The gravitational force or weight of the magnet due to Earth (pointing vertically downward). 3. \(F_{g_{p}}\): The gravitational force or weight of the photograph due to Earth (also pointing vertically downward). 4. \(F_{f}\): The frictional force between the magnet and the photograph (pointing vertically upward). [Replace this text with a simple drawing of the FBD. Label each force with the correct symbol mentioned above.]

Long-range forces: 1. \(F_{m}\): The magnetic force is a long-range force, as it acts over a distance without direct physical contact. 2. \(F_{g_{m}}\) and \(F_{g_{p}}\): The gravitational forces are also long-range forces, as they act over a distance without physical contact. Contact forces: 1. \(F_{f}\): The frictional force is a contact force, as it acts only when the two surfaces (magnet and photograph) are in direct contact.

We're given the magnitudes of the magnetic force \(F_{m}\) and the weights of the magnet \(F_{g_{m}}\) and photograph \(F_{g_{p}}\): 1. \(F_{m} = 2.10 \mathrm{N}\) 2. \(F_{g_{m}} = 0.14 \mathrm{N}\) 3. \(F_{g_{p}} = 0.030 \mathrm{N}\) To find the magnitude of the frictional force \(F_{f}\), we apply Newton's second law, according to which the net force acting on an object in equilibrium is zero. Since the magnet is not moving vertically, the net vertical force must be zero, which means: \(F_{f} - F_{g_{m}} - F_{g_{p}} = 0\) Solving for \(F_{f}\), we get: \(F_{f} = F_{g_{m}} + F_{g_{p}} = 0.14 \mathrm{N} + 0.030 \mathrm{N} = 0.170 \mathrm{N}\) So, the magnitudes of all the forces acting on the magnet are: 1. \(F_{m} = 2.10 \mathrm{N}\) 2. \(F_{g_{m}} = 0.14 \mathrm{N}\) 3. \(F_{g_{p}} = 0.030 \mathrm{N}\) 4. \(F_{f} = 0.170 \mathrm{N}\)