Use the diagram to answer question; Camphor is a chemical that sub-limes and interacts with water reducing the surface tension where it is put. Explain why ...

What the diagram shows. The pointed toy boat floats on water. Camphor is fixed at the flat rear end. The label \(F_1\) is a leftward (backward) arrow acting at the rear where the camphor is, and \(F_2\) is a rightward (forward) arrow acting at the pointed front. \(F_1\) and \(F_2\) represent the surface-tension pulls of the water on the two ends of the boat.

Key idea: forces due to surface tension. The free surface of water behaves like a stretched elastic skin. Wherever this skin touches the edge of the floating boat, it pulls the boat outward (away from the boat) along the surface. So the water pulls the rear of the boat backward with force \(F_1\) and pulls the front of the boat forward with force \(F_2\). The size of each pull is proportional to the surface tension of the water touching that edge.

Before the camphor is placed (boat stationary). The surface tension of the water is the same all around the boat, so the backward pull equals the forward pull:

The two pulls cancel, the resultant horizontal force is zero, and the boat stays at rest.

After the camphor is placed at the back. The camphor sublimes and dissolves into the water immediately behind the boat, and it reduces the surface tension of the water there. The water in front of the boat still has its full (higher) surface tension. Therefore the backward pull becomes small while the forward pull stays large:

There is now an unbalanced resultant force acting forward:

By Newton's second law this net force accelerates the boat, so it moves forward (in the direction of \(F_2\)). Equivalently, the higher-tension water in front pulls harder than the weakened-tension water behind, and the boat is dragged toward the region of greater surface tension. As soon as the camphor is used up and the surface tension becomes uniform again, \(F_1\) once more equals \(F_2\) and the boat stops.