In this experiment, you will investigate the resultant of two forces acting at an angle. You attaches three strings to a small ring. Two strings pass over p...

(a) All forces and angles are recorded correctly in Table 47.1 as given. [1]

(b) For the first row, both forces are equal (\(F = 2.0\) N) and the angle between the strings is \(\theta = 60°\). The resultant of two equal forces at an angle is given by:

\[ R = 2F\cos\left(\frac{\theta}{2}\right) = 2 \times 2.0 \times \cos\left(\frac{60}{2}\right) = 4.0 \times \cos(30°) \]

[1]

\[ R = 4.0 \times 0.866 = 3.46 \text{ N} \]

(Accept 3.5 N to 2 significant figures.) [1]

The formula works because when two equal forces act at an angle, the resultant bisects that angle. Using the cosine of the half-angle gives the component of each force along the resultant direction, and doubling gives the total.

(c) The percentage difference between the calculated resultant (3.46 N) and the measured equilibrant (3.5 N):

\[ \text{Difference} = |3.5 - 3.46| = 0.04 \text{ N} \]

\[ \text{Percentage difference} = \frac{0.04}{3.46} \times 100 = 1.2\% \]

(Accept 0% to 2%.) This small percentage confirms good agreement between theory and experiment. [1]

(d) When both forces are equal, as the angle between them increases, the resultant decreases. At 0° (parallel forces), the resultant equals 2F (the maximum). At 90°, it equals \(F\sqrt{2}\). At 180° (opposite forces), the resultant is zero. [1]

(e) The graph below shows the equilibrant force \(F_3\) (y-axis) plotted against the angle between the strings (x-axis) for the first three rows. Axes are labelled with units, three data points are plotted, and a smooth curve is drawn. [3]

(f) To ensure the ring is in equilibrium before recording: wait until the ring is completely stationary and not oscillating or drifting. All three strings must be taut. Gently tap the ring and check that it returns to the same position, confirming true equilibrium rather than the ring being stuck due to friction. [1]

(g) The equilibrant is the single force that keeps the system in equilibrium. For the ring to remain stationary, the net force must be zero. The resultant of \(F_1\) and \(F_2\) is a single force that represents their combined effect. The equilibrant must exactly cancel this resultant, so it is equal in magnitude to the resultant but acts in the opposite direction. [1]

(h) One difficulty: friction at the pulleys may prevent the strings from adjusting freely. If there is significant friction, the forces transmitted to the ring are slightly less than the weights hanging from the strings, causing inaccurate readings. Measuring the angles precisely with a protractor is also difficult because the strings have finite thickness and may not be perfectly straight near the ring. [1]