Tre cani stanno addentando un frisbee. Un cane sta ti- rando con una forza di \( 5,0 \mathrm{~N} \) verso est, un altro cane con una forza di \( 5,0 \mathrm{~N} \) in direzione nord. Se il frisbee rimane fermo, qual è l'intensità della for- za del terzo cane? Che angolo forma con la direzione ovest? [7,1 N; \( 45^{\circ} \) al di sotto]

We start by noting that the frisbee remains at rest, so the net force on it must be zero. Two dogs apply forces of 5.0 N—one toward the east and one toward the north. Representing these forces as vectors:  • F₁ = 5.0 N east = (5.0 N, 0)  • F₂ = 5.0 N north = (0, 5.0 N) The sum of these vectors is F₁ + F₂ = (5.0, 5.0) N. For equilibrium, the third force F₃ must exactly cancel this sum. Hence:  F₃ = –(5.0 N, 5.0 N) = (–5.0 N, –5.0 N) Now, calculate the magnitude of F₃:  |F₃| = √[ (–5.0)² + (–5.0)² ]     = √(25 + 25)     = √50     ≈ 7.07 N ≈ 7.1 N Next, we determine the angle that F₃ makes with the west direction. The west direction corresponds to the negative x-axis. Since F₃ has components (–5.0 N, –5.0 N), its direction is obtained by:  tan θ = (magnitude of the south component) / (magnitude of the west component)     = 5.0 N / 5.0 N     = 1  ⟹ θ = 45° Because the force has a southward (negative y) component, this angle is 45° below the horizontal (westward) direction. Thus, the third dog must exert a force of approximately 7.1 N that is directed 45° below west.