Question

a 0.5 kg basketball sits motionless inside a v - shaped wedge. the walls of the wedge make an angle of 36°, as shown in the figure. what is the magnitude of the force that the left side of the wedge exerts on the basketball?
○ 2.5 n
○ 2.6 n
○ 4.2 n
○ 3.0 n
○ 7.9 n

Explanation:

Step1: Calculate weight of basketball

$W = mg = 0.5\ \text{kg} \times 9.8\ \text{m/s}^2 = 4.9\ \text{N}$

Step2: Define angle for force balance

Let $\theta = \frac{36^\circ}{2} = 18^\circ$, the angle between normal force and vertical.

Step3: Solve for normal force $F_N$

Vertical equilibrium: $2F_N\cos\theta = W$
$F_N = \frac{W}{2\cos\theta} = \frac{4.9\ \text{N}}{2\cos(18^\circ)}$
$\cos(18^\circ) \approx 0.9511$, so $F_N \approx \frac{4.9}{2\times0.9511} \approx 2.576\ \text{N} \approx 2.6\ \text{N}$

Answer:

2.6 N (Option: ○ 2.6 N)