(7) A \( 7,6 \mathrm{~kg} \) object is at rest on an inclined plane. If the inclined plane makes an angle with the horizontal of \( 33^{\circ} \), what is the normal force acting on the object?

To find the normal force acting on the object resting on an inclined plane, we can use the formula for the normal force \( N \). The gravitational force acting on the object can be given by \( F_g = mg \), where \( m = 7.6 \, \text{kg} \) and \( g \approx 9.81 \, \text{m/s}^2 \). Thus, \( F_g = 7.6 \, \text{kg} \times 9.81 \, \text{m/s}^2 \approx 74.5 \, \text{N} \). The normal force can be calculated using \( N = F_g \cdot \cos(\theta) \), where \( \theta = 33^\circ \). Therefore, \( N \approx 74.5 \, \text{N} \cdot \cos(33^\circ) \approx 74.5 \, \text{N} \cdot 0.838 = 62.5 \, \text{N} \). So, the normal force acting on the object is approximately \( 62.5 \, \text{N} \! \).