To solve this problem, we need to follow these steps:
1. Understand the given values and the problem setup:
- The mass of the box ([tex]\( m \)[/tex]) is 45.0 kg.
- The angle of the incline ([tex]\( \theta \)[/tex]) is 35.0°.
- The acceleration due to gravity ([tex]\( g \)[/tex]) is 9.81 m/s².
- The box is at rest on the inclined plane, meaning the frictional force balances the component of the gravitational force pulling the box down the ramp.
2. Convert the angle from degrees to radians:
- Since trigonometric functions in many mathematical formulas use the angle in radians, we need to convert 35.0° to radians.
- The formula to convert degrees to radians is: [tex]\( \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \)[/tex]
3. Calculate the component of the gravitational force acting down the ramp:
- The gravitational force acting directly downwards is [tex]\( F_{\text{gravity}} = m \times g \)[/tex].
- We are interested in the component of this force along the incline, which is given by [tex]\( F_{\text{gravity down ramp}} = m \times g \times \sin(\theta_{\text{radians}}) \)[/tex].
4. Determine the frictional force:
- Since the box is at rest, the frictional force must exactly oppose the component of the gravitational force along the ramp. Therefore, [tex]\( F_{\text{friction}} = F_{\text{gravity down ramp}} \)[/tex].
Following these steps, let's write out the calculations and results:
1. Conversion to radians:
[tex]\[ \theta_{\text{radians}} = 35.0 \times \frac{\pi}{180} \approx 0.6108652381980153 \text{ radians} \][/tex]
2. Component of the gravitational force down the ramp:
[tex]\[ F_{\text{gravity down ramp}} = 45.0 \times 9.81 \times \sin(0.6108652381980153) \approx 45.0 \times \ 9.81 \times 0.574 \approx 253.2053178271693 \text{ N} \][/tex]
3. Frictional force:
[tex]\[ F_{\text{friction}} = 253.2053178271693 \text{ N} \][/tex]
Thus, the frictional force acting on the box, which keeps it at rest on the inclined plane, is approximately [tex]\( 253.2053178271693 \)[/tex] newtons.
