Questions: A plane makes an angle of 30.0° with the horizontal. The force required to cause a 15 kg box to slide up the plane with an acceleration of 1.2 m / s^2.

Solution Steps

We need to consider the forces acting on the box along the inclined plane. These include:

• The gravitational force component parallel to the plane: \( F_{\text{gravity, parallel}} = mg \sin \theta \)
• The gravitational force component perpendicular to the plane: \( F_{\text{gravity, perpendicular}} = mg \cos \theta \)
• The normal force: \( F_{\text{normal}} \)
• The applied force: \( F_{\text{applied}} \)
• The frictional force (if any): \( F_{\text{friction}} \)

Given:

• Mass of the box, \( m = 15 \, \text{kg} \)
• Angle of the incline, \( \theta = 30.0^\circ \)
• Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)

Calculate the parallel and perpendicular components of the gravitational force: \[ F_{\text{gravity, parallel}} = mg \sin \theta = 15 \times 9.81 \times \sin 30.0^\circ = 15 \times 9.81 \times 0.5 = 73.575 \, \text{N} \] \[ F_{\text{gravity, perpendicular}} = mg \cos \theta = 15 \times 9.81 \times \cos 30.0^\circ = 15 \times 9.81 \times 0.8660 = 127.4 \, \text{N} \]

Given:

• Acceleration down the plane, \( a = 1.2 \, \text{m/s}^2 \)

Using Newton's second law along the plane: \[ F_{\text{net}} = ma = 15 \times 1.2 = 18 \, \text{N} \]

The net force is the difference between the applied force and the gravitational force component parallel to the plane: \[ F_{\text{applied}} - F_{\text{gravity, parallel}} = F_{\text{net}} \] \[ F_{\text{applied}} - 73.575 = 18 \] \[ F_{\text{applied}} = 18 + 73.575 = 91.575 \, \text{N} \]

Final Answer