Question

examine the forces acting on a car on a ramp. select the forces or force components that need to be added together to calculate the net force in the x-direction. choose two correct answers. options: $f_n$, $-f_g$, $f_g(cos 15^circ)$, $f_g(sin 15^circ)$, $-f_f$ (diagram: car on a 15° ramp with forces $f_n$ (normal), $f_f$ (friction), $f_g$ (gravity) and its components)

Explanation:

To find the net force in the \( x \)-direction (along the ramp), we analyze the forces:

• The gravitational force component along the ramp is \( F_g(\sin 15^\circ) \) (down the ramp, positive or negative depending on sign convention).
• The frictional force \( F_f \) acts opposite to the direction of motion (or tendency), so it is \( -F_f \) if \( F_g(\sin 15^\circ) \) is positive down the ramp.
• \( F_N \) is normal (perpendicular to the ramp), so it has no \( x \)-component. \( -F_y \) is vertical, not \( x \)-direction. \( F_g(\cos 15^\circ) \) is the vertical component of gravity, not \( x \)-direction.

Thus, the forces/components in the \( x \)-direction are \( F_g(\sin 15^\circ) \) (down-ramp) and \( -F_f \) (up-ramp, opposing the gravitational component).

Answer:

• \( F_g(\sin 15^\circ) \)
• \( -F_f \)