Question

You're standing on a ramp that is inclined at a 25 degree angle. The weight (Fg) experienced is 50N. What is the parallel force (Fx) and the perpendicular force (Fy)?

Answer 1

The parallel component of force acting on person is,

`F_x=F_g\sin \theta`

Substitute the known values,

`\begin{gathered} F_x=(50\text{ N)s}in25 \\ =(50\text{ N)(}0.423) \\ \approx21.15\text{ N} \end{gathered}`

The perpendicular component of force acting on person is,

`F_y=F_g\cos \theta`

Substitute the known values,

`\begin{gathered} F_y=(50\text{ N)}\cos 25 \\ =(50\text{ N)(}0.906) \\ \approx45.3\text{ N} \end{gathered}`

Thus, the parallel force acting on person is 21.15 N and the perpendicular force is 45.3 N.