What is the component of gravity that acts perpendicular to the inclined plane?

What is the component of gravity that acts perpendicular to the inclined plane?

• A. m g
• B. m g sin theta
• C. m g cos theta
• D. F_N

Answer: (C)

When you analyze gravity on an incline, you break the weight into two components: one perpendicular to the surface and one parallel to it. The weight points straight down, and the angle between that weight and the line perpendicular to the plane is the incline angle, theta. In this decomposition, the component that points perpendicular to the plane has magnitude mg cos theta, because it is the adjacent side of the right triangle formed by the weight vector and the plane’s normal. The component along the plane has magnitude mg sin theta. So the part of gravity that acts perpendicular to the surface is mg cos theta. The normal force from the plane balances this perpendicular component (in the absence of other forces causing vertical acceleration), giving F_N = mg cos theta. Quick checks: if the plane is flat (theta = 0), the perpendicular component is mg, as expected; if the plane is vertical (theta = 90°), the perpendicular component is zero, and gravity acts entirely along the plane.

When you analyze gravity on an incline, you break the weight into two components: one perpendicular to the surface and one parallel to it. The weight points straight down, and the angle between that weight and the line perpendicular to the plane is the incline angle, theta. In this decomposition, the component that points perpendicular to the plane has magnitude mg cos theta, because it is the adjacent side of the right triangle formed by the weight vector and the plane’s normal. The component along the plane has magnitude mg sin theta.

So the part of gravity that acts perpendicular to the surface is mg cos theta. The normal force from the plane balances this perpendicular component (in the absence of other forces causing vertical acceleration), giving F_N = mg cos theta. Quick checks: if the plane is flat (theta = 0), the perpendicular component is mg, as expected; if the plane is vertical (theta = 90°), the perpendicular component is zero, and gravity acts entirely along the plane.