Tyler and Kelly decides to rollerskate. They come to a hill that is inclined at 8.5 degrees. Tyler pushes Kelly halfway up the hill and then holds her there. If Kelly weighs 58.0 lbs., find the magnitude of the force Tyler must push with to keep Kelly from rolling down the hill.
2006-10-08 10:58:16 · 3 answers · asked by boredX_x 2 in Science & Mathematics ➔ Mathematics
Ignoring friction, there are 3 forces acting on Kelly: mass*gravity going downwards, the normal force pushing up perpendicular to the ground, and Tyler pushing Kelly up the hill preventing her from rolling down. The vector sum of the 3 forces must be zero:
gravity force = mass * gravity = m * g = 58 * 32.17 foot-pounds
normal force = m * g * cos (8.5)
Tyler force = m * g * sin (8.5)
See link below under "physics inclined plane problem" for diagram.
2006-10-08 13:12:26 · answer #1 · answered by Joe C 3 · 0⤊ 0⤋
It's been a long time since Ive done trigonometry, but I thought it would be fun to try. Okay, the angle is 8.5 degrees. look at it as a right angle triangle. What I think you need to do is find horizontal component of the force compared to the vertical component of it. Since this is a ratio, no units are necessary. Fix the vertical component at 1. Therefore, to find the horizontal component,
Tan8.5 degrees = 1/x
x = 1/tan8.5 degrees = .6.69
The vertical component is 1, and the horizontal component is 6.69; therefore, the force required to hold 58.0 lbs. is
1/6.69(58) = 8.67 lbs
2006-10-08 11:19:20 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Convert lbs to kg
calculate F = kg*9.8*sin(8.5) to get the required force
Doug
2006-10-08 11:02:46 · answer #3 · answered by doug_donaghue 7 · 1⤊ 0⤋