Perpendicular Tube
Physics Question?
Up near Truckie, the Schmidt boys are inner tubing down a snow covered creek bed. Fred and his tube weigh 225 pounds together. The creek bed is a 20 degree incline. How fast will Fred be gping at the end of a 30 meter run? Find force parallel, force perpendicular, and acceleration as well.
Please show/explain all work. Please show all formulas used.
Let:
m be the mass of the Scmidt boys and their tube,
R be the normal reaction of the incline,
t be the inclination of the run to the horizontal,
a be the acceleration down the inclne,
g be the acceleration due to gravity,
s be the length of the run,
v be the velocity at the end of the run.
Provided there is no friction, resolving parallel and perpendicular to the incline:
mg sin(t) = ma ...(1)
mg cos(t) = R ...(2)
The final speed is given by:
v^2 = 2as ...(3)
The left hand side of (1) gives the force parallel to the incline:
225 * 9.81 * sin(20)
= 754.9 N.
The acceleration is:
a = 9.81 * sin(20)
= 3.36 m/s^2.
The LHS of (2) gives the force perpendicular:
225 * 9.81 * cos(20)
= 2074.1 N.
(3) gives:
v = sqrt(2 * 3.36 * 30)
= 14.2 m/s.
![]() |
![]() 10 X 300 4 TR87 Tube Bent Perpendicular Stem $4.49 Time Remaining: 14d 11h 31m Buy It Now for only: $4.49 |
![]() 13 x 500 6 TR87 Tube Bent Perpendicular Stem $5.95 Time Remaining: 14d 11h 31m Buy It Now for only: $5.95 |
![]() 410 x 350 5 TR87 Tube Bent Perpendicular Stem $5.49 Time Remaining: 14d 11h 32m Buy It Now for only: $5.49 |
![]() 410 x 350 6 TR87 Tube Bent Perpendicular Stem $5.49 Time Remaining: 14d 11h 32m Buy It Now for only: $5.49 |
![]() 8 X 300 TR87 Tube Bent Perpendicular Stem $4.49 Time Remaining: 14d 11h 32m Buy It Now for only: $4.49 |





